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Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Probability Solutions Exercise 9(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Probability Solutions Exercise 9(a)

I. In the experiment of throwing a die, consider the following events:

Question 1.
A = {1, 3, 5}, B = {2, 4, 6}, C = {1, 2, 3} Are these events equally likely?
Solution:
Since events A, B, C has an equal chance to occur, hence they are equally likely events.

Question 2.
In the experiment of throwing a die, consider the following events:
A = {1, 3, 5}, B = {2, 4}, C = {6}
Are these events mutually exclusive?
Solution:
Since the happening of one of the given events A, B, C prevents the happening of the other two, hence the given events are mutually exclusive.
Otherwise A ∩ B = φ, B ∩ C = φ, C ∩ A = φ
Hence they are mutually exclusive events.

Question 3.
In the experiment of throwing a die, consider the events.
A = (2, 4, 6}, B = {3, 6}, C = {1, 5, 6}
Are these events exhaustive?
Solution:
A = {2, 4, 6}, B = {3, 6}, C = {1, 5, 6}
Let S be the sample space for the random experiment of throwing a die
Then S = {1, 2, 3, 4, 5, 6}
∵ A ⊂ S, B ⊂ S and C ⊂ S, and A ∪ B ∪ C = S
Hence events A, B, C are exhaustive events.

II.

Question 1.
Give two examples of mutually exclusive and exhaustive events.
Solution:
Examples of mutually exclusive events:
(i) The events {1, 2}, {3, 5} are disjoint in the sample space S = {1, 2, 3, 4, 5, 6}
(ii) When two dice are thrown, the probability of getting the sums of 10 or 11.
Examples of exhaustive events:
(i) The events {1, 2, 3, 5}, (2, 4, 6} are exhaustive in the sample space S = {1, 2, 3, 4, 5, 6}
(ii) The events {HH, HT}, {TH, TT} are exhaustive in the sample space S = {HH, HT, TH, TT} [∵ tossing two coins]

Question 2.
Give examples of two events that are neither mutually exclusive nor exhaustive.
Solution:
(i) Let A be the event of getting an even prime number when tossing a die and let B be the event of getting even number.
∴ A, B are neither mutually exclusive nor exhaustive.
(ii) Let A be the event of getting one head tossing two coins.
Let B be the event of getting atleast one head tossing two coins.
∴ A, B are neither mutually exclusive nor exhaustive.

Question 3.
Give two examples of events that are neither equally likely nor exhaustive.
Solution:
(i) Two coins are tossed
Let A be the event of getting an one tail and
Let B be the event of getting atleast one tail.
∴ A, B are neither equally likely nor exhaustive.
(ii) When a die is tossed
Let A be the event of getting an odd prime number and
Let B be the event of getting odd number.
∴ B are are neither equally likely nor exhaustive.

Students get through AP Inter 2nd Year Physics Important Questions 7th Lesson Moving Charges and Magnetism which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 7th Lesson Moving Charges and Magnetism

Very Short Answer Questions

Question 1.
What is the importance of Oersted’s experiment ? [T.S. Mar. 17]
Answer:
Importance of Oersted’s experiment is every current carrying conductor produces a magnetic field around it and which is perpendicular to current carrying conductor.

Question 2.
State Ampere’s law and Biot-Savart’s law.
Answer:
Ampere’s law : The line integral of the intensity of magnetic induction around a closed path is equal to g0 times the total current enclosed in it.
∴ \(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d}} l\) = μ₀ i.
Biot – Savart’s laws : Biot – Savart’s law states that the intensity of magnetic induction (dB) due to a small element is directly proportional to the
i) current (i)
ii) length of the element (dZ)
iii) sine angle between radius vector (r) and dl and inversely proportional to the square of the point from current element.
∴dB ∝ \(\frac{\mathrm{i} \mathrm{dl} \sin \theta}{\mathrm{r}^2}\)
dB = \(\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{i} \mathrm{d} l \sin \theta}{\mathrm{r}^2}\)

Question 3.
Write the expression for the magnetic induction at any point on the axis of a circular current-carrying coil. Hence, obtain an expression for the magnetic induction at the centre of the circular coil.
Answer:

1. Intensity of magnetic induction field on the axis of the circular coil B = \(\frac{\mu_0 \mathrm{ni} \mathrm{r}^2}{2\left(\mathrm{r}^2+\mathrm{x}^2\right)^{3 / 2}}\)
2. At the centre of the coil B = \(\frac{\mu_0 \mathrm{ni}}{2 \mathrm{r}}\)

Question 4.
A circular coil of radius T having N turns carries a current “i”. What is its magnetic moment ?
Answer:
Magnetic moment (M) = N i A
M = N i (πr²) (∵ A = πr²)
∴ M = π N i r²

Question 5.
What is the force on a conductor of length L carrying a current “i” placed in a magnetic field of induction B ? When does it become maximum ?
Answer:

1. Force on a conductor (F) = B i L sinθ
2. If θ = 90°,F_Max = BiL
   i.e., the direction of current and magnetic field are perpendicular to,each other, then force is maximum.

Question 6.
What is the force on a charged particle of charge “q” moving with a velocity “v” in a uniform magnetic field of induction B ? When does it become maximum ?
Answer:

1. Force on a charged particle (F) = B q v sin θ.
2. If θ = 90°, F_Max = B q v.

Question 7.
Distinguish between ammeter and voltmeter. [A.P. Mar. 17; A.P. Mar. 15]
Answer:
Ammeter

1. It is used to measure current.
2. Resistance of an ideal ammeter is zero.
3. It is connected in series in the circuits.

Voltmeter

1. It is used to measure RD between two points.
2. Resistance of ideal voltmeter is infinity.
3. It is connected in parallel in the circuits.

Question 8.
What is the principle of a moving coil galvanometer ?
Answer:
Moving coil galvanometer is based on the fact that when a current carrying coil is placed in a uniform magnetic field, it experiences a torque.
∴ current in the coil (i) ∝ deflecting angle (θ).

Question 9.
What is the smallest value of current that can be measured with a moving coil galvanometer ?
Answer:
Moving coil galvanometer is sensitive galvanometer, it is used to measure very small current upto 10^-9 A.

Question 10.
How do you convert a moving coil galvanometer into an ammeter ?
Answer:
A small resistance is connected in parallel to the moving coil galvanometer, then it converts to ammeter.

S = \(\frac{\mathrm{G}}{\frac{\mathrm{i}}{\mathrm{i}_{\mathrm{g}}}-1}\)

Question 11.
How do you convert a moving coil galvanometer into a voltmeter ? [T.S. Mar. 16, 15, 14; A.P. Mar. 16]
Answer:
A high resistance is connected in series to the moving coil galvanometer, then it converts to voltmeter.

R = \(\frac{\mathrm{v}}{\mathrm{i}_g}\) – G

Question 12.
What is the relation between the permittivity of free space e₀, the permeability of free space m₀ and the speed of light In vaccum?
Answer:
Speed of light in vaccum (C) = \(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)
Here μ₀ = m₀ = permeability in vaccum
ε₀ = permittivity in vaccum.

Question 13.
A current carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic Held be set up in such a manner that the loop turns about the vertical axis ?
Answer:
Torque (τ) = \(\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}=\mathrm{i} \overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\) (M = n i A)
where i is current, \(\overrightarrow{\mathrm{A}}\) is area vector, \(\overrightarrow{\mathrm{B}}\) is magnetic field. Area vector \(\overrightarrow{\mathrm{A}}\) acts normal to the loop, so torque \(\vec{\tau}\) cannot act along the vertical axis. The magnetic field is not set up to turn the loop around itself.

Question 14.
A current carrying circular loop is placed in a uniform external magnetic field. If the loop is free to turn, what is its orientation when it achieves stable equilibrium?
Answer:
The plane of the loop is perpendicular to the direction of magnetic field because the torque on the loop in this orientation is zero.

Question 15.
A wire loop of irregular shape carrying current is placed in an external magnetic field. If the wire is flexible, what shape will the loop change to ? Why ?
Answer:
For a given perimeter, a circle has maximum area among all geometrical shapes. So to maximise the magnetic flux through it will assume a circular shape with its plane normal to the field.

Question 16.
Consider a tightly wound 100 turn coil of radius 10 cm, carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil ?
Solution:
Since the coil is tightly wound we may take each circular element to have the same radius R = 10 cm = 0.1 m. The number of turns N = 100. The magnitude of the magnetic field is (From Eq.),
B₀ = \(\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}} \hat{\mathrm{i}}\)
B = \(\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}=\frac{4 \pi \times 10^{-7} \times 10^2 \times 1}{2 \times 10^{-1}}\) = 2π × 10^-4 = 6.28 × 10^-4 T

Question 17.
A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid ?
Solution:
The number of turns per unit length is,
n = \(\frac{500}{0.5}\) = 1000 turns / m
The length l = 0.5m and radius r = 0.01 m. Thus, l/a = 50 i.e., l >> a.
Hence, we can use the long solenoid formula, namely, Eq. (B = μ₀nI)
B = μ₀ n I
= 4π × 10^-7 × 10³ × 5 = 6.28 × 10^-3 T

Question 18.
A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil ?
Solution:
Here, n = 100, r = 8cm = 8 × 10^-2 m and I = 0.40 A
The magnetic field B at the centre
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \pi \mathrm{In}}{\mathrm{r}}=\frac{10^{-7} \times 2 \times 3.14 \times 0.4 \times 100}{8 \times 10^{-2}}\) = 3.1 × 10^-4 T
The direction of magnetic field depends on the direction of current if the direction of current is anticlockwise. According to Maxwell’s right hand rule, the direction of magnetic field at the centre of coil will be perpendicular outwards to the plane of paper.

Question 19.
What is the magnitude of magnetic force per unit length on a wire carrying a current of 8A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T ?
Solution:
According to the question
I = 8 A, 6 = 30°, B = 0,15 T, l = 1 m

The magnitude of magnetic force
f = I (l × B) = I l B sin θ
= 8 × 1 × 0.15 × sin 30°
= \(\frac{8 \times 0.15}{2}\) = 4 × 0.15 = 0.6 N/m

Short Answer Questions

Question 1.
State and explain Biot-Savart’s law.
Answer:
Consider a very small element of length dl of a conductor carrying current (i). Magnetic induction due to small element at a point P distance r from the element.
Magnetic induction (dB) is directly proportional to i) current (i) ii) Length of the element (di) iii) sine angle between r and dl and inversely proportional to the square of the distance from small element to point P.

Question 2.
State and explain Ampere’s law.
Answer:
Ampere’s law : The line integral of the intensity of magnetic induction field around closed path is equal to μ₀ times the net current (i) enclosed by the path.
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = μ₀i
Proof: Consider a long straight conductor carrying current i as shown in figure. Magnetic induction at a distance r from the conductor is given by

B = \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\) (From Biot-Savart’s law)
The value of B is same at all points on the circle.
\(\oint \overrightarrow{\mathrm{B}} . \overrightarrow{\mathrm{d} l}=\oint \mathrm{B} \mathrm{d} l \cos \theta\)
= \(\mathrm{B} \oint \mathrm{d} l\) = B × 2π
= \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\) × 2πr
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = μ₀i
This proves Ampere’s circuital laws.

Question 3.
Find the magnetic induction due to a long current carrying conductor.
Answer:
Consider a long straight conductor carrying a current i. Let P be a point at a distance r from the conductor. Let r be the radius of the circle passing through point p.

Magnetic induction is same at all points on the circle. Consider a small element of length dl.
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}=\oint \mathrm{Bd} l \cos \theta\)
Angle between B and dl is zero i.e. θ = 0
= \(\mathrm{B} \oint \mathrm{d} l\)
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = B (2πr) ………………. (1)
According to Ampere’s laws
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = μ₀i ……………. (2)
From equations (1) and (2), B (2πr) = μ₀i
= \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\)

Question 4.
Derive an expression for the magnetic induction at the centre of a current carrying circular coil using Biot-Savart’s law.
Answer:
Consider a circular coil of radius r and carry a current! Consider a small element ‘dl’. Let O is the centre of the coil. By using Biot – Savart’s law,

dB = \(\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{id} l \sin \theta}{\mathrm{r}^2}\)
Here angle \(\overrightarrow{\mathrm{d} l}\) and \(\overrightarrow{\mathrm{r}}\) is 90° (i.e., θ = 90°)
dB = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{idl}}{\mathrm{r}^2}\) …………… (1)
As the field due to all elements of the circular loop have the same direction. The resultant magnetic field can be obtained by integrating equation (1)

Question 5.
Derive an expression for the magnetic induction of a point on the axis of a current carrying circular coil using Biot-Savart’s law.
Answer:

Consider a circular coil of radius R and carrying a current i. Let P is a point on the axis at a distance x from the centre O. Let r be the distance of small element (dl) from P.
From Biot – savart’s law
dB = \(\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{id} l \sin \theta}{\mathrm{r}^2}=\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{id} l}{\mathrm{r}^2}\) ……………….. (1)
(∵ θ = 90° Angle between \(\overrightarrow{\mathrm{d} l}\) and \(\overrightarrow{\mathrm{r}}\))
dB can be resolved into two components dB cosθ and dB sinθ. If we consider another This also resolved into dB cosθ and dB sinθ.
The components along the axis will add up and perpendicular to the axis will cancel.
∴ Resultant magnetic induction at P is

Question 6.
Explain how crossed E and B fields serve as a velocity selector.
Answer:
When a charged particle q moving with a velocity v in presence of both electric and magnetic fields.
The force experienced due to electric field F_E = q\(\overrightarrow{\mathrm{E}}\)
The force experienced due to magnetic field F_B = q \((\vec{v} \times \vec{B})\)
Consider electric and magnetic fields are perpendicular to each other and also perpendicular to the velocity of the particle.

E = E\(\hat{\mathrm{j}}\),B = B\(\hat{\mathrm{k}}\), v = υ\(\hat{\mathrm{i}}\)
F_E = qE\(\hat{\mathrm{j}}\), F_B =q(v\(\hat{\mathrm{i}}\) × B\(\hat{\mathrm{k}}\)) = – qvB\(\hat{\mathrm{j}}\)
∴ F = F_E + F_B
F = q (E – υB)\(\hat{\mathrm{j}}\)
Thus electric and magnetic forces are in opposite directions.
We adjust E and B such that, the forces are equal
F_E = F_B
qE = q υ B
υ = \(\frac{E}{B}\)
This condition can be used to selefct charged particles of a particular velocity. The crossed field E and B serve as a velocity selector.

Question 7.
What are the basic components of a cyclotron ? Mention its uses ?
Answer:
Cyclotron is a device used to accelerate positively charged particles like protons, α – particles, deutrons etc.

Cyclotron mainly consists of

1. Two hollow D-shaped metallic chambers D₁ and D₂
2. High-frequency oscillator
3. Strong electro magnet
4. Vaccum chamber.

Uses of cyclotron :

1. It is used for producing radioactive material for medical purposes i.e. diagnostics and treatment of chronic diseases. “
2. It is used to improve the quality of solids by adding ions.
3. It is used to synthesise fresh substances.
4. It is used to bombard the atoms with highly accelerated particles to study the nuclear reactions.

Question 8.
Derive an expression for the magnetic dipole moment of a revolving electron. [A.P. Mar. 16]
Solution:
Consider an electron revolving in a circular orbit of radius r with speed v and frequency υ. If the electron cross a point P on the circle in every revolution, then distance travelled by electron to complete one revolution = 2πr.
No. of revolutions in one second (υ) = \(\frac{\mathrm{v}}{2 \pi \mathrm{r}}\)
The electric current (i) = \(\frac{\text { Charge }}{\text { Time }}\) = charge × frequency
i = e × \(\frac{\mathrm{v}}{2 \pi \mathrm{r}}\)
∴ Magentic dipole moment (M) = iA (∵ N = 1)
M = \(\frac{\mathrm{ev}}{2 \pi \mathrm{r}}\) × πr² (∵ A = πr²)
M = \(\frac{\mathrm{evr}}{2}\)

Long Answer Questions

Question 1.
Deduce an expression for the force on a current carrying conductor placed in a magnetic field. Derive an expression for the force per unit length between two parallel current-carrying conductors.
Answer:
Expression for the Force acting on a current carrying conductor :
Consider a straight conductor (wire) of length T, area of cross section ’A1, carrying a current T, which is placed in a uniform magnetic field of induction ’B’ as shown in fig.

We know the external magnetic field exerts a force on the conductor.
The electrons in effect move with an average velocity called drift velocity ‘V_d‘. The direction of conventional current will be opposite to the direction of drift velocity.

Let us assume that the current flows through the conductor from left ‘B’ in the plane of the paper makes an angle ‘θ’ with the direction of current ‘i’ as shown in fig.
If F’ is the force acting on the charge ‘q’ in B.
∴ F’ = q V_d B sin θ
If ‘n’ represents number of moving electrons per unit volume (∵ n = \(\frac{N}{V}\))
∴ Current i = nq V_d A
If ‘N’ is the number of electrons in the length ‘l’
N = nlA
Total force on conductor F = F’.N (∵N = nV = n × A × l)
= (q V_d B sin θ) (nlA)
(nqV_dA) (lB sin θ)
∴ F = ilB sin θ
Case (i) : If θ = 0°, F_Min = 0
Case (ii) : If θ = 90°, F_Max = Bil
Expression for the force between two Parallel conductors carrying conductors :
Consider two straight parallel conductors AB and ‘CD’ carrying currents ‘ix’ and ‘i2’ and which are separated by a distance ‘r’ as shown in fig.

If B₁ and B₂ are magnetic inductions produced by the current carrying conductors AB and CD. Magnetic induction Bx at a distance ‘r’ from the conductor ‘AB’ can be written as B₁ = \(\frac{\mu_0 \mathrm{i}_1}{2 \pi \mathrm{r}}\)
If ‘F’ is forœ acting on ‘CD’ clue to magnetic induction ‘B₁‘ then
F_CD = i₂lB₁
Where l = length of the conductor
F_CD = i₂l \(\left(\frac{\mu_0 i_1}{2 \pi \mathrm{r}}\right)=\frac{\mu_0 i_1 i_2 l}{2 \pi r}\) ……………. (1)
The direction of the force can be determined by using Flemings left hand rule.
Similarly we can find the force acting on the A conductor AB due to magnetic induction B₂.
F_AB = i₁lB₂
∴ F_AB = i₁l \(\left(\frac{\mu_0 i_2}{2 \pi r}\right)\) ………….. (2) [∵B₂ = \(\left(\frac{\mu_0 i_2}{2 \pi r}\right)\)]
From the equations (1) and (2) F_AB = F_CD = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 l}{2 \pi \mathrm{r}}\)
∴ Force between two parallel, straight conductors carrying currents,
F = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 l}{2 \pi \mathrm{r}}\)
Force per unit length \(\frac{\mathrm{F}}{l}=\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2}{2 \pi \mathrm{r}}\)

Question 2.
Obtain an expression for the torque on a current carrying loop placed in a uniform ‘ magnetic field. Describe the construction and working of a moving coil galvanometer.
Answer:
Torque acting on a coil carrying a current kept in a uniform magnetic field : Let a rectangular current loop ABCD of length l = AB = CD and width b = AD – BC carrying a current “i” be suspended in a magnetic field of flux density B.
The normal ON drawn to the plane of the coil makes an angle ‘θ’ with the magnetic field B.

Force on arm AD = \(\mathrm{i} \overline{\mathrm{b}} \times \overline{\mathrm{B}}\) acting upwards along the axis of suspension
Force on arm BC = \(\mathrm{i} \overline{\mathrm{b}} \times \overline{\mathrm{B}}\) acting downwards along the axis of suspension
Hence these two forces cancel.

Force on arm AB = ilB acting perpendicular to the plane as shown.
Force on arm CD = ilB acting perpendicular to the plane as shown.
These two forces constitute a couple on the coil.
Moment of the couple = (Force) × (Perpendicular distance between the forces) = ilB (PQ sin θ)
Torque = ilB b sinθ
But l × b = Area of coil
∴ Torque = iAB sin θ
If the loop has ‘n’ turns the torque on the coil
τ = n i AB sin θ
If ‘Φ’ is the deflection of the coil, that is the angle between the plane of the coil and magnetic field B
τ = n i AB cos Φ

Moving coil galvanometer:
Principle : When a current carrying coil is placed in the uniform magnetic field, it experiences a torque.
Construction :

1. It consists of a coil wound on a non metallic frame.
2. A rectangular coil is suspended between two concave shaped magnetic poles with the help of phosphour Bronze wire.
3. The lower portion of the coil is connected to a spring.
4. A small plane mirror M is fixed to the phosphour Bronze wire to measure the deflection of the coil.
5. A small soft iron cylinder is placed within the coil without touching the coil. The soft iron cylinder increases the induction field strength.
6. The concave shaped magnetic poles render the field radial. So maximum torque acting on it.
7. The whole of the apparatus is kept inside a brass case provided with a glass window.
   

Theory:
Consider a rectangular coil of length l and breadth b and carrying current i suspended in the induction field strength B.
Deflecting torque (τ) = B i A N …………….. (5)
where A = Area of the coil
N = Total number of turns.
The restoring torque developed in the suspension = C θ …………….. (2)
Where C is the couple per unit twist and 9 is the deflection made by the coil.
When the coil is in equilibrium position
Deflecting torque = Restoring torque
B i A N = Cθ
i = \(\left(\frac{\mathrm{C}}{\mathrm{BAN}}\right) \theta\)
Where K = \(\frac{\mathrm{C}}{\mathrm{BAN}}\) = Galvanometer constant.
i = K θ ……………… (3)
i ∝ θ
Thus deflection of the coil is directly proportional to the current flowing through it. The deflection in the coil is measured using lamp and scale arrangement.

Question 3.
How can a galvanometer be converted to an ammeter ? Why is the parallel resistance smaller that the galvanometer resistance ? A moving coil galvanometer can measure a current of 10-6 A. What is the resistance of the shunt required if it is to measure 1A ?
Answer:
Conversion of Galvanometer into Ammeter :
Galvanometer is converted into an ammeter by connecting a suitable resistance is parallel to it:

This arrangement decreases the effective resistance.

Ammeter is used for measuring the current in an electric circuit and it is connected in series in circuit. The inclusion of the ammeter in the circuit should not alter the current or total resistance of the circuit so it has very low resistance.

The resistance of An ideal Ammeter is zero.
Let G and S be the Galvanometer and shunt resistances respectively.
Let ‘i’ be the total current, divided at A into i_g and i_s as shown in fig.
From Kirchhoffs I^st law, i = i_g + i_s
As ‘G’ and ‘S’ are parallel P.D. across
Galvanometer = P.D. across shunt
i_gG = i_sS
S = \(\frac{\mathrm{i}_{\mathrm{g}}}{\mathrm{i}_{\mathrm{s}}} \mathrm{G}\)
= \(\frac{\mathrm{Gi}_{\mathrm{g}}}{\mathrm{i}-\mathrm{i}_{\mathrm{g}}}\) [∵ i_s = i – i_g]
S = \(\frac{\mathrm{G}}{\frac{\mathrm{i}}{\mathrm{i}_{\mathrm{g}}}-1}\)
If \(\frac{\mathrm{i}}{\mathrm{i}_{\mathrm{g}}}\) = n ⇒ i_g = \(\frac{\mathrm{i}}{\mathrm{n}}\)
∴ The current flowing through the galvanometer be \(\left(\frac{1}{n}\right)^{\text {th }}\) of total current.
∴ S = \(\frac{\mathrm{G}}{\mathrm{n}-1}\)
If ‘R’ is the effective resistance between points ‘A’ and ‘B’ then

Hence current through galvanometer is proportional to the total current. Since ‘S’ is small major portion of the current flows through it and a small portion of current flows through G. So shunt protects the galvanometer from high currents. Parallel resistance is smaller than Galvanometer resistance because to protect the Galvanometer from high (large) current (or) to pass. Large currents through shunt and small current passes through the galvanometer.
Solution for the problem : Current in the circuit i = 1A;
Current through the galvanometer, i_g = 10^-6A
Shunt resistance, S = \(\frac{G}{n-1}=\frac{G}{\frac{i}{i_g}-1}=\frac{G}{10^6-1}=\frac{G}{99.999} \Omega\)

Question 4.
How can a galvanometer be converted to a voltmeter ? Why is the series resistance greater that the galvanometer resistance ? A moving coil galvanometer of resistance 5Ω can measure a current of 15mA. What is the series resistance required if it is to measure 1.5V ?
Answer:
Conversion of Galvanometer into Voltmeter : A galvanometer is converted into voltmeter by connecting a high resistance (R) in series with it. Voltmeter is used to measure the P.D. between any two points in circuit and it is connected in parallel to the component of the circuit.

Let ‘V’ be the potential difference to be measured between the points ‘A’ and ‘B’.
∴ V = (R + G) i_g [∴ V = iR]
i_g = Current passing through the galvanometer
\(\frac{\mathrm{V}}{\mathrm{i}_{\mathrm{g}}}\) = R + G
R = \(\frac{\mathrm{V}}{\mathrm{i}_{\mathrm{g}}}\) – G ……………. (1)
The value of ‘R’ can be calculated by using the above formula. If V_g is the maximum P.D. across the galvanometer then V_g = i_g G
∴ i_g = \(\frac{V_g}{G}\) …………….. (2)
Substitute ‘i_g‘ in Equ (1)
R = \(\frac{V_G}{V_g}\) – G = \(\left(\frac{V}{V_g}-1\right)\)
If \(\frac{V}{V_g}\) = n ⇒ R = G(n – 1)
Note : n = \(\frac{V}{V_g}\) is the ratio of maximum voltage to be measured to the maximum voltage across the galvanometer.

Series resistance is greater than galvanometer resistance because the current in external resistance and potential difference will be decreased and to increase the resistance of the galvanometer.
Solution for the problem:

Problems

Question 1.
Two long and parallel straight wires A and B canying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
Solution:
Given I₁ = 8A, I₂ = 5A and r = 4 cm = 0.4m
F = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{I}_1 \mathrm{I}_2}{\mathrm{r}}=\frac{10^{-7} \times 2 \times 8 \times 5}{0.04}\) = 2 × 10^-4 N
The force on A of length 10 cm is F¹ = F × 0.1 (∵ 1 m = 100 cm)
F¹ = 2 × 10^-4 × 0.1
F¹ = 2 × 10^-5 N.

Question 2.
A current of 10A passes through two very long wires held parallel to each other and separated by a distance of 1m. What is the force per unit length between them ? [A.P. & T.S. Mar. 15]
Answer:
i₁ = i₂ = 10A
r = 1m
\(\frac{\mathrm{F}}{l}=\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2}{2 \pi \mathrm{r}}\)
= \(\frac{4 \pi \times 10^{-7} \times 10 \times 10}{2 \pi \times 1}\)
\(\frac{\mathrm{F}}{l}\) = 2 × 10^-5 Nm^-1.

Question 3.
A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B (Fig.). What is the magnitude of the magnetic field ?

Solution:
From Eq F = Il × B we find that there is an upward force F, of magnitude IlB,. For mid-air suspension, this must be balanced by the force due to gravity.
m g = I lB mg
B = \(\frac{\mathrm{mg}}{\mathrm{I} l}\)
= \(\frac{0.2 \times 9.8}{2 \times 1.5}\) = 0.65 T
Note that it would have been sufficient to specify mll, the mass per unit length of the wire. The earth’s magnetic field is approximately 4 × 10^-5 T and we have ignored it.

Question 4.
The horizontal component of the earth’s magnetic field at a certain place is 3.0 × 10^-5 T and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1 A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is
(a) east to west;
(b) south to north ?
Solution:
F = Il × B
F = IlB sin θ
The force per unit length is
f = F/l = IB sinθ
a) When the current is flowing from east to west,
θ = 90°
Hence,
f = IB
= 1 × 3 × 10^-5 = 3 × 10^-5 Nm^-1
This is larger than the value 2 × 10^-7 Nm^-1 quoted in the definition of the ampere. Hence it is important to eliminate the effect of the earth’s magnetic field and other stray fields while standardising the ampere.
The direction of the force is downwards. This direction may be obtained by the directional property of cross product of vectors.

b) When the current is flowing from south to north, .
θ = 0°
f = 0
Hence there is no force on the conductor.

Textual Examples

Question 1.
A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal niagnetic field B (Fig.). What is the magnitude of the magnetic field ?

Solution:
From Eq F = Il × B we find that there is an upward force F, of magnitude IlB,. For mid-air suspension, this must be balanced by the force due to gravity, m g = I lB .
B = \(\frac{\mathrm{mg}}{\mathrm{I} l}\)
= \(\frac{0.2 \times 9.8}{2 \times 1.5}\) = 0.65 T
Note that it would have been sufficient to specify m/l, the mass per unit length of the wire. The earth’s magnetic field is approximately 4 × 10^-5 T and we have ignored it.

Question 2.
If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis (Fig.), which way would the Lorentz force be for (a) an electron (negative charge), (b) a proton (positive charge).

Solution:
The velocity v of particle is along the x-axis, while B, the magnetic field is along the y-axis, so v × B is along the z-axis (screw rule or right-hand thumb rule). So, (a) for electron it will be along -z axis, (b) for a positive charge (proton) the force is along +z axis.

Question 3.
What is the radius of the path of ah electron (mass 9 × 10^-31 kg and charge 1.6 × 10^-19 C) moving at a speed of 3 × 10⁷ m/s in a magnetic field of 6 × 10^-4 T perpendicular to it ? What is its frequency ? Calculate its energy in keV. (1 eV = 1.6 × 10^-19 J).
Solution:
Using Eq. r = mυ/qb we find
r = mυ/(qB) = 9 × 10^-31 kg × 3 × 10⁷ m s^-1 / (1.6 × 10^-19 C × 6 × 10^-4 T)
= 26 × 10^-2 m = 26 cm
v = υ / (2 πr) = 2 × 10⁶ s^-1 = 2 × 10⁶ Hz = 2MHz.
E = (1/2)mυ² = (1/2) 9 × 10^-31 kg × 9 × 10¹⁴ m2/s = 40.5 × 10^-17 J
≈ 4 × 10^-16 J = 2.5 keV.

Question 4.
A cyclotron’s oscillator frequency is 10 MHz. What should be the operating magnetic field for accelerating protons ? If the radius of its does is 60 cm, what is the kinetic energy (in MeV) of the proton beam produced by the accelerator.
(e = 1.60 × 10^-19 C, m_p = 1.67 × 10^-27 kg, 1 MeV = 1.6 × 10^-13 J).
Solution:
The oscillator frequency should be same as proton’s cyclotron frequency.
Using Eqs. r = mυ/qb and ω = 2πυ = \(\frac{\mathrm{qB}}{\mathrm{M}}\) we have
B = 2π m υ/q = 6.3 × 1.67 × 10^-27 × 10⁷ / (1.6 × 10^-19) = 0.66 T
Final velocity of protons is
υ = r × 2π v = 0.6 m × 6.3 × 10⁷ = 3.78 × 10⁷ m/s.
E = 1/2 mv² = 1.67 × 10^-27 × 14.3 × 10¹⁴/ (2 × 1.6 × 10^-13) = 7 MeV

Question 5.
element ∆1 = ∆x\(\hat{\mathrm{i}}\) is placed at the origin and carries a large current I = 10 A (Fig.). Wat is the magnetic field on the y-axis at a distance of 0.5 m. ∆x = 1 cm.

Solution:
|dB| = \(\frac{\mu_0}{4 \pi} \frac{I \mathrm{dl} \sin \theta}{\mathrm{r}^2}\)
dl = ∆x = 10^-2 m, I = 10 A, r = 0.5 m = y, μ₀/4π = 10^-7 \(\frac{\mathrm{Tm}}{\mathrm{A}}\)
θ = 90°; sin θ = 1
|dB| = \(\frac{10^{-7} \times 10 \times 10^{-2}}{25 \times 10^{-2}}\) = 4 × 10^-8 T
The direction of the field is in the +z-direction. This is so since,
dl × r = \(\Delta \mathrm{x} \hat{\mathrm{i}} \times \mathrm{y} \hat{\mathrm{j}}=\mathrm{y} \Delta \mathrm{x}(\hat{\mathrm{i}} \times \hat{\mathrm{j}}) \mathrm{y} \Delta \mathrm{x} \hat{\mathrm{k}}\)
We remind you of the following cyclic property of cross-products,
\(\hat{\mathrm{i}} \times \hat{\mathrm{j}}=\mathrm{k} ; \hat{\mathrm{j}} \times \hat{\mathrm{k}}=\hat{\mathrm{i}} ; \hat{\mathrm{k}} \times \hat{\mathrm{i}}=\hat{\mathrm{j}}\)
Note that the field is small in magnitude.

Question 6.
A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2.0 cm as shown in Fig. Consider the magnetic field B at the centre of the arc. (a) What is the magnetic field due to the straight segments ? (b) In what way the contribution to B from the semicircle differs from that of a circular loop and in what way does it resemble ? (c) Would your answer be different if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown in Fig. (b) ?

Solution:
a) dl and r for each element of the straight segments are parallel. Therefore, dl × r = 0. Straight segments do not contribute to |B|.

b) For all segments of the semicircular arc, dl × r are all parallel to each other (into the plane of the paper). All such contributions add up in magnitude. Hence direction of B for a semicircular arc is given by the right-hand rule and magnitude is half that of a circular loop. Thus B is 1.9 × 10^-4 T normal to the plane of the paper going into it.

c) Same magnitude of B but opposite in direction to that in (b).

Question 7.
Consider a tightly wound 100 turn coil of radius 10 cm, carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil?
Solution:
Since the coil is tightly wound we may take each circular element to have the same radius R = 10 cm = 0.1 m. The number of turns N = 100. The magnitude of the magnetic field is (From Eq.),
B₀ = \(\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}} \hat{\mathrm{i}}\)
B = \(\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}=\frac{4 \pi \times 10^{-7} \times 10^2 \times 1}{2 \times 10^{-1}}\) = 2π × 10^-4 = 6.25 × 10^-4 T

Question 8.
Magnetic field due to a long current-carrying wire Oersted’s experiments showed that there is a magnetic field around a current-carrying wire. We determine the magnitude of magnetic field at some distance from a long striaght wire carrying a current I.
Solution:
The direction of the field is given by the right hand rule. The figure shows an element dl of the current-carrying wire. The point P, where the field is to be determined is at a perpendicular distance ‘S’ from the wire. The position vector of P from dl is r.

The magnitude dB of the magnetic field due to dl is given by Biot-Savart law to be

now from the figure S = r cosθ which gives l/r² = cos²θ/s²
and l’ = S tanθ which gives dl’ = S sec²θ dθ = S dθ/cos²θ
thus dB = \(\frac{\mu_0 I \cos \theta}{4 \pi S}\) dθ
we integrate this to get B at P If the wire is very long then the limits for 0 would be -π/2 to π/2
thus B = \(\frac{\mu_0 l}{2 \pi S}\) (emerging from the paper at P)

Question 9.
Find \(\oint\) B . dl for the paths shown in (a) and (b)

Solution:
a) Going around theKpath in the anticlockwise direction, I₁ is taken as positive while I₃ is negative. Currents I₂ and I₄ do not matter as they are NOT enclosed by the path.
\(\oint\) B . dl = μ₀(I₁ – I₃)
Note : Currents I₂ and I₄ create magnetic fields all around them and B due to them on any element of the path would be non-zero. However, the sum B.dZ due to them would be zero, b) Calculation of B. dl for the entire path can be broken up into two separate calculations, one covering all contributions from an anti-clockwise traversal around I₁ and the other covering all contributions from a clockwise traversal around I₃. Thus
\(\int_1 \mathrm{~B} . \mathrm{d} l\) = μ₀I₁ for all elements around I₁ traversed in an anti-clockwise direction
\(\int_2 \mathrm{~B} . \mathrm{d} l\) = μ₀I₃ for all elements around I₃ traversed in a clockwise direction; I₃ taken as positive because it is flowing into the plane. Thus the total \(\oint\) B.dl = μ₀ (I₁ – I₃)

Question 10.
Figure shows the circular cross-section of a long straight wire of radius a carrying steady current I. The current I is uniformly distributed across this cross-section. Calculate the magnetic field in the region r < a (dashed inner circle) and r > a (dashed outer circle).

Solution:
a) Consider the case r > a. The Amperian loop, labelled 2, is a circle concentric with the cross-section. For this loop,
L = 2πr
I_e = Current enclosed by the loop = I
The result is the familiar expression for a long straight wire B(2πr) = μ₀I
B = \(\frac{\mu_0 I}{2 \pi \mathrm{r}}\) ……………… (1)
b) Consider the case r < a. The Amperian loop is a circle labelled 1 . For this loop, taking the radius of the circle to be r,
L = 2 π r
Now the current enclosed I_e is not I (because r < a), but is less than this value. Since the current distribution is uniform, the current enclosed is,

Figure shows a plot of the magnitude of B with distance r from the centre (axis) of the wire. The direction of the field is tangential to the respective circular loop (1 or 2) and given by the right-hand rule described earlier in this section.
This example possesses the required symmetry so that Ampere’s law can be applied readily.

Question 11.
A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid ?
Solution:
The number of turns per unit length is,
n = \(\frac{500}{0.5}\) = 1000 turns / m
The length l = 0.5m and radius r = 0.01 m. Thus, l/a = 50 i.e., l > >a.
Hence, we can use the long solenoid formula, namely, Eq. (B = μ₀nI)
B = μ₀ n I
= 4π × 10^-7 × 10³ × 5 = 6.28 × 10^-3 T

Question 12.
The horizontal componet of the earth’s magnetic field at a certain place is 3.0 × 10^-5 T and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is
(a) east to west;
(b) south to north ?
Solution:
F = Il × B
F = IlB sin θ
The force per unit length is
f = F/l = IB sinθ
a) When the current is flowing from east to west,
θ = 90°
Hence,
f = IB
= 1 × 3 × 10^-5 = 3 × 10^-5 Nm^-1
This is larger than the value 2 × 10^-7 Nm^-1 quoted in the definition of the ampere. Hence it is important to eliminate the effect of the earth’s magnetic field and other stray fields while standardising the ampere.
The direction of the force is downwards. This direction may be obtained by the directional property of cross product of vectors.

b) When the current is flowing from south to north,
θ = 0°
f = 0
Hence there is no force on the conductor.

Question 13.
A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2A. (a) What is the field at the centre of the coil ? (b) What is the magnetic moment of this coil ?
The coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 2T in the horizontal direction exists such that initially the axis of the coil is in the direction of the field. The coil rotates through an angle of 90° under the influence of the magnetic field, (c) What are the magnitudes of the torques on the coil in the initial and final position ? (d) What is the angular speed acquired by the coil when it has rotated by 90° ? The moment of inertia of the coil is 0.1 kg m². [A.P. Mar. 19]
Solution:
a) From B = \(\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}\)
Here, N = 100; I = 3.2 A, and R = 0.1 m. Hence,
B = \(\frac{4 \pi \times 10^{-7} \times 10^2 \times 3.2}{2 \times 10^{-1}}=\frac{4 \times 10^{-5} \times 10}{2 \times 10^{-1}}\) (using π × 3.2 = 10)
= 2 × 10^-3 T
The direction is given by the right-hand thumb rule.

b) The magnetic moment is given by
m = N I A = N I π r² = 100 × 3.2 × 3.14 × 10^-2 = 10 A m²
The direction is once again given by the right hand thumb rule.

c) τ = |m × B|
= mBsin θ
Initially, θ = 0. Thus initial torque τ_i = 0. Finally, θ = π/2 (or 90°).
Thus, final torque τ_f = m B = 10 × 2 = 20 N m.

d) From Newton’s second law.
I\(\frac{\mathrm{d} \omega}{\mathrm{dt}}\) = mBsin θ
where I is the moment of inertia of the coil. From chain rule,
\(\frac{\mathrm{d} \omega}{\mathrm{dt}}=\frac{\mathrm{d} \omega}{\mathrm{d} \theta} \frac{\mathrm{d} \theta}{\mathrm{dt}}=\frac{\mathrm{d} \omega}{\mathrm{d} \theta} \omega\)
Using this,
Iω dω = m B sinθ dθ
Integrating from θ = 0 to θ = π/2,
\(\mathrm{I} \int_0^{\omega \mathrm{I}} \omega \mathrm{d} \omega=\mathrm{mB} \int_0^{\pi / 2} \sin \theta d \theta\)
\(\mathrm{I} \frac{\omega_{\mathrm{I}}^2}{2}=-\left.\mathrm{mB}[\cos \theta]\right|_0 ^{\pi / 2}=\mathrm{mB}\)
ω_f = \(\left(\frac{2 \mathrm{mB}}{\mathrm{I}}\right)^{1 / 2}=\left(\frac{2 \times 20}{10^{-1}}\right)^{1 / 2}\) = 20 s^-1

Question 14.
a) A current-carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns around it self (i.e. turns about the vertical axis).
Solution:
No, because that would require τ to the in the vertical direction. But τ = I A × B, and since A of the horizontal loop is in the vertical direction, τ would be in the plane of the loop for any B.

b) A current-carrying circular loop is located in a uniform external magnetic field. If the loop is free to turn, what is its. orientation of stable equilibrium ? Show that in this orientation, the flux of the total field (external field + field produced by the loop) is maximum.
Solution:
Orientation of stable equilibrium is one where the area vector A of the loop is in the direction of external magnetic field. In this orientation, the magnetic field produced by the loop is in the same direction as external field, both normal to the plane of the loop, thus giving rise to maximum flux of the total field.

c) A loop of irregular shape carrying current is located in an external magnetic field. If the wire is flexible, why does it change to a circular shape ? What could be the sense of current in the loop and the direction of magnetic field ?
Solution:
It assumes circular shape with its plane normal to the field to maximize flux, since for a given perimeter, a circle encloses greater area than any other shape.

Question 15.
In the circuit (Fig.) the current is to be measured. What is the value of the current if the ammeter shown (a) is a galvanometer with a resistance R_G = 60.00 Ω; (b) is a galvanometer described in (a) but converted to an ammeter by a shunt resistance r_s = 0.02 Ω; (c) is an ideal ammeter with zero resistance ?

Solution:
a) Total resistance in the circuit is,
R_G + 3 = 63 Ω. Hence I = 3/63 = 0.048 A.

b) Resistance of the galvanometer converted to an ammeter is,
\(\frac{\mathrm{R}_{\mathrm{G}} \mathrm{r}_{\mathrm{s}}}{\mathrm{R}_{\mathrm{G}}+\mathrm{r}_{\mathrm{s}}}=\frac{60 \Omega \times 0.02 \Omega}{(60+0.02) \Omega}\) = 0.02 Ω
Total resistance in the circuit is,
0.02 Ω + 3 Ω = 3.02 Ω. Hence I = 3/3.02 = 0.99 A.

c) For the ideal ammeter with zero resistance,
1 = 3/3 = 1.00 A

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Measures of Dispersion Solutions Exercise 8(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Measures of Dispersion Solutions Exercise 8(a)

I.

Question 1.
Find the mean deviation about the mean for the following data.
(i) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Solution:
The arithmetic mean of the given data
\(\bar{x}=\frac{38+70+48+40+42+55+63+46+54+44}{10}\)
= \(\frac{500}{10}\)
= 50
The absolute values of the deviations are
\(\left|x_i-\bar{x}\right|\), |50 – 38|, |50 – 70|, |50 – 48|, |50 – 40|, |50 – 42|, |50 – 55|, |50 – 63|, |50 – 46|, |50 – 54|, |50 – 44|
= 12, +20, 2, 10, 8, +5, +13, 4, +4, 6
∴ The mean deviation from the mean

(ii) 3, 6, 10, 4, 9, 10
Solution:
The arithmetic mean of the given data
\(\bar{x}=\frac{3+6+10+4+9+10}{6}\)
= \(\frac{42}{6}\)
= 7
The absolute values of the deviations are
\(\left|x_i-\bar{x}\right|\), |3 – 7|, |16 – 7|, |10 – 7|, |4 – 7|, |9 – 7|, |10 – 7|
= 4, 1, 3, 3, 2, 3
∴ The mean deviation from the mean

Question 2.
Find the mean deviation about the median for the following data.
(i) 13, 17, 16, 11, 13, 10, 16, 11, 18, 12, 17
Solution:
Given ungrouped data are
13, 17, 16, 11, 13, 10, 16, 11, 18, 12, 17
Expressing the data in the ascending order of magnitude, we have
10, 11, 11, 12, 13, 13, 16, 16, 17, 17, 18
∴ Median = 13 = b(say)
The absolute values are
|13 – 10|, |13 – 11|, |13 – 11|, |13 – 12|, |13 – 13|, |13 – 13|, |13 – 16|, |13 – 16|, |13 – 17|, |13 – 17|, |13 – 18|
= 3, 2, 2, 1, 0, 0, 3, 3, 4, 4, 5
∴ Mean deviation from the median

(ii) 4, 6, 9, 3, 10, 13, 2
Solution:
Given ungrouped data are 4, 6, 9, 3, 10, 13, 2
Expressing the data in the ascending order of magnitude, we have
2, 3, 4, 6, 9, 10, 13
∴ Median = 6 = b (say)
The absolute values are
|6 – 2|, |6 – 3|, |6 – 4|, |6 – 6|, |6 – 9|, |6 – 10|, |6 – 13|
= 4, 3, 2, 0, 3, 4, 7
∴ Mean deviation from the median = \(\frac{\sum_{i=1}^7\left|x_i-b\right|}{7}\)
= \(\frac{4+3+2+0+3+4+7}{7}\)
= 3.285

Question 3.
Find the mean deviation about the mean for the following distribution.
(i)

Solution:

(ii)

Solution:

Question 4.
Find the mean deviation about the median for the following frequency distribution.

Solution:

\(\frac{N}{2}\) = 13
∴ Median = 7
Hence mean deviation from the median = \(\frac{1}{N} \sum f_i \mid x_i-\text { Median } \mid\)
= \(\frac{1}{26}\) (84)
= 3.23

II.

Question 1.
(i) Find the mean deviation about the median for the following continuous distribution.

Solution:
Construct the table

\(\frac{N}{2}\) = 25
Observation lies in the class-interval 20-30.
This is the median class.

(ii)

Solution:
Construct the table
\(\frac{N}{2}=\frac{100}{2}\) = 50
Observation lies in the interval 40-50
This is the median class
∴ Median = \(L+\left\{\frac{\frac{N}{2}-\text { P.C.f }}{f}\right\} \times i\)
= 40 + \(\left\{\frac{50-32}{28}\right\} \times 10\)
= 40 + 6.43
= 46.43

∴ Median deviation about median = \(\frac{1}{N} \sum_{i=1}^8 f_i \mid x_i-\text { median } \mid\)
= \(\frac{1}{100}\) (1428.6)
= 14.286

Question 2.
Find the mean deviation about the mean for the following continuous distribution.

Solution:
Taking the assumed mean a = 130 and h = 10 construct the table

\(\bar{x}=a+\left(\frac{\sum f_i d_i}{N}\right) h\)
= 130 + \(\left(\frac{-47}{100}\right) 10\)
= 130 – 4.7
= 125.3
Mean deviation about the mean = \(\frac{1}{N} \sum_{i=1}^6 f_i\left|x_i-\bar{x}\right|\)
= \(\frac{1}{100}\) (1428.8)
= 14.288

Question 3.
Find the variance for the discrete data given below.
(i) 6, 7, 10, 12, 13, 4, 8, 12
Solution:
Given data are 6, 7, 10, 12, 13, 4, 8, 12
Mean \(\bar{x}=\frac{6+7+10+12+13+4+8+12}{8}\)
= \(\frac{72}{8}\)
= 9
Construct the table

Variance σ² = \(\frac{1}{n} \Sigma\left(x_i-\bar{x}\right)^2\)
= \(\frac{1}{8}\) (74)
= 9.25

(ii) 350, 361, 370, 373, 376, 379, 385, 387, 394, 395
Solution:
Given data are 350, 361, 370, 373, 376, 379, 385, 387, 394, 395
Mean \(\bar{x}=\frac{350+361+370+373+376+379+385+387+394+395}{10}\) = 377
Construct the table

Variance = \(\frac{1}{n} \sum_{i=1}^{10}\left(x_i-\bar{x}\right)^2\)
= \(\frac{1}{10}\) (1832)
= 183.2

Question 4.
Find the variance and standard deviation of the following frequency distribution.

Solution:

III.

Question 1.
Find the mean and variance using the step deviation method, of the following tabular data, giving the age distribution of 542 members.

Solution:
If we take assumed mean A = 55 then y_i = \(\frac{x_i-55}{10}\)
Here h = 10
Construct the table

Question 2.
The coefficient of variation of the two distributions are 60 and 70 and their standard deviations are 21 and 16 respectively. Find their arithmetic means.
Solution:
Given coefficient of variance C.V = 60
Standard deviation s = 21
We know C.V = \(\frac{\sigma}{x} \times 100\)
⇒ \(\bar{x}=\frac{\sigma}{C . V} \times 100\)
= \(\frac{21}{60}\) × 100
= 35
Arithmetic mean = 35
Given co-efficient of variance C.V = 70
Standard deviation σ = 16
We know C.V = \(\frac{\sigma}{x} \times 100\)
⇒ \(\bar{x}=\frac{\sigma}{C . V} \times 100\)
= \(\frac{16}{70}\) × 100
= 22.85

Question 3.
From the prices of Shares X and Y given below, for 10 days of trading, find out which share is more stable.

Solution:
For share x
\(\bar{x}=\frac{35+54+52+53+56+58+52+50+51+49}{10}\)
= \(\frac{510}{10}\)
= 51
For share y
\(\bar{y}=\frac{108+107+105+105+106+107+104+103+104+101}{10}\)
= \(\frac{1050}{10}\)
= 105

C.V of share x = \(\frac{\sigma_{\mathrm{x}}}{\mathrm{x}} \times 100\)
= \(\frac{5.92}{51}\) × 100
= 11.61
C.V of share y = \(\frac{\sigma_{\mathrm{y}}}{\mathrm{y}} \times 100\)
= \(\frac{2}{105}\) × 100
= 1.91
since \(\bar{x}<\bar{y}\) and C.V of share x > C.V of shares y
∴ Y share is more stable.

Question 4.
The mean of 5 observations is 4.4. Their variance is 8.24. If three of the observations are 1, 2, and 6. Find the other two observations.
Solution:
Let x and y be the required two observations.
Given, that the mean of 5 observations = 4.4
⇒ \(\frac{1+2+6+x+y}{5}\) = 4.4
⇒ 9 + x + y = 22
⇒ x + y = 13 ……..(1)
Also thin variance = 8.24
⇒ \(\frac{1}{5}\) [(1 – 4.4)² + (2 – 4.4)² + (6 – 4.4)² + (x – 4.4)² + (y – 4.4)²] = 8.24
⇒ (-3.4)² + (-2.4)² + (1.6)² + x² – 8.8x + 19.36 + y² – 8.8y + 19.36 = 41.2
⇒ x² + y² – (8.8) (x + y) = 41.2 – 11.56 – 5.76 – 2.56 – 19.36 – 19.36
⇒ x² + y² – (8.8) (13) = -17.4
⇒ x² + y² = -17.4 + 114.4
⇒ x² + y² = 97
We know (x + y)² = x² + y² + 2xy
⇒ 169 = 97 + 2xy
⇒ 2xy = 72
⇒ xy = 36
(x – y)² = (x + y)² – 4xy
⇒ (x – y)² = 169 – 144 = 25
⇒ x – y = 5 ……..(2)
(1) + (2) ⇒ 2X = 18
⇒ X = 9
(1) – (2) ⇒ 2Y = 8
⇒ Y = 4
∴ The other two observations are 4, 9.

Question 5.
The arithmetic mean and standard deviation of a set of 9 items are 43 and 5 respectively. If an item of value 63 is added to that set, find the new mean and standard deviation of 10 item set given.
Solution:
Given the arithmetic mean of 9 items is 43
∴ \(\bar{x}=\frac{1}{9} \sum_{i=1}^9 x_i=43\)
⇒ Σx_i = 387
If an item of value 63 is added to that set then
Σy_i = Σx_i + 63
= 387 + 63
= 450
∴ \(\bar{y}=\frac{1}{10} \sum y_i\)
= \(\frac{1}{10}\) (450)
= 45
∴ New mean = 45
Given the standard deviation of 9 items is 5
⇒ σ = 5
⇒ σ² = 25
⇒ \(\frac{1}{9} \sum_{i=1}^9\left(x_i-\bar{x}\right)^2\)
⇒ \(\sum_{i=1}^9\left(x_i-43\right)^2\) = 225 ……(1)
New variance σ² = \(\frac{1}{10} \sum_{i=1}^{10}\left(y_i-45\right)^2\)
= \(\frac{1}{10}\)[(x₁ – 45)² + (x₂ – 45)² + …….. + (x₉ – 45)² + (63 – 45)²]
Since first 9 items are same i.e, x_i = y_i, i = 1, 2, 3, …….., 9
= \(\frac{1}{10}\) [(x_i – 43 – 2)² + (x₂ – 43 – 2)² + …… + (x₉ – 43 – 2)² + 324]
= \(\frac{1}{10}\) [(x_i – 43)² + 4 – 2 . 2 . (x₁ – 43) + (x₂ – 43)² + 4 – 2 . 2 . (x₂ – 43) + …. (x₉ – 43)² + 4 – 2 . 2 . (x₉ – 43) + 324]
= \(\frac{1}{10}\) [(xi – 43)² + (x₂ – 43)² + ……. + (x₉ – 43)² + 36 – 4{x₁ – 43 + x₂ – 43 + …….. + x₉ – 43} + 324]
= \(\frac{1}{10}\) [225 + 36 – 4(x₁ + x₂ + …. + x₉ – 387) + 324]
= \(\frac{1}{10}\) [225 + 36 – 4(387 – 387) + 324]
= \(\frac{1}{10}\) [585]
= 58.5
New standard deviation σ = √58.5 = 7.65

Students get through AP Inter 2nd Year Physics Important Questions 8th Lesson Magnetism and Matter which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 8th Lesson Magnetism and Matter

Very Short Answer Questions

Question 1.
A magnetic dipole placed in a magnetic field experiences a net force. What can you say about the nature of the magnetic field ?
Answer:
The nature of the magnetic field is uniform, magnetic dipole (bar magnet) experiences a net force (or torque).

Question 2.
Do you find two magnetic field lines intersecting ? Why ?
Answer:
Two magnetic field lines never intersect. If they intersect, at the point of intersection the field can have two directions. This is impossible. So, two field lines never intersect.

Question 3.
What happens to the compass needles at the Earth’s poles ? [T.S. Mar. 17; IPE 2014]
Answer:
At the Earth poles, the magnetic field lines are converging or diverging,vertically so that the horizontal component is negligible. Hence, the compass needle can point along any direction.

Question 4.
What do you understand by the ‘magnetisation’ of a sample ? Give its SI unit. [IPE 2016 (AP)]
Answer:
Magnetisation (M) of a sample is equal to its net magnetic moment per unit volume.
Magnetisation, M = \(\frac{m_{n e t}}{V}\), SI unit of magnetisation is Am^-1.

Question 5.
What is the magnetic moment associated with a solenoid ? [IPE 2016 (TS)]
Answer:
The magnitude of magnetic moment of the solenoid is, M = n × 2l × i × πa²
where, ‘n’ is number of turns, ‘2l’ is length of the solenoid, ‘i’ is current passing through coil, and ‘a’ is area of cross section of solenoid.

Question 6.
What are the units of magnetic moment, magnetic induction and magnetic field ? [IPE 2016 (AP), (TS)]
Answer:
The unit of magnetic moment (M) is ampere-meter² (Am²).
The unit of magnetic induction (B) is tesla (T) or N/A-m.
The unit of magnetic field (B) is tesla.

Question 7.
Magnetic lines form continuous closed loops. Why ? [T.S. 2017; IPE 2016(AP)]
Answer:
Magnetic lines are imaginary lines. Within the magnet, they move from south pole to north pole and outside the magnet they move from north pole to south pole. Hence, magnetic lines form continuous closed loops.

Question 8.
Define magnetic declination. [IPE 2016 (TS)]
Answer:
Magnetic declination at a place is the angle between magnetic meridian and geographic meridian at that place.

Question 9.
Define magnetic inclination or angle of dip. [A.P. Mar. 17; A.P. & T.S. 2015 (TS), 14]
Answer:
Magnetic inclination at a place is the angle between direction of total strength of earth’s magnetic field and horizontal line in magnetic meridian.

Question 10.
Classify the following materials with regard to magnetism: Manganese, Cobalt, Nickel, Bismuth, Oxygen, Copper. [T.S. 2015, 2016 (TS); A.P. Mar. 17]
Answer:
Manganese …………. Paramagnetic
Cobalt ………….. Ferromagnetic
Nickel …………….. Ferromagnetic
Bismuth ……………. Diamagnetic
Oxygen ………………. Paramagnetic
Copper …………………. Diamagnetic

Question 11.
The force between two magnet poles separated by a distance ‘d’ in air is ‘F’. At what distance between them does the force become doubled ?
Answer:
Force between two magnetic poles, F₁ = F;
Distance between two magnetic poles, d₁ = d
Force between two magnetic poles increased by double F₂ = 2F
Distance between two magnetic poles, d₂ = ?
From Coulombs law, F₁d₁² = F₂ d₂²
Fd² = 2 F d₂²
⇒ d₂² = \(\frac{\mathrm{d}^2}{2}\)
d₂ = \(\frac{\mathrm{d}}{\sqrt{2}}\)

Question 12.
If B is the magnetic field produced at the centre of a circular coil of one turn of length L carrying current I then what is the magnetic field at the centre of the same coil which is made into 10 turns ?
Answer:
For first circular coil; B₁ = B, n₁ = 1; I₁ = I; a₁ = \(\frac{\mathrm{L}}{2 \pi}\)
For second circular coil, B₂ = ? n₂ = 10; I₂ = I; a₂ = \(\frac{\mathrm{L}}{2 \pi}\)
As B = \(\frac{\mu_0 \mathrm{n} \mathrm{Ia}^2}{2 \mathrm{r}}\), B ∝ n.
\(\frac{\mathrm{B}_2}{\mathrm{~B}_1}=\frac{\mathrm{n}_2}{\mathrm{n}_1}\)
\(\frac{\mathrm{B}_2}{\mathrm{~B}}=\frac{10}{1}\)
∴ B₂ = 10 B

Question 13.
If the number of turns of a solenoid is doubled, keeping the other factors constant, how does the magnetic field at the axis of the solenoid change ?
Answer:
B₁ = B (say); n₁ = n; n₂ = 2n; B₂ = ?
Magnetic field at the centre of a solenoid is given by B = \(\frac{\mu_0 \mathrm{nI} \mathrm{a}^2(2 l)}{2 \mathrm{r}^3}\)
As I, a², 2l and r are constants, B ∝ n
⇒ \(\frac{\mathrm{B}_2}{\mathrm{~B}_1}=\frac{\mathrm{n}_2}{\mathrm{n}_1} \Rightarrow \frac{\mathrm{B}_2}{\mathrm{~B}}=\frac{2 \mathrm{n}}{\mathrm{n}}\)
∴ B₂ = 2B

Question 14.
A closely wound solenoid of 800 turns and area of cross section 2.5 × 10^-4 m² carries a current of 3.0A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment ?
Answer:
Here n = 800, a = 2.5 × 10^-4 m², I = 3.0 A
A magnetic field develop along the axis of the solenoid. Therefore the current carrying solenoid behaves like a bar magnet
m = N IA = 800 × 3.0 × 2.5 × 10^-4
= 0.6 Am² along the axis of solenoid.

Short Answer Questions

Question 1.
Compare the properties of para, dia and ferromagnetic substances.
Answer:
Diamagnetic substances
a) When these materials placed in a magnetic field, they are magnetised feebly in the opposite direction to the applied external field.
b) When a rod of diamagnetic material is suspended freely in a uniform magnetic field, it comes to rest in the perpe-ndicular direction to the magnetic field.
c) When they are kept in a non-uniform magnetic field, they move from the region of greater field strength to the region of less field strength.
d) The relative permeability is less than 1. μ_r < 1 and negative.
e) The susceptibility (χ) is low and negative.
E.g.: Copper, Silver, Water, Gold, Antimony, Bismuth, Mercury, Quartz Diamond etc.

Paramagnetic substances
a) When these materials placed in a magnetic field, they are magnetised feebly in the direction of the applied magnetic field.
b) When a rod of paramagnetic material is suspended freely in a uniform magnetic field, it comes to rest in the direction of the applied magnetic field.
c) When they are kept in a non-uniform magnetic field„they move from the region of less field strength to the region of greater field strength.
d) The relative permeability is greater than 1. μ_r > 1 and positive.
e) The susceptibility (χ) is small and positive.
E.g.: Aluminium, Magnesium, Tungsten, Platinum, Mang-anese, liquid oxygen, Ferric chloride, Cupric chloride etc.

Ferromagnetic substances
a) When these materials placed in a magnetic field,they are magnetised strongly in the direction of the applied external field.
b) When a rod of ferromagnetic material is suspended freely in a uniform magnetic field, it comes to rest in the direction of the applied magnetic field.
c) When they are kept in a non-uniform magnetic field they move from the regions of lesser (magnetic field) strength to the regions of stronger (magnetic field) strength.
d) The relative permeability is much greater than 1. μ_r >> 1 and positive.
e) The susceptibility (χ) is high and positive.
E.g.: Iron, Cobalt, Nickel, Gadolinium and their alloys.

Question 2.
Explain the elements of the Earth’s magnetic field and draw a sketch showing the relationship between the vertical component, horizontal component and angle of dip.
Answer:
The magnetic field of the earth at a point on its surface can be specified by the declination D, the angle of dip or the inclination I and the horizontal component of the earth’s field HE. These are known as the elements of the earth’s magnetic field.

Explanation:

1. The total magnetic field at P can be resolved into a horizontal component H_E and a vertical component Z_E.
2. The angle that B_E makes with H_E is the angle of dip, I.
3. Representing the vertical component by Z_E, we have
   Z_E= B_E Sin I
   H_E = B_E Cos I
   Which gives Tan I = \(\frac{\mathrm{Z}_{\mathrm{E}}}{\mathrm{H}_{\mathrm{E}}}\)

Question 3.
Define magnetic susceptibility of a material. Name two elements one having positive susceptibility and other having negative susceptibility.
Answer:

1. Susceptibility: When a material is placed in a magnetic field, the ratio of the intensity of magnetization acquired by it to the intensity of the applied magnetic field is called its susceptibility.
   Suspectibility χ = \(\frac{\text { Intensity of magnetisation, } \mathrm{I}}{\text { Applied magnetic field, } \mathrm{H}}\)
2. The susceptibility of a material represents its ability to get magnetism.
3. Susceptibility is a dimension less quantity.
4. Relation between μ_r and χ :
   a) Suppose that material is placed in a magnetic field of intensity H. Let I be the intensity of magnetisation acquired by it.
   b) Then the magnetic induction within the material is
   B = μ₀H + μ₀I ⇒ \(\frac{\mathrm{B}}{\mathrm{H}}\) = μ₀[1 + \(\frac{\mathrm{I}}{\mathrm{H}}\)]
   ⇒ μ = μ₀[1 + χ] ⇒ \(\frac{\mu}{\mu_0}\) = 1 + χ
   ∴ μ_r = 1 + χ [∵μ_r = \(\frac{\mu}{\mu_0}\)]
5. Negative susceptibility (χ) of diamagnetic elements are Bismuth (-1.66 × 10^-5) and copper (-9.8 × 10^-6).
6. Positive susceptibility of paramagnetic elements are Aluminium (2.3 × 10^-5) and oxygen at STP (2.1 × 10^-6).
7. Large and positive susceptibility of Ferromagnetic elements are Cobalt and Nickel.

Question 4.
Derive an expression for magnetic field induction on the equatorial line of a barmagnet. [Board Model Paper]
Answer:
At a point on equatorial line: Let us consider a point ‘P’ at a distance ‘d’ on the equatorial line from the centre of a bar magnet.

Qeustion 5.
What do you understand by “hysteresis” ? How does this propertry influence the choice of materials used in different appliances where electromagnets are used ?
Answer:

1. Cycle of magnetisation : When a ferromagnetic specimen is slowly magnetised, the intensity of magnetisation varies with magnetic field through a cycle is called cycle of magnetisation.
2. Hysterisis : The lagging of intensity of magnetisation (I) and magnetic induction (B) behind magnetic field intensity (H) when a magnetic specimen is subjected to a cycle of magnetisation is called hysterisis.
3. Retentivity : The value of I for which H = 0 is called retentivity or residual magnetism.
4. Coercivity: The value of magnetising force required to reduce I is zero in reverse direction of H is called coercive force or coercivity.
5. Hysterisis curve : The curve represents the relation between B or I of a ferromagnetic material with magnetising force or magnetic intensity H is known as Hysterisis curve.
6. Explanation of hysterisis loop or curve :
   a) In fig, a closed curve ABCDEFA in H – I plane, called hysteris loop is shown in fig.
   
   b) When ferromagnetic specimen is slowly magnetised, I increases with H.
   c) Part OA of the curve shows that I increases with H.
   d) At point A, the value I becomes constant is called saturation value.
   e) At B, I has some value while H is zero.
   f) In fig. BO represents retentivity. and OC represents coercivity.
7. Uses : The properties of hysterisis curve, i.e., saturation, retentivity, coercivity and hysterisis loss help us to choose the material for specific purpose.
   1. Permanent magnets : A permanent magnet should have both large retentivity and large coercivity. Permanent magnets are used in galvanometers, voltmeres, ammeters, etc.
   2. An electromagnet core : The electromagnet core material should have maximum induction field B even with small fields H, low hysterisis loss and high initial permeability.
   3. Transformer cores, Dynamocore, Chokes, Telephone diaphragms: The core material should have high initial permeability, low hysterisis loss and high specific resistance to reduce eddy currents. Soft iron is the best suited material.

Question 6.
Prove that a bar magnet and a solenoid produce similar fields.
Answer:
Bar magnet produce similar field of Solenoid :

1. We know that the current loop acts as a magnetic dipole. According to Ampere’s all magnetic phenomena can be explained in terms of circulating currents.
2. Cutting a bar magnet is like a solenoid. We get two similar solenoids with weaker magnetic properties.
3. The magnetic field lines remain continuous, emerging from one face of solenoid and entering into other face of solenoid.
4. If we were to move a small compass needle in the neighbourhood of a bar magnet and a current carrying solenoid, we would find that the deflections of the needle are similar in both cases as shown in diagrams.
   
   
   The axial field of a finite solenoid in order to demonstrate its similarity to that of a bar magnet
5. The magnetic field at point P due to bar magnet in the form of solenoid is B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 m}{r^3}\)
6. The total magnetic field, at a point P due to solenoid is given by
   B = \(\frac{\mu_0 \mathrm{n} \mathrm{I}}{2} \frac{\mathrm{a}^2}{\mathrm{r}^3}(2 l)=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{n}(2 l) \mathrm{I} \pi \mathrm{a}^2}{\mathrm{r}^3}\)
7. The magnitude of the magnetic moment of the solenoid is, m = n(2l) I (πa²).
   ∴ B = \(\frac{\mu_0}{4 \pi} \frac{2 \mathrm{~m}}{\mathrm{r}^3}\)
8. Therefore, magnetic moment of a bar magnet is equal to magnetic moment of an equivalent solenoid that produces the same magnetic field.

Question 7.
A small magnetic needle is set into oscillations in a magnetic field B obtain an expression for the time period of oscillation.
Answer:
Expression for time period of oscillation :

1. A small compass needle (magnetic dipole) of known magnetic moment m and moment of Inertia i is placing in uniform magnetic field B and allowing it to oscillate in the magnetic field.
2. This arrangement is shown in Figure.
3. The torque on the needle is τ = m × B
4. In magnitude τ = mB sin θ.
   Here τ is restoring torque and θ is the angle between m and B.
5. Therefore, in equilibrium i \(\frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}\) = – mB sinθ. Negative sign with mB sin0 implies that restoring torque is in opposition to deflecting torque.
   
6. For small values of o in radians, we approximate sinθ ≃ θ and get \(i \frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}\) ≃ – mBθ
   \(\frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2} \approx \frac{-\mathrm{mB}}{\mathcal{j}} \theta\) …………….. (1)
   This represents a simple harmonic motion. .
7. From defination of simple harmonic motion, we have \(\frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}\) = – ω²θ …………… (2)
   From equation (I) and (II), we get ⇒ ω² = \(\frac{\mathrm{mB}}{\mathcal{J}}\)
   ∴ ω = \(\sqrt{\frac{\mathrm{mB}}{\mathcal{J}}}\)
8. Therefore, the time period is T = \(=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathcal{J}}{\mathrm{mB}}}\)

Qeustion 8.
A bar magnet, held horizontally, is set into angular oscillations in the Earth’s magnetic field. It has time periods T₁ and T₂ at two places, where the angles of dip are θ₁ and θ₂ respectively. Deduce an expression for the ratio of the resultant magnetic fields at the two places.
Answer:

1. Suppose, the resultant magnetic fields is to be compared at two places A and B.
2. A barmagnet, held horizontally at A and which is set into angular oscillations in the Earth’s magnetic field.
3. Let time period of a bar magnet at place A’ is T₁ and angular displacement or angle of dip is θ₁.
4. As the bar magnet is free to rotate horizontally, it does nqt remain vertical component (B₁ sin θ₁). It can have only horizontal component (B₁ cosθ₁).
5. The time period of a bar magnet in uniform magnetic field is given by T = 2π \(\sqrt{\frac{\mathrm{I}}{\mathrm{mB}_{\mathrm{H}}}}\)
   
6. Now, in this case T = T₁ and B_H = B₁Cosθ₁
7. Therefore time period of a bar magnet at place ‘A’ is given by
   T₁ = 2π \(\sqrt{\frac{\mathrm{I}}{\mathrm{mB}_1 \cos \theta_1}}\) …………… (1) Where I is moment of Inertia of a barmagnet and m is magnitude of magnetic moment.
8. Similarly, the same bar magnet is placed at B and which is set into angular oscillations in the earth’s magnetic field.
9. Let time period of a bar magnet at place B is T₂ and angle of dip is θ₂.
10. Since horizontal component of earths field at B is B_H = B₂ cos θ₂, time period,
    T₂ = 2π \(\sqrt{\frac{\mathrm{I}}{\mathrm{mB}_2 \cos \theta_2}}\) ………………… (2)
11. Dividing equation (1) by equation (2), we get \(\frac{T_1}{T_2}=\sqrt{\frac{\mathrm{mB}_2 \cos \theta_2}{\mathrm{mB}_1 \cos \theta_1}}\)
    Squaring on both sides, we have \(\frac{\mathrm{T}_1^2}{\mathrm{~T}_2^2}=\frac{\mathrm{B}_2 \cos \theta_2}{\mathrm{~B}_1 \cos \theta_1}\)
12. But B₁ = μ₀H₁, and B₂ = μ₀H₂
    \(\frac{\mathrm{T}_1^2}{\mathrm{~T}_2^2}=\frac{\mu_0 \mathrm{H}_2 \cos \theta_2}{\mu_0 \mathrm{H}_1 \cos \theta_1} \)
13. Therefore, \(\frac{\mathrm{H}_1}{\mathrm{H}_2}=\frac{\mathrm{T}_2^2 \cos \theta_2}{\mathrm{~T}_1^2 \cos \theta_1}\)
14. By knowing T₁, T₂ and θ₁, θ₂ at different places A and B, we can find the ratio of resultant magnetic fields.

Question 9.
Obtain Gauss’ Law for magnetism and explain it.
Answer:
Gauss law for Magnetism :

1. According to Gauss’s law for magnetism, the net magnetic flux (Φ_B) through any closed surface is always zero.
2. The law implies that the no. of magnetic field lines leaving any closed surface is always equal to the number of magnetic field lines entering it.
3. Suppose a closed surface S is held in a uniform magnetic field B. Consider a small vector area element ∆S of this surface as shown in figure.
4. Magnetic flux through this area element is defined as ∆Φ_B = B. ∆S. Then the net flux Φ_B, is,
   Φ_B = \(\sum_{\text {all }} \Delta \phi_B=\sum_{\text {all }} \text { B. } \Delta \mathrm{S}=0\)
5. If the area elements are really small, we can rewrite this equation as
   Φ_B = \(\oint\)B.ds = 0 …………………. (I)
   
6. Comparing this equation with Gauss’s law of electrostatics i.e., electric flux through a closed surface S is given by
   Φ_E = \(\oint \text { E. } \Delta S=\frac{q}{\varepsilon_0}\) …………….. (II) Where q is the electric charge enclosed by the surface.
7. In an electric dipole were enclosed by the surface equal and opposite charges in the dipole add upto zero. Therefore, Φ_E would be zero.
8. The fact that Φ_B = 0 indicates that the simplest magnetic element is a dipole or current loop.
9. The isolated magnetic poles, called magnetic monopoles are not known to exist.
10. All magnetic phenomena can be explained interms of an arrangement of magnetic dipoles and /or current loops.
11. Thus corresponding to equation (II) of Gauss’s theorem in electrostatics, we can visualize equation (I) as
    Φ_E = \(\int_S\) B . dS = μ₀ (m) + μ₀ (-m) = 0 where m is strength of N-pole and -m is strength of
    S – pole of same magnet.
12. The net magnetic flux through any closed surface is zero.

Question 10.
What are ferromagnetic materials ? Give examples. What happens to a ferromagnetic material at curie temperature ? [IPE 2015 (TS)]
Answer:
Ferro magnetic substances : (a) These are strongly attracted by magnet, (b) Susceptibility is large, positive and temperature dependent, (c) Relative permeability, μ_r > > 1 (d) Atoms have permanent dipole moments which are organised in domains. Ex: Iron, Cobalt, Nickel

Curie temperature : The temperature above which a ferro magnetic substance changes in to para magnetic substance changes in to para magnetic substance is called curie temperature.

Problems

Question 1.
A coil of 20 turns has an area of 800 mm² and carries a current of 0.5A. If it is placed in a magnetic field of intensity 0.3T with its plane parallel to the field, what is the torque that it experiences ? ,
Answer:
n = 20; A = 800 mm² = 800 × 10^-6 m²; i = 0.5A; B = 0.3T; θ = 0°.
When the plane parallel to the field,
T = Bin A cos θ = 0.3 × 0.5 × 20 × 800× 10^-6 × cos 0°
∴ τ = 2.4 10^-3 Nm

Question 2.
In the Bohr atom model the electrons move around the nucleus in circular orbits. Obtain an expression the magnetic moment (p) of the electron in a Hydrogen atom in terms of its angular momentum L.
Answer:
Consider an electron of charge e, moves with constant speed v in a circular orbit of radius ‘r’ in Hydrogen atom as shown in fig.

The current constitute by revolving electron in circular motion around a nucleus, I = \(\).
Time period of orbiting electron, T = \(\frac{2 \pi \mathrm{r}}{\mathrm{v}}\) ⇒ I = \(\frac{e}{\frac{2 \pi r}{v}}=\frac{e v}{2 \pi r}\)
orbital magnetic moment, μ = IA = I (πr²)
⇒ μ = \(\frac{\mathrm{ev}}{2 \pi \mathrm{r}}\) (πr²) = \(\frac{\mathrm{evr}}{2}\)
μ = \(\frac{\mathrm{e}}{2 \mathrm{~m}}\) (mvr) [∵ Multiplying and dividing with ‘m’ on right side]
∴μ = \(\frac{\mathrm{e}}{2 \mathrm{~m}}\) L where L = mvr = angular momentum.

Qeustion 3.
A bar magnet of length 0.1m and with a magnetic moment of 5Am² is placed in a uniform a magnetic field of intensity 0.4T, with its axis making an angle of 60° with the field. What is the torque on the magnet ?
Answer:
Given, 2l = 0.1m; m = 5A – m²; B = 0.4T; θ = 60°.
Torque, T = mB sin θ = 5 × 0.4 × sin 60° = 2 × \(\frac{\sqrt{3}}{2}\)
∴ T = 1.732 N – m

Question 4.
A solenoid of length 22.5 cm has a total of 900 turns and carries a current of 0.8 A. What is the magnetising field H near the centre and far away from the ends of the solenoid ?
Answer:
l = 22.5 cm = 22.5 × 10^-2 m = \(\frac{45}{2}\) × 10^-2m
N = 900; I = 0.8A; H = ?
H = \(\frac{\mathrm{NI}}{l}=\frac{900 \times 0.8}{\left(\frac{45}{2}\right) \times 10^{-2}}\)
H = \(\frac{900}{45}\) × 0.8 × 10² × 2
∴ H = 3200 Am^-1

Qeustion 5.
The horizontal component of the earth’s magnetic field at a certain place is 2.6 × 10^-5T and the angle of dip is 60°. What is the magnetic field of the earth at this location ?
Answer:
Given H_E = 2.6 × 10^-5T;
D (or) δ = 60°
B_E = \(\frac{\mathrm{H}_{\mathrm{E}}}{\cos \mathrm{D}}=\frac{2.6 \times 10^{-5}}{\cos 60^{\circ}}=\frac{2.6 \times 10^{-5}}{(1 / 2)}\) = 5.2 × 10^-5 T
∴ B_E = 5.2 × 10^-5 T

Question 6.
In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26 G and the dip angle is 60°. What is the magnetic field of the earth at this location ?
Solution:
It is given that H_E = 0.26 G. From Fig., we have

The earth’s magnetic field, B_E, its horizontal and vertical components. H_E and Z_E. Also shown are the declination, D and the inclination or angle of dip, I.
cos 60° = \(\frac{\mathrm{H}_{\mathrm{E}}}{\mathrm{B}_{\mathrm{E}}}\)
B_E = \(\frac{\mathrm{H}_{\mathrm{E}}}{\cos 60^{\circ}}\)
= \(\frac{0.26}{(1 / 2)}\) = 0.52 G

Qeustion 7.
What is the magnitude of the equatorial and axial fields due to a bar magnet of length 8.0 cm at a distance of 50 cm from its mid-point ? The magnetic moment of the bar magnet is 0.40 A m².
Solution:
From Eq.

Question 8.
The earth’s magnetic field at the equator is approximately 0.4 G. Estimate the earth’s dipole moment. .
Solution:
The equatorial magnetic field is,
We are given that B_E ~ 0.4 G = 4 × 10^-5 T. For r, we take the radius of the earth 6.4 × 10⁶ m.
Hence,
m = \(\frac{4 \times 10^{-5} \times\left(6.4 \times 10^6\right)^3}{\mu_0 / 4 \pi}\) = 4 × 10² × (6.4 × 10⁶)³ (μ₀/4π = 10^-7)
= 1.05 × 10²³ A m²
This is close to the value 8 × 10²² A m² quoted in geomagnetic texts.

Textual Examples

Question 1.
In Fig, the magnetic needle has magnetic moment 6.7 × 10^-2 Am² and moment of inertia i = 7.5 × 10^-6 kg m². It performs 10 complete oscillations in 6.70 s. What is the magnitude of the magnetic field ?
Solution:
The time period of oscillation is, :
T = \(\frac{6.70}{10}\) = 0.67 s

The axial field of a finite solenoid in order to demonstrate its similarity to that of a bar magnet.
From Eq. B =
= \(\frac{4 \times(3.14)^2 \times 7.5 \times 10^{-6}}{6.7 \times 10^{-2} \times(.067)^2}\)
= 0.01 T

Question 2.
A short bar magnet placed with its axis at 30° with an external field of 800 G experiences a torque of 0.016 Nm.
(a) What is the magnetic moment of the magnet ?
(b) What is the work done in moving it from its most stable to most unstable position ?
(c) The bar magnet is replaced by a solenoid of cross-sectional area 2 × 10^-4 m² and 1000 turns, but of the same magnetic moment. Determine the current flowing through the solenoid.
Solution:
a) From Eq., τ = m B sin θ, θ = 30°, hence sin θ = 1/2.
Thus, 0.016 = m × (800 × 10^-4 T) × (1/2)
m = 160 × 2/800 =0.40 Am²

b) From Eq . U_m = -m. B, the most stable position is θ = 0° and the most unstable position is q = 180°. Work done is given by
W = U_m (θ = 180°) – U_m (θ = 0°)
= 2 m B = 2 × 0.40 × 800 × 10^-4 = 0.064 J

c) From Eq., m_s = NIA. From part (a), m_s = 0.40 Am²
0.40 = 1000 × I × 2 × 10^-4
I = 0.40 × 10⁴/(1000 × 2) = 2A

Question 3.
a) What happens if a bar magnet is cut into two pieces :
(i) transverse to its length,
(ii) along its length ?
b) A magnetised needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to a torque. Why ?
c) Must every magnetic configuration have a north pole and a south pole ? What about the field due to a toroid ?
d) Two identical looking iron bars A and B are given, one of which is definitely known to be magnetised. (We do not know which one.) How would one ascertain whether or not both are magnetised ? If only one is magnetised how does one ascertain which one ? (Use nothing else but the bars A and B].
Solution:
a) In either case, one gets two magnets, each with a north and south pole.

b) No force if the field is uniform. The iron nail experiences a non-uniform field due to the bar magnet. There is induced magnetic moment in the nail, therefore, it experiences both force and torque. Then net force is attractive because the induced south pole (say) in the nail is closer to the north pole of magnet than induced north pole.

c) Not necessarily. True only if the source of the field has a net nonzero magnetic moment. This is not so for a toroid or even for a straight infinite conductor.

d) Try to bring different ends of the bars closer. A repulsive force in some situation establishes that both are magnetised. If it is always attractive, then one of them is not magnetised. In a bar magnet the intensity of the magnetic field is the strongest at the two ends (poles) and weakest at the central region. This fact may be used to determine whether A or B is the magnet. In this case, to see which one of the two bars is magnet, pick up one, (say, A) and lower one of its end first one of the ends of the other (say, B) and then on the middle of B. If you notice that in the middle of B, A experiences no force, then B is magnetised. If you do not notice any change from the end to the middle of B, then A is magnetised.

Question 4.
What is the magnitude of the equatorial and axial fields due to a bar magnet of length 8.0 cm at a distance of 50 cm from its mid-point ? The magnetic moment of the bar magnet is 0.40 A m², the same as in Example – 2.
Solution:
From Eq.

Question 5.
Figure shows a small magnetised needle P placed at a point O. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle Q.

a) In which configuration the system is not in equilibrium ?
b) In which configuration is the system in (i) stable, and (ii) unstable equilibrium ?
c) Which configuration corresponds to the lowest potential energy among all the configurations shown ?
Solution:
Potential energy of the configuration arises due to the potential energy of one dipole (say, Q) in the magnetic field due to other (P). Use the result that the field due to P is given by the expression. *
B_P = –\(\frac{\mu_0}{4 \pi} \frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{r}^3}\) (on the normal bisector)
B_P = \(\frac{\mu_0 2}{4 \pi} \frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{r}^3}\) (on axis)
where mp is the magnetic moment of the dipole P.
Equilibrium is stable when m_Q is parallel to B_P, and unstable when it is anti-parallel to B_P. For instance for the configuration Q₃ for which Q is along the perpendicular bisector of the dipole P, the magnetic moment of Q is parallel to the magnetic field at the position 3. Hence Q₃ is stable. Thus,
a) PQ₁ and PQ₂
b) (i) PQ₃, PQ₆ (stable); (ii) PQ₅, PQ₄ (unstable)
c) PQ₆.

Question 6.
Many of the diagrams given in Fig. show magnetic field lines (thick lines in the figure) wrongly. Point out what is wrong with them. Some of them may describe electrostatic field lines correctly. Point out which ones.

Solution:

a) Wrong: Magnetic field lines can never emanate from a point, as shown in figure. Over any closed surface, the net flux of B must always be zero, i.e., pictorially as many field lines should seem to enter the surface as the number of lines leaving it. The field lines shown, in fact, represent electric field of a long positively charged wire. The correct magnetic field lines are circling the straight conductor.

b) Wrong: Magnetic field lines (like electric lines) can never cross each other, because otherwise the direction of field at the point of intersection is ambiguous. There is further error in the figure. Magnetostatic field lines can never form closed loops around empty space. A closed loop of static magnetic field line must enclose a region across which a current is passing. By contrast, electrostatic field lines can never form closed loops, neither in empty space, nor when the loop encloses charges.

c) Right: Magnetic lines are completely confined within a toroid: Nothing wrong here in field lines forming closed loops, since each loop encloses a region across which a current passes. Note, for clarity of figure, only a few field lines within the toroid have been shown. Actually, the entire region enclosed by the windings contains magnetic field.,

d) Wrong: Field lines due to a solenoid at its ends and outside cannot be so completely straight and confined; such a thing violates Ampere’s law. The lines should curve out at both ends, and meet eventually to form closed loops.

e) Right: These are field lines outside and inside a bar magnet. Note carefully the direction of field lines inside. Not all field lines emanate out of a north pole (or converge into a south pole). Around both the N-pole, and the S-pole, the next flux of the field is zero.

f) Wrong: These field lines cannot possibly represent a magnetic field. Look at the upper region. All the field lines seem to emanate out of the shaded plate. The net flux through a surface surrounding the shaded plate is not zero. This is impossible for a magnetic field. The given field lines, in fact, show the electrostatic field lines around a positively charged upper plate and a negatively charged lower plate. The difference between Fig. [(e) and (f)] should be carefully grasped.

g) Wrong: Magnetic field lines between two pole pieces cannot be precisely straight at the ends. Some fringing of lines is inevitable. Otherwise, Ampere’s law is violated. This is also true for electric field lines.

Question 7.
a) Magnetic field lines show the direction (at every point) along which a small magnetised needle aligns (at the point). Do the magnetic field line’s also represent the lines of force on a moving charged particle at every point ?
b) Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why ?
c) If magnetic monopoles existed, how would the Gauss’s law of magnetism be modified ?
d) Does a bar magnet exert a torque on itself due to its own field ? Does one element of a current – carrying wire exert a force on another element of the same wire ?
e) Magnetic field arises due to charges in motion. Can a system have magnetic moments even though its net charge is zero ?
Solution:
a) No. The magnetic force is always normal to B (remember magnetic force = qv × B). It is misleading to call magnetic field lines as lines of force.

b) If field lines were entirely confined between two ends of a straight solenoid, the flux through the cross-section at each end would be non-zero. But the flux of field B through any closed surface must always be zero. For a toroid, this difficulty is absent because it has no ‘ends’.

c) Gauss’s law of magnetism states that the flux of,B thrugh any closed surface is always
zero \(\int_s B \cdot d s\) = o.
If monopoles existed, the right hand side would be equal to the monopole (magnetic charge) qm enclosed by S. [Analogous to Gauss’s law of electrostatics, \(\int_s B \cdot d s\) = μ₀q_m
where q_m is the (monopole) magnetic charge enclosed by S.]

d) No. There is no force or torque on an element due to the field produced by that element itself. But there is a force (or torque) on an element of the same wire. (For the special case of a straight wire, this force is zero).

e) Yes. The average of the cahrge in the system may be zero. Yet, the mean of the magnetic moments due to various current loops may not be zero. We will come across such examples in connection with paramagnetic material where atoms have net dipole moment through their net charge is zero.

Question 8.
The earth’s magnetic field at the equator is approximately 0.4 G. Estimate the earth’s dipole moment.
Solution:
The equatorial magnetic field is,
B_E = \(\frac{\mu_0 \mathrm{~m}}{4 \pi \mathrm{r}^3}\)
We are given that B_E ~ 0.4 G = 4 × 10^-5 T. For r, we take the radius of the earth 6.4 × 10⁶ m. Hence,
m = \(\frac{4 \times 10^{-5} \times\left(6.4 \times 10^6\right)^3}{\mu_0 / 4 \pi}\) = 4 × 10² × (6.4 × 10⁶)³ (μ₀/4π = 10^-7)
= 1.05 × 10²³ A m²
This is close to the value 8 × 10²² A m² quoted in geomagnetic texts.

Question 9.
In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26 G and the dip angle is 60°. What is the magnetic field of the earth at this location ?
Solution:
It is given that H_E = 0.26 G. From Fig., we have

The earth’s magnetic field, B_E, its horizontal and vertical components. H_E and Z_E. Also shown are the declination, D and the inclination or angle of dip, I.
cos 60° = \(\frac{\mathrm{H}_{\mathrm{E}}}{\mathrm{B}_{\mathrm{E}}}\)
B_E = \(\frac{\mathrm{H}_{\mathrm{E}}}{\cos 60^{\circ}}\)
= \(\frac{0.26}{(1 / 2)}\) = 0.52 G

Question 10.
A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) H, (b) M, (c) B and (d) the magnetising current I_m.
Solution:
a) The field H is dependent of the material of the core, and is
H = nI = 1000 × 2.0 = 2 × 10³ A/m

b) The magnetic field B is given by
B = μ_rμ₀H
= 400 × 4π × 10^-7 (N/A³) × 2 × 10³ (A/m) = 1.0 T

c) Magnetisation is given by
M = (B – μ₀ H)/μ₀
= (μ_rμ₀H – μ₀H)/μ₀ = (μ_r – 1) H = 399 × H ≃ 8 × 10⁵ A/m

d) The magnetising current IM is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the core. Thus B = μ_rn₀ (I + I_M). Using I = 2A, B = 1 T, we get I_M = 794A.

Question 11.
A domain in ferromagnetic iron is in the form of a cube of side length 1 μm. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetisation of the domain. The molecular mass of iron is 55 g/mole and its density is 7.9 g/cm³. Assume that each iron atom has a dipole moment of 9.27 × 10^-24 A m³.
Solution:
The volume of the cubic domain is
V = (10^-6 m)³ = 10^-18 m³ = 10^-12 cm³
Its mass is volume × density = 7.9 g cm^-3 × 10^-12 cm³ = 7.9 × 10^-12 g
It is given that Afagadro number (6.023 × 10²³) of iron atoms have a mass of 55g. Hence,the number of atoms in the domain is
N = \(\frac{7.9 \times 10^{-12} \times 6.023 \times 10^{23}}{55}\)
= 8.65 × 10¹⁰ atoms
The maximum possible dipole moment m_max is achieved for the (unrealistic) case when all the atomic moments are perfectly aligned. Thus,
m_max = (8.65 × 10¹⁰) × (9-27 × 10^-24)
= 8.0 × 10^-13 Am²
The consequent magnetisation is
M_max = m_max/DomainVolume :
= 8.0 × 10^-13 Am²/10^-18 m³
= 8.0 × 10⁵ Am^-1.

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(d) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(d)

Question 1.
Find the coefficient of x³ in the power series expansion of \(\frac{5 x+6}{(x+2)(1-x)}\) specifying the region in which the expansion is valid.
Solution:

Question 2.
Find is the coefficient of x⁴ in the power series expansion of \(\frac{3 x^2+2 x}{\left(x^2+2\right)(x-3)}\) specifying the interval in which the expansion is valid.
Solution:
Let \(\frac{3 x^2+2 x}{\left(x^2+2\right)(x-3)}=\frac{A}{x-3}+\frac{B x+C}{x^2+2}\)
Multiplying with (x² + 2) (x – 3)
3x² + 2x = A(x² + 2) + (Bx + C) (x – 3)
x = 3
⇒ 27 + 6 = A(9 + 2)
⇒ 33 = 11A
⇒ A = 3
Equating the coefficients of x²
3 = A + B
⇒ B = 3 – A = 3 – 3 = 0
Equating the constants,
2A – 3C = 0
⇒ 3C = 2A = 6
⇒ C = 2

Question 3.
Find the coefficient of xⁿ in the power series expansion of \(\frac{x-4}{x^2-5 x+6}\) specifying the region in which the expansion is valid.
Solution:
Let \(\frac{x-4}{x^2-5 x+6}=\frac{A}{x-2}+\frac{B}{x-3}\)
Multiplying with (x – 2) (x – 3)
x – 4 = A(x – 3) + B(x – 2)
x = 2
⇒ -2 = A(2 – 3) = -A
⇒ A = 2
x = 3
⇒ -1 = B(3 – 2) = B
⇒ B = -1

Question 4.
Find the coefficient of xⁿ in the power series expansion of \(\frac{3 x}{(x-1)(x-2)^2}\)
Solution:

∴ 3x = A(x – 2)² + B(x – 1) (x – 2) + C(x – 1) ……..(1)
putting x = 1,
3 = A(1 – 2)²
⇒ A = 3
putting x = 2,
6 = C(2 – 1)
⇒ C = 6
Now equating the co-efficient of x² terms in (1)
0 = A + B
⇒ B = -A
⇒ B = -3

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(c)

Resolve the following into partial fractions.

Question 1.
\(\frac{x^2}{(x-1)(x-2)}\)
Solution:
Let \(\frac{x^2}{(x-1)(x-2)}=1+\frac{A}{x-1}+\frac{B}{x-2}\)
Multiplying with (x – 1) (x – 2)
x² = (x – 1) (x – 2) + A(x – 2) + B(x – 1)
Put x = 1, 1 = A(-1) ⇒ A = -1
Put x = 2, 4 = B(1) ⇒ B = 4
∴ \(\frac{x^2}{(x-1)(x-2)}=1-\frac{1}{x-1}+\frac{4}{x-2}\)

Question 2.
\(\frac{x^3}{(x-1)(x+2)}\)
Solution:

Question 3.
\(\frac{x^3}{(2 x-1)(x-1)^2}\)
Solution:
Let \(\frac{x^3}{(2 x-1)(x-1)^2}\) = \(\frac{1}{2}+\frac{A}{2 x-1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}\)
Multiplying with 2(2x – 1) (x – 1)²
2x³ = (2x – 1) (x – 1)² + 2A(x – 1)² + 2B(2x – 1) (x – 1) + 2C(2x – 1)
Put x = \(\frac{1}{2}\),
⇒ 2(\(\frac{1}{8}\)) = 2A(\(\frac{1}{4}\))
⇒ A = \(\frac{1}{2}\)
Put x = 1,
⇒ 2(1) = 2C(1)
⇒ C = 1
Put x = 0,
0 = (-1) (1) + 2A(1) + 2B(-1) (-1) + 2C(-1)
⇒ 2A + 2B – 2C = 1
⇒ 2B = 1 + 2C – 2A
⇒ 2B = 1 + 2 – 1 = 2
⇒ B = 1
∴ \(\frac{x^3}{(2 x-1)(x-1)^2}\) = \(\frac{1}{2}+\frac{1}{2(2 x-1)}+\frac{1}{(x-1)}+\frac{1}{(x-1)^2}\)

Question 4.
\(\frac{x^3}{(x-a)(x-b)(x-c)}\)
Solution:
Let \(\frac{x^3}{(x-a)(x-b)(x-c)}\) = \(1+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}\)
Multiplying with (x – a)(x – b) (x – c),
x³ = (x – a)(x – b) (x – c) + A(x – b) (x – c) + B(x – a) (x – c) + C(x – a) (x – b)
Put x = a,
a³ = A(a – b) (a – c)
⇒ A = \(\frac{a^3}{(a-b)(a-c)}\)
Put x = b,
b³ = B(b – a) (b – c)
⇒ B = \(\frac{b^3}{(b-a)(b-c)}\)
Put x = c, c³ = C(c – a) (c – b)
⇒ C = \(\frac{c^3}{(c-a)(c-b)}\)
∴ \(\frac{x^3}{(x-a)(x-b)(x-c)}\) = \(1+\frac{a^3}{(a-b)(a-c)(x-a)}+\frac{b^3}{(b-a)(b-c)(x-b)}\) + \(\frac{c^3}{(c-a)(c-b)(x-c)}\)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(b)

Resolve the following into partial fractions.

Question 1.
\(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}\)
Solution:
Let \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{A}{x-1}+\frac{B x+C}{x^2+2}\)
Multiplying with (x – 1) (x² + 2)
2x² + 3x + 4 = A(x² + 2) + (Bx + C) (x – 1)
x = 1
⇒ 2 + 3 + 4 = A(1 + 2)
⇒ 9 = 3A
⇒ A = 3
Equating the coefficients of x²
2 = A + B
⇒ B = 2 – A = 2 – 3 = -1
Equating constants
4 = 2A – C
⇒ C = 2A – 4 = 6 – 4 = 2
∴ \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{3}{x-1}+\frac{-x+2}{x^2+2}\)

Question 2.
\(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}\)
Solution:
Let \(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}=\frac{A}{2+x}+\frac{B x+C}{1-x+x^2}\)
Multiplying with (2 + x) (1 – x + x²)
3x – 1 = A(1 – x + x²) (Bx + C) (2 + x)
x = -2
⇒ -7 = A(1 + 2 + 4) = 7A
⇒ A = -1
Equating the coefficients of x²
0 = A + B ⇒ B = -A = 1
Equating the constants
-1 = A + 2C
⇒ 2C = -1 – A = -1 + 1 = 0
⇒ C = 0
∴ \(\frac{3 x-1}{\left(1-x+x^2\right)(2+x)}=-\frac{1}{2+x}+\frac{x}{1-x+x^2}\)

Question 3.
\(\frac{x^2-3}{(x+2)\left(x^2+1\right)}\)
Solution:
Let \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{A}{x+2}+\frac{B x+C}{x^2+1}\)
Multiplying with (x + 2) (x² + 1)
x² – 3 = A(x² + 1) + (Bx + C) (x + 2)
x = -2
⇒ 4 – 3 = A(4 + 1)
⇒ 1 = 5A
⇒ A = \(\frac{1}{5}\)
Equating the coefficients of x²
1 = A + B
⇒ B = 1 – A = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
Equating the constants
-3 = A + 2C
⇒ 2C = -3 – A
⇒ 2C = -3 – \(\frac{1}{5}\)
⇒ 2C = \(-\frac{16}{5}\)
⇒ C = \(-\frac{8}{5}\)
∴ \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{1}{5(x+2)}+\frac{4 x-8}{5\left(x^2+1\right)}\)

Question 4.
\(\frac{x^2+1}{\left(x^2+x+1\right)^2}\)
Solution:
Let \(\frac{x^2+1}{\left(x^2+x+1\right)^2}=\frac{A x+B}{x^2+x+1}+\frac{C x+D}{\left(x^2+x+1\right)^2}\)
Multiplying with (x² + x + 1)²
x² + 1 = (Ax + B) (x² + x + 1) + (Cx + D)
Equating the coefficients of x³,
A = 0
Equating the coefficients of x²,
A + B = 1 ⇒ B = 1
Equating the coefficients of x,
A + B + C = 0
⇒ 1 + C = 0
⇒ C = -1
Equating the constants,
B + D = 1
⇒ D = 1 – B = 1 – 1 = 0
∴ Ax + B = 1, Cx + D = -x
∴ \(\frac{x^2+1}{\left(x^2+x+1\right)^2}=\frac{1}{x^2+x+1}-\frac{x}{\left(x^2+x+1\right)^2}\)

Question 5.
\(\frac{x^3+x^2+1}{(x-1)\left(x^3-1\right)}\)
Solution:

∴ x³ + x² + 1 = A(x – 1) (x² + x + 1) + B(x² + x + 1) + (Cx + D) (x – 1)² …….(2)
Put x = 1 in (2)
1 + 1 + 1 = A(0) + B(1 + 1 + 1) + (C(1) + D) (0)
⇒ 3B = 3
⇒ B = 1
Equating the coefficients of x³ in (2)
1 = A + C ………(3)
Equating the coefficients of x² in (2)
1 = A(1 – 1) + B(1) + C(-2) + D(1)
⇒ 1 = B – 2C + D
∵ B = 1,
⇒ 1 = 1 – 2C + D
⇒ 2C = D ………(4)
Put x = 0 in (2)
1 = A(-1)(1) + B(1) + D(-1)²
⇒ -A + B + D = 1
⇒ -A + 1 + D = 1
⇒ A = D ………(5)
From (3), (4) and (5)

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 3rd Lesson Wave Optics Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 3rd Lesson Wave Optics

Very Short Answer Questions

Question 1.
What is Fresnel distance ?
Answer:
“Fresnel distance is the minimum distance a beam of light has to travel before its deviation from straight line path becomes significant”.

Fresnel distance (Z_F) = \(\frac{\mathrm{a}^2}{\lambda}\) ; Where a = width of the aperture, λ = wave length.

Question 2.
Give the justification for validity of ray optics.
Answer:
The distances much smaller than Z_F, the spreading due to diffraction is smaller compared to the size of the beam.

When the distance is approximately Z_F, and much more than Z_F, the spreading due to diffraction dominates over that due to ray optics (size of the aperature (a)]
Z = \(\frac{\mathrm{a}^2}{\lambda}\)
From this equation ray optics is valid in the limit of wave length tending to zero.

Question 3.
What is polarisation of light ?
Answer:
The vibrations of the light confined only one direction. This phenomenon is called polarisation.
(or)
The phenomenon due to which the transverse vibrations of electric field vector of a light wave become confined to one plane, is called polarisation.

Question 4.
What is Malus’ law.
Answer:
Malus’ law : It states that the intensity of polarised light transmitted through the analyser varies as a square of cosine of the angle between the plane of transmission of analyser and polariser.
I ∝ cos² θ; I = I₀ cos² θ.

Question 5.
Explain Brewster’s law.
Answer:
Brewster’s law : The tangent of the polarising angle is equal to the refractive index of the medium.
μ = tan i_B, where i_B = polarising angle and μ = refractive index. Note : r + i_B = 90°

Question 6.
When does a monochromatic beam of light incident on a reflective surface get completely transmitted ?
Answer:
Let the light emitted by laser source passes through polariser, and incident on the surface of the reflective surface with Brewster’s angle (i_B). Now rotate the polariser at particular alignment the light incident on the surface is completely transmitted.

Short Answer Questions

Question 1.
Explain Doppler effect in light. Distinguish between red shift and blue shift. (T.S. Mar. ’18, ’16)
Answer:
Dopper effect in light: The change in the apparent frequency of light, due to relative motion between source of light and observer. This phenomenon is called Doppler effect.

The apparent frequency of light increases when the distance between observer and Source of light is decreasing and the apparent frequency of light decreases, if the distance between source of light and observer increasing.

Doppler shift can be expressed as \(\frac{\Delta v}{v}\) = \(\frac{-v_{\text {radial }}}{\mathrm{C}}\)

Applications of Doppler effect in light:

1. It is used in measuring the speed of a star and speed of galaxies.
2. Measuring the speed of rotation of the sun.

Red shift: The apparent increase in wave length in the middle of the visible region of the spectrum moves towards the red end of the spectrum is called red shift.

Blue shift: When waves are received from a source moving towards the observer, there is an apparent decrease in wave length, this is called blue shift.

Question 2.
What is total internal reflection. Explain the phenomenon using Huygen’s principle.
Answer:
Total internal reflection: When a light ray travels from denser medium to rarer medium, the angle of incidence is greater than critical angle then it reflects into the same medium. This phenomenon is called total internal reflection.

Huygen’s principle : According to Huygen’s principle, every point on the wavefront AB is a source of secondary wavelets and time during which wavelets from B reaches at C, Let \(\tau\) be the time taken by the wave front to advance from B to C.

Then distance BC = υ\(\tau\)
In order to construct the reflected wavefront, we draw a sphere of radius ux from point A.
Let CE represent the tangent drawn from C to this sphere.
AE = BC = υ\(\tau\)

Consider the triangles EAC and BAC are congruent.
∴ Angles i and r would be equal. This is the law of reflection.

Question 3.
Derive the expression for the intensity at a point where interference of light occurs. Arrive at the conditions for maximum and zero intensity. (A.P. & T.S. Mar. ’16)
Answer:
Let y₁ and y₂ be the displacements of the two waves having same amplitude a and Φ is the phase difference between them.
y₁ = a sin ωt …… (1)
y₂ = a sin (ωt + ϕ)) …… (2)
The resultant displacement y = y₁ + y₂
y = a sin ωt + a sin (ωt + ϕ)
y = a sin ωt + a sin ωt cos ϕ + a cos ωt sin ϕ
y = a sin ωt [1 + cos ϕ] + cos ωt (a sin ϕ)

Let R cos θ = a(1 + cos ϕ) —– (4)
R sin θ = a sin ϕ —– (5)
y = R sin ωt . cos θ + R cos ωt . sin θ
y = R sin(ωt + θ) —– (6)
where R is the resultant amplitude at P, squaring equations (4) and (5), then adding
R² (cos² θ + sin θ] = a²[1 + cos² ϕ + 2 cos ϕ + sin² ϕ]
R²[1] = a²[1 + 1 + 2 cos ϕ]
I = R² = 2a² [1 + cos ϕ] = 2a² × 2 cos² \(\frac{\phi}{2}\); I = 4a² cos² \(\frac{\phi}{2}\) —— (7)

i) Minimum intensity (I_max)
c0s² \(\frac{\phi}{2}\) = 1
ϕ = 2nπ Where n = 0, 1, 2, 3 …….
ϕ = 0, 2π, 4π, 6π
∴ I_max = 4a².

ii) Minimum intensity (I_min)
cos² \(\frac{\theta}{2}\) = 0
ϕ = (2n + 1)π where n = 0, 1, 2, 3, …….
ϕ = π, 3π, 5π, 7π
I_min = 0

Question 4.
Does the principle of conservation of energy hold for interference and diffraction phenomena ? Explain briefly. (Mar. ’14)
Answer:
Yes, law of conservation of energy is obeyed. In case of constructive interference, intensity becomes maximum. Hence bright fringes are formed on the screen where as in the case of destructive interference, intensity becomes minimum. Hence dark fringes are formed on the screen.

This establishes that in the interference and diffraction pattern, the intensity of light is simply being redistributed i.e., energy is being transferred from dark fringe to bright fringe. No energy is being created (or) destroyed in the process. Hence energy is redistributed.
Thus the principle of conservation of energy is being obeyed in the process of interference and diffraction.

Question 5.
How do you determine the resolving power of your eye ? (A.P. Mar. ’19)
Answer:
Make black strips of equal width separated by white strips. All the black strips having same width, while the width of white strips should increase from left to right.
Now watch the pattern with one eye. By moving away (or) closer to the wall, find the position where you can just see some two black strips as separate strips.

All black strips to the left of this strips would merge into one another and would not be distinguishable on the other hand, the black strips to the right of this would be more and more clearly visible.

Note the width d of the white strips and measure the distance D of the wall from eye.
Then resolution of your eye = \(\frac{\mathrm{d}}{\mathrm{D}}\).

Question 6.
Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids.
Answer:
Let I₀ be the intensity of polarised light after passing through the first polariser P₁. Then the intensity of light after passing through second polariser P₂ will be I = I₀cos²θ.
Where θ is the angle between pass axes P₁ and P₂. Since P₁ and P₂ are crossed the angle between the pass axes of P₂ and P₃ will be \(\left(\frac{\pi}{2}-\theta\right)\)
Hence the intensity of light emerging from P₃ will be
I = I₀cos² θ. cos² \(\left(\frac{\pi}{2}-\theta\right)\)
= I₀ cos² θ . sin² θ
I = \(\frac{I_0}{4}\) sin² 2θ
The transmitted intensity will be maximum when θ = \(\frac{\pi}{4}\)

Long Answer Questions

Question 1.
What is Huygen’s Principle? Explain the optical phenomenon of refraction using Huygen’s principle.
Answer:
Huygens principle: Every point on a wave front is the source of secondary wavelets.
Refraction of a plane wave using Huygen’s principle:
Let the surface PP’ separating the two medium of refractive index μ₁ and μ₂. Let υ₁ and υ₂ be the velocities of light in medium 1 and medium 2.

According to Huygen’s principle, every point on incident wave front AB is a source of secondary wavelets. By the time wavelét from point B reaches at point C, the wavelet from point A would have reached at point E. Let t be the time taken from B to C is equal to time taken from A to D.

μ₁ sin i = μ₂ sin r
This is the Snell’s law of refraction.

Second law of refraction: Since incident ray, refracted ray and the normal all the lie on the same plane PP’ at the point of incidence. This proves the second law of refraction.

Question 2.
Distinguish between Coherent and Incoherent addition of waves. Develop the theory of constructive interferences.
Answer:
Coherent sources : The two sources which maintain zero (or) any constant phase relation between themselves are known as Coherent sources.
Incoherent sources : If the phase difference changes with time, the two sources are known as incoherent sources.

Theory of constructive and destructive interference :
Let the waves of two coherent sources be
y₁ = a sin ωt —— (1)
y₂ = a sin (ωt + ϕ) —– (2)
where a is amplitude and ϕ is the phase difference between two displacements.
According to superposition principle, y = y₁ + y₂
y = a sin ωt + a sin (ωt + ϕ)= a sin ωt + a sin ωt cos ϕ + a cos ωt sin ϕ
y = a sin ωt [1 + cos ϕ] + cos ωt [a sin ϕ] —– (3)
Let A cos θ = a(1 + cos ϕ] —– (4)
A sin θ = a sin ϕ —– (5)
Substituting equations (4) and (5) in equation (3)
y = A sin ωt. cos θ + A cos ωt sin θ
y = A sin (ωt + θ) —— (6)
Where A is resultant amplitude. Squaring equations (4) and (5), then adding
A² [cos² θ + sin² θ] = a²[1 + cos² ϕ + 2 cos ϕ + sin² ϕ]
A² [1] = a² [1 + 1 + 2 cos ϕ]
I = A² = 2a² [1 + cos ϕ] ( ∵ I = A²)
I = 2a² × 2 cos² \(\frac{\phi}{2}\)
I = 4a² cos² \(\frac{\phi}{2}\)
I = 4I₀ cos² \(\frac{\phi}{2}\) —— (7) (∵ I₀ = a²)

Case (i) For constructive interference : Intensity should be maximum.
cos \(\frac{\phi}{2}\) = 1 ⇒ ϕ = 2nπ
Where n = 0, 1, 2, 3….. ⇒ ϕ = 0, 2π, 4π, 6π ….. I_max = 4I₀
Case (ii) For destructive interference : Intensity should be minimum
i.e., cos ϕ = 0 ⇒ ϕ = (2n + 1) π ; where n = 0, 1, 2, 3……. ; ϕ = π, 3π, 5π ⇒ I_min = 0.

Question 3.
Describe Young’s experiment for observing interference and hence arrive at the expression for ‘fringe width’. ‘
Answer:
Interference : The modification of intensity obtained by the super position of two (or) more light waves is called interference.

1. Thomas Young experimentally observed the phenomenon of interference of light using two coherent sources.
2. A small pin hole ‘S’ illuminated by monochromatic source of light which produces a spherical wave.
3. S₁ and S₂ are two narrow pin holes equidistant from S.
4. Screen is placed at a distance D.
5. The points at which any two crests (or) any two troughs are superimposed, constructive interference takes place bright fringe will be observed on the screen.
6. The points at which crest of one wave and trough of another wave are super imposed, destructive interference takes place dark fringe will be observed on the screen.
7. Thus on the screen alternately bright and dark frings are observed.

Expression for fringe width :

i) It is the distance between two successive bright (or) dark fringes, denoted by p.

ii) The path difference (δ) = d sin θ

If θ is very small then from figure sin θ ≈ tan θ = \(\frac{x}{D}\)

iii) For bright fringes path difference S₂P – S₁P = nλ
∴ d sin θ = nλ,
d × \(\frac{x}{D}\) = nλ
x = \(\frac{n \lambda \mathrm{D}}{\mathrm{d}}\) —– (1) where n = 0, 1, 2, 3,…….
This equation represents the position of bright fringe.
When n = 0, x₀ = 0
n = 1, x₁ = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\) and n = 2, x₂ = \(\frac{2 \lambda \mathrm{D}}{\mathrm{d}}\)

The distance between any two consecutive bright fringes is
x₂ – x₁ = \(\frac{2 \lambda \mathrm{D}}{\mathrm{d}}-\frac{\lambda \mathrm{D}}{\mathrm{d}}\) ⇒ β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\) ——- (2)

iv) For dark fringes path difference S₂P – S₁P = (2n + 1) \(\frac{\lambda}{2}\) ∴ d sin θ = (2n + 1)\(\frac{\lambda}{2}\)
d × \(\frac{x}{D}\) = (2n + 1) \(\frac{\lambda}{2}\) = \(\frac{(2 \mathrm{n}+1) \lambda \mathrm{D}}{2 \mathrm{~d}}\) ——- (3) where n =0, 1, 2, 3, ………
This equation (3) represents, position of dark fringe.
When n = 0, x₀ = \(\frac{\lambda \mathrm{D}}{2 \mathrm{~d}}\) ⇒ n = 1, x₁ = \(\frac{3 \lambda \mathrm{D}}{2 \mathrm{~d}}\) ; n = 2, x₂ = \(\frac{5 \lambda \mathrm{D}}{2 \mathrm{~d}}\) ……
The distance between any two consecutive dark fringes is x₂ – x₁ = \(\frac{5 \lambda \mathrm{D}}{2 \mathrm{~d}}-\frac{3 \lambda \mathrm{D}}{2 \mathrm{~d}}\) = \(\frac{5 \lambda \mathrm{D}-3 \lambda \mathrm{D}}{2 \mathrm{~d}}\)
β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\) —– (4)
Hence fringe width is same for bright and dark fringes.

Question 4.
What is diffraction ? Discuss diffraction pattern obtainable from a single slit.
Answer:
Diffraction : The phenomenon of bending of light at the edges of an obstacle and light enters into the geometrical shadow is known as diffraction of light.
Example : The silver lining surrounding the profile of a mountain just before sunrise.

Diffraction of light at a single slit:

1. Consider a narrow slit AB of width d. A parallel beam of light of wave length λ falling normally on a single slit.
2. Let the diffracted light be focussed by means of a convex lens on a screen.
3. The secondary wavelets travelling normally to the slit, i.e., along the direction of OP₀.
   Thus P₀ is a bright central image.
   
4. The secondary wavelets travelling at an angle θ with the normal are focussed at a point P₁ on the screen.
5. In order to find out intensity at P₁, draw a perpendicular AC on BR.
6. The path difference between secondary wavelets = BC
   = AB sin θ = a sin θ (∵ sin θ = 0)
   Path difference (λ) ≈ a θ —– (1)
7. Experimental observations shown in figure, that the intensity has a central maximum at θ = 0 and other secondary maxima at θ ≈ \(\left(\mathrm{n}+\frac{1}{2}\right) \frac{\lambda}{\mathrm{a}}\) and has minima at θ = \(\approx \frac{n \lambda}{a}\)
8. From equation (1), θ = \(\frac{\lambda}{\mathrm{a}}\). Now we divide the slit into two equal halves, each of size \(\frac{a}{2}\).
9. We can show that the intensity is zero for θ = \(\frac{n \lambda}{a}\) where n = 1, 2, 3….
10. It is alsó to see why there are maxima at θ = \(\left(\mathrm{n}+\frac{1}{2}\right) \frac{\lambda}{\mathrm{a}}\)
11. Consider θ = \(\frac{3 \lambda}{2 \mathrm{a}}\) which is midway between two of the dark fringes.
12. If we take the first two thirds of the slit, the path difference between two ends is
    \(\frac{2}{3} a \times \theta\) = \(\frac{2 \mathrm{a}}{3} \times \frac{3 \lambda}{2 \mathrm{a}}\) = λ —– (2)
    
13. The first two third of the slit can be divided into two halves which have a \(\frac{\lambda}{2}\) path difference. The contribution of two halves cancel and only remaining one third of the slit contributes to the intensity minima.

Question 5.
What is resolving power of Optical Instruments? Derive the condition under which images are resolved.
Answer:
Resolving power : The resolving power of a lens is its ability to resolve two points that are to each other.

Resolving power of optical instruments:

1. Consider a parallel beam of light falling on a convex lens. Due to diffraction effect, the beam focussed to a spot of finite area.
2. Taking into account the effects duè to diffraction, the pattern on the focal plane would consist of a central bright region (circular) surrounded by a concentric dark and bright rings.
3. The radius of the central bright region is given by r₀ = \(\frac{1.22 \lambda \mathrm{f}}{2 \mathrm{a}}\) = \(\frac{0.61 \lambda \mathrm{f}}{\mathrm{a}}\)
   where f is focal length of the lens
   2a = diameter of the lens.

Derive the condition under which images are resolved : The size of the spot is very small, it plays an important role in determining the limit of resolution.
For the two stars to be Just resolved
f Δ θ ≈ r₀ ≈ \(\frac{0.61 \lambda \mathrm{f}}{\mathrm{a}}\)
Δ θ ≈ \(\frac{0.61 \lambda}{\mathrm{a}}\) —– (1)
Thus Δθ will be small, if the diameter (2a) of the objective is large. This implies that the telescope will have better resolving power if a is large.
In case of microscope, the object is placed slightly beyond f. The corresponding minimum seperation (d_Min) between the object and the objective lens is given by
d_Min = \(\frac{1.22 \lambda}{2 \mu \sin \beta}\)
Where μ = Refractive index
μ sin β = Numerical aperture.

Textual Exercises

Question 1.
Monochromatic light of wavelength. 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light ? Refractive index of water is 1.33.
Solution:
λ = 589 nm = 589 × 10^-9 m

a) Reflected light: (Wavelength, frequency, speed same as incident light)
λ = 589 nm, v = 5.09 × 10¹⁴ Hz
c = 3 × 10⁸ m/s ⇒ υ = \(\frac{c}{\lambda}=\frac{3 \times 10^8}{589 \times 10^{-9}}\) = 5.093 × 10¹⁴ Hz.

b) Refracted light: (frequency same as the incident frequency)
y = 5.093 × 10¹⁴ Hz
υ = \(\frac{\mathrm{c}}{\mu}=\frac{3 \times 10^8}{1.33}\) = 2.256 × 10⁸ m/s ⇒ λ = \(\frac{v}{v}=\frac{2.26 \times 10^8}{5.09 \times 10^{14}}\) = 443 m.

Question 2.
What is the shape of the wavefront in each of the following cases :
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant intercepted by the Earth.
Soution:
a) It is spherical wavefront.
b) It is plane wavefront.
c) Plane wavefront (a small area on the surface of a large sphere is nearly planar.

Question 3.
(a) The refractive index of glass is 1.5. What is the speed of light in glass ? (Speed of light in vacuum is 3.0 × 10⁸ m s^-1)
(b) Is the speed of light in glass independent of the colour of light ? If not, which of the two colours red and violet travels slower in a glass prism ?
Solution:
a) Here, µ = \(\frac{\mathbf{c}}{v}\) ⇒ υ = \(\frac{\mathrm{c}}{\mu}\) = \(\frac{3 \times 10^8}{1.5}\) = 2 × 10⁸ m/s
b) No, the refractive index and speed of light in a medium depend on wavelength i.e. colour of light. We know that µ_v > µ_r.
Therefore v_violet < v_red. Hence violet component of white light travels slower than the red component.

Question 4.
In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Solution:
d = 0.28 mm = 0.28 × 10^-3 m, D = 1.4 m, β = 1.2 × 10^-2 m, n = 4

⇒ λ = 600 nm.

Question 5.
In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3 ?
Solution:
Let I₁ = I₂ = I. If ϕ is phase difference between the two light waves, then resultant intensity,
I_R = I₁ + I₂ + \(2 \sqrt{\mathrm{I}_1 \mathrm{I}_2}\) . cos ϕ
When path difference = λ, Phase difference ϕ = 0° ∴ I_R = I + I + \(2 \sqrt{I I}\). cos 0° = 4I = k
When path difference = \(\frac{\lambda}{3}\), phase difference ϕ = \(\frac{2 \pi}{3}\) rad.
∴ I_R = I + I + \(2 \sqrt{\mathrm{II}} \cdot \cos \frac{2 \pi}{3}\) ⇒ I’_R = 2I + 2I\(\left(\frac{-1}{2}\right)\) = I = \(\frac{\mathrm{k}}{4}\)

Question 6.
A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both’the wavelengths coincide ?
Solution:
Here λ₁ = 650 nm = 650 × 10^-9 m, λ₂ = 520 nm = 520 × 10^-9 m .
Suppose d = Distance between two slits; D = Distance of screen from the slits
a) For third bright fringe, n = 3 ⇒ x = nλ, \(\frac{\mathrm{D}}{\mathrm{d}}\) = 3 × 650 \(\frac{\mathrm{D}}{\mathrm{d}}\) nm
b) Let n^th fringe due to λ₂ = 520 nm coincide with (n – 1)^th bright fringe due to λ₁ = 650 nm
∴ nλ₂ = (n – 1) λ₁ ; n × 520 = (n – 1) 650; 4n = 5n – 5 or n = 5 .
∴ The least distance required, x = nλ₂ \(\frac{\mathrm{D}}{\mathrm{d}}\) = 5 × 520 \(\frac{\mathrm{D}}{\mathrm{d}}\) = 2600 \(\frac{\mathrm{D}}{\mathrm{d}}\) nm.

Question 7.
In a double-slit experiment the angular width of the fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water ? Take refractive index of water to be 4/3.
Solution:
Here, θ₁ = 0.2°, D = 1m, λ₁ = 600 nm, θ₂ = ?, μ = 4/3

Question 8.
What is the Brewster angle for air to glass transition ? (Refractive index of glass = 1.5.)
Solution:
Here, i_p = ? μ = 1.5; As tan i_p = μ = 1.5 ∴ i_p = tan^-1 (1.5); i_p = 56.3

Question 9.
Light of wavelength 5000 A falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light ? For what angle of incidence is the reflected ray normal to the incident ray ?
Solution:
Given λ = 5000 A = 5 × 10^-7 m
The wavelength and frequency of reflected light remains the same.
∴ Wavelength of reflected light, λ = 5000 A
Frequency of reflected light, υ = \(\frac{\mathrm{c}}{\lambda}=\frac{3 \times 10^8}{5 \times 10^{-7}}\) = 6 × 10¹⁴ Hz
The reflected ray is normal to incident if angle of incidence i = 45.

Question 10.
Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.
Solution:
Here, a = 4 mm = 4 × 10^-3 m; λ = 400 nm = 400 × 10^-9 m = 4 × 10^-7 m
Ray optics is good approximation upto distances equal to Fresnels’ distance (Z_F)
Z_F = \(\frac{a^2}{\lambda}=\frac{\left(4 \times 10^{-3}\right)^2}{4 \times 10^{-7}}\) = 40 m

Additional Exercises

Question 11.
The 6563 A Hα line emitted by hydrogen in a star is found to be red-shifted by 15 A. Estimate the speed with which the star is receding from the Earth.
Solution:
Given λ’ – λ= 15A = 15 × 10^-10 m; λ = 6563 A = 6563 × 10^-10 m; v = ?
Since λ’ – λ = \(\frac{v \lambda}{c}\) ∴ v = \(\frac{c\left(\lambda^{\prime}-\lambda\right)}{\lambda}\) ⇒ v = \(\frac{3 \times 10^8 \times 15 \times 10^{-10}}{6563 \times 10^{-10}}\) = 6.86 × 10⁵ m/s

Question 12.
Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water ? If not, which alternative picture of light is consistent with experiment ?
Solution:
In Newton’s corpuscular picture of refraction, particles of light incident from a rarer to a denser medium experience a force of attraction normal to the surface. This results in an increase in the normal component of velocity but the component along the surface remains unchanged. This means
c sin i = v sin r or \(\frac{\mathrm{v}}{\mathrm{c}}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}\) = μ; Since μ > 1, ∴ v > c
The prediction is opposite to the experimental result: (v < c) . The wave picture of light is consistent with experiment.

Question 13.
You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.
Solution:
In figure, P is a point object placed at a distance r from a plane mirror M₁ M₂. with P as centre and OP = r as radius, draw a spherical arc; AB. This is the spherical wave front from the object, incident on M₁ M₂. If mirrors were not present, the position of wave front AB would be A’B’ where PP’ = 2r. In the presence of the mirror, wave front AB would appear as A”PB”, according to Huygen’s construction. As is clear from the fig. A’B’ and

A”B” are two spherical arcs located symmetrically on either side of M₁ M₂. Therefore, A’PB’ can be treated as reflected image of A”PB”. From simple geometry, we find OP = OP’, which was to be proved.

Question 14.
Let us list some of the factors, which could possibly influence the speed of wave propagation :

1. nature of the source
2. direction of propagation
3. motion of the source and/or observer
4. wavelength
5. intensity of the wave On which of these factors, if any, does

(a) the speed of light in vacuum
(b) the speed of light in a medium (say, glass or water), depend ?
Solution:
a) The speed of light in vacuum is a universal constant, independent of all the factorslisted and anything else.
b) Dependence of the speed of light in a medium

1. Does not depend on the nature of the source.
2. Independent of the direction of propagation for isotropic media.
3. Independent of the motion of the source relative to the medium but depends on the motion of the Observer relative to the medium.
4. Depends on wavelength.
5. Independent of intensity.

Question 15.
For sound waves, the Doppler formula for frequency shift differs slightly between the two situations :

(i) source at rest; observer moving and
(ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium ?

Solution:

Sound waves require a material medium for propagation. That is why situation (i) and (ii) are not identical physically though relative motion between the source and the observer is the same in the two cases. Infact, relative motion of the observer relative to the medium is different in two situations. That is why Doppler’s formulae for sound are different in the two cases.

For light waves travelling in vacuum, there is nothing to distinguish between the two situations. That is why the formulae are strictly identical.
For light propagating in a medium, situation (i) and (ii) are not identical. The formulae governing the two situations would obviously be different.

Question 16.
In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between the two slits ?
Solution:
Here λ = 600 nm = 6 × 10^-7 m, θ = 0.1° = \(\frac{0.1^{\circ}}{180^{\circ}} \times \pi \mathrm{rad}\), d = ?
Since θ = \(\frac{\lambda}{\mathrm{d}}\) ⇒ d = \(\frac{\lambda}{\theta}\) = \(\frac{6 \times 10^{-7} \times 180^{\circ}}{0.1^{\circ} \times \pi}\) = 343 × 10^-4 m.

Question 17.
Answer the following questions :
(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and density of the central diffraction band ?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment ?
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why ?
(d) Two students are separated by a 7m partition wall in a room 10 m high. If both light and sound waves Can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.
(e) Ray optics is based on the assumption that light travels in a straight line.fDiffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification ?
Solution:
a) The size of centred diffraction band reduces by half according to the relation : size \(\frac{\lambda}{\mathrm{d}}\). Intensity increase four fold.

b) The intensity of interference fringes in a double slit arrangement is modulated by diffraction pattern Of each slit.

c) Waves diffracted from the edge of the circular obstacle interfer constructively at the centre of the shadow producing a bright spot.

d) For diffraction the size of the obstacle should be comparable to wavelength if the size of the obstacle is much too large compared to wavelength, diffraction is by a small angle. Here the size partition of wall is of the order of few metres. The wavelength of light is about 5 × 10^-7 m, while sound waves of say 1 kHZ frequency have wavelength of about 0.3 m. Thus sound waves can bend around the partition while light waves cannot.

e) Typical sizes of apertures involved in ordinary optical instruments are much larger than the wavelength.

Question 18.
Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects ?
Solution:
We want \(\frac{(5.0)^2}{\lambda}\) > > \(\frac{40,000}{2}\) ⇒ i.e. λ = < < \(\frac{(5.0)^2}{20,000}\) = \(\frac{250}{20,000}\) = 0.125 m = 12.5 cm

Question 19.
A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.
Solution:
Here λ = 500 nm = 5 × 10^-7 m, D = 1 m, y = 2.5 mm = 2.5 × 10^-3 m, d = ?
sin θ = \(\frac{\lambda}{d}=\frac{y}{D}\) ∴ d = \(\frac{\lambda \mathrm{D}}{\mathrm{y}}=\frac{5 \times 10^{-7} \times 1}{2.5 \times 10^{-3}}\) = 2 × 10^-4 m = 0.2 mm

Question 20.
Answer the following questions :
(a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.
(b) As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle ?
Solution:
a) Interference of the direct signal received by the antenna with the (weak) signal reflected by the passing air craft.
b) Super position principle follows from the linear character of the equation governing wave motion. It is true so ions as wave have small amplitude.

Question 21.
In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of nλ/a. Justify this by suitable dividing the slit to bring out the cancellation.
Solution:
Divide the signal slit into n smaller slits of width a’ = \(\frac{\mathrm{a}}{\mathrm{n}}\). Each of the smaller slits sends zero intensity in the direction θ. The combination gives zero intensity as well.

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(a)

Resolve the following into partial fractions.

I.

Question 1.
\(\frac{2 x+3}{(x+1)(x-3)}\)
Solution:
Let \(\frac{2 x+3}{(x+1)(x-3)}=\frac{A}{x+1}+\frac{B}{x-3}\)
Multiplying with (x + 1) (x – 3)
2x + 3 = A(x – 3) + B(x + 1)
x = -1 ⇒ 1 = A(-4) ⇒ A = \(-\frac{1}{4}\)
x = 3 ⇒ 9 = B(4) ⇒ B = \(\frac{9}{4}\)
\(\frac{2 x+3}{(x+1)(x-3)}=\frac{-1}{4(x+1)}+\frac{9}{4(x-3)}\)

Question 2.
\(\frac{5 x+6}{(2+x)(1-x)}\)
Solution:
Let \(\frac{5 x+6}{(2+x)(1-x)}=\frac{A}{2+x}+\frac{B}{1-x}\)
Multiplying with (2 + x) (1 – x)
5x + 6 = A(1 – x) + B(2 + x)
Put x = -2,
-10 + 6 = A(1 + 2)
⇒ A = \(-\frac{4}{3}\)
Put x = 1,
5 + 6 = B(2 + 1)
⇒ B = \(\frac{11}{3}\)
∴ \(\frac{5 x+6}{(2+x)(1-x)}=-\frac{4}{3(2+x)}+\frac{11}{3(1-x)}\)

II.

Question 1.
\(\frac{3 x+7}{x^2-3 x+2}\)
Solution:
\(\frac{3 x+7}{x^2-3 x+2}=\frac{3 x+7}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}\)
Multiplying with x² – 3x + 2
3x + 7 = A(x – 2) + B(x – 1)
x = 1 ⇒ 10 = -A ⇒ A = -10
x = 2 ⇒ 13 = B ⇒ B = 13
∴ \(\frac{3 x+7}{x^2-3 x+2}=\frac{-10}{x-1}+\frac{13}{x-2}\)

Question 2.
\(\frac{x+4}{\left(x^2-4\right)(x+1)}\)
Solution:
\(\frac{x+4}{\left(x^2-4\right)(x+1)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x-2}\)
Multiplying with (x² – 4) (x + 1)
x + 4 = A(x² – 4) + B(x + 1) (x – 2) + C(x + 1) (x + 2)
x = -1
⇒ 3 = A(1 – 4)
⇒ 3 = -3A
⇒ A = -1
x = -2
⇒ 2 = B(-2 + 1) (-2 – 2)
⇒ 2 = 4B
⇒ B = \(\frac{1}{2}\)
x = 2
⇒ 6 = C(2 + 1)(2 + 2)
⇒ 6 = 12C
⇒ C = \(\frac{1}{2}\)
∴ \(\frac{x+4}{\left(x^2-4\right)(x+1)}=-\frac{1}{x+1}+\frac{1}{2(x+2)}\) + \(\frac{1}{2(x-2)}\)

Question 3.
\(\frac{2 x^2+2 x+1}{x^3+x^2}\)
Solution:
Let \(\frac{2 x^2+2 x+1}{x^3+x^2}=\frac{2 x^2+2 x+1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}\)
Multiplying with x²(x + 1)
2x² + 2x + 1 = Ax(x + 1) + B(x + 1) + Cx²
Put x = 0, 1 = B
Put x = -1, 2 – 2 + 1 = C(1) ⇒ C = 1
Equating the coefficients of x²,
2 = A + C
⇒ A = 2 – C = 2 – 1 = 1
∴ \(\frac{2 x^2+2 x+1}{x^3+x^2}=\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x+1}\)

Question 4.
\(\frac{2 x+3}{(x-1)^3}\)
Solution:
\(\frac{2 x+3}{(x-1)^3}\)
Put x – 1 = y ⇒ x = y + 1
⇒ \(\frac{2 x+3}{(x-1)^3}=\frac{2(y+1)+3}{y^3}=\frac{2 y+5}{y^3}\)
⇒ \(\frac{2 x+3}{(x-1)^3}\) = \(\frac{2}{y^2}+\frac{5}{y^3}=\frac{2}{(x-1)^2}+\frac{5}{(x-1)^3}\)
∴ \(\frac{2 x+3}{(x-1)^3}=\frac{2}{(x-1)^2}+\frac{5}{(x-1)^3}\)

Question 5.
\(\frac{x^2-2 x+6}{(x-2)^3}\)
Solution:
Let x – 2 = y then x = y + 2

III.

Question 1.
\(\frac{x^2-x+1}{(x+1)(x-1)^2}\)
Solution:
Let \(\frac{x^2-x+1}{(x+1)(x-1)^2}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}\)
Multiplying with (x + 1) (x – 1)²
x² – x + 1 = A(x – 1)² + B(x + 1) (x – 1) + C(x + 1)
Put x = -1,
1 + 1 + 1 = A(4)
⇒ A = \(\frac{3}{4}\)
Put x = 1,
1 – 1 + 1 = C(2)
⇒ C = +\(\frac{1}{2}\)
Equating the coefficients of x²,
A + B = 1
⇒ B = 1 – A
⇒ B = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)
∴ \(\frac{x^2-x+1}{(x+1)(x-1)^2}=\frac{3}{4(x+1)}+\frac{1}{4(x-1)}\) + \(\frac{1}{2(x-1)^2}\)

Question 2.
\(\frac{9}{(x-1)(x+2)^2}\)
Solution:
Let \(\frac{9}{(x-1)(x+2)^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}\)
Multiplying with (x – 1) (x + 2)²
9 = A(x + 2)² + B(x – 1) (x + 2) + C(x – 1)
x = 1
⇒ 9 = 9A
⇒ A = 1
x = -2
⇒ 9 = -3C
⇒ C = -3
Equating the coefficients of x²
A + B = 0 ⇒ B = -A = -1
∴ \(\frac{9}{(x-1)(x+2)^2}=\frac{1}{x-1}-\frac{1}{x+2}-\frac{3}{(x+2)^2}\)

Question 3.
\(\frac{1}{(1-2 x)^2(1-3 x)}\)
Solution:
Let \(\frac{1}{(1-2 x)^2(1-3 x)}=\frac{A}{1-3 x}+\frac{B}{1-2 x}+\frac{C}{(1-2 x)^2}\)
Multiplying with (1 – 2x)² (1 – 3x)
1 = A(1 – 2x)² + B(1 – 3x) (1 – 2x) + C(1 – 3x)

Question 4.
\(\frac{1}{x^3(x+a)}\)
Solution:
Let \(\frac{1}{x^3(x+a)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x+a}\) = \(\frac{A \cdot x^2(x+a)+B(x)(x+a)+C(x+a)+D x^3}{x^3(x+a)}\)
∴ 1 = A (x²) (x + a) + Bx (x + a) + C(x + a) + Dx³ ……..(1)
Put x = 0 in (1)
1 = A(0) + B(0) + C(0 + a) + D(0)
⇒ 1 = C(a)
⇒ C = \(\frac{1}{a}\)

Question 5.
\(\frac{x^2+5 x+7}{(x-3)^3}\)
Solution:
Let x – 3 = y ⇒ x = y + 3
\(\frac{x^2+5 x+7}{(x-3)^3}=\frac{(y+3)^2+5(y+3)+7}{y^3}\)

Question 6.
\(\frac{3 x^3-8 x^2+10}{(x-1)^4}\)
Solution:
Put x – 1 = y ⇒ x = y + 1

Students can go through AP Inter 2nd Year Accountancy Notes 3rd Lesson Consignment will help students in revising the entire concepts quickly.

AP Inter 2nd Year Accountancy Notes 3rd Lesson Consignment

→ Consignment means goods are consigned or sent to the agent or consignee for the purpose of sale.

→ There are two parties in the consignment. Consignor or the owns who send goods on consignment. Consignee or the agent who receives the goods sent by the consignor and sells them for consideration.

→ The main difference between the consignment and sale is that in consignment ownership of goods remain with consignor and possession is transferrable. In sale both ownership and possession are transferred to buyer.

→ Proforma invoice means a statement forwarded by the consignor along with the goods giving a description of goods consigned, the weight, quantity, price and other relevant data. Proforma invoice price is the minimum price at which the consignee is expected to sell the goods.

→ Account sales is a document or statement sent by the consignee to the consignor showing the details of gross sale proceeds, the various expenses incurred by him and the commission due is deducted. Any advance payment to the consignor is deducted and net amount due is shown.

→ Consignee is remunerated by a commission which is calculated as a fixed percentage on gross sale proceeds. Sometimes, consignor in order to avoid the risk or loss of bad debts, provides additional commission known as Del Credre Commission. Over riding commission is allowed Over the normal commission. This is offered when the agent is required to workhard if a new product is introduced in the market.

→ కన్నానార్ కనానికి అమ్మడానికి, సరుకును పంపడాన్ని కన్సెన్మెంటు అంటారు. సరుకును పంపే వ్యక్తిని కన్పైనార్ అని, ఎవరికయితే సరుకు పంపబడుతుందో అతనిని కన్సైనీ అంటారు.

→ ప్రొఫార్మా ఇన్వాయిస్ అనేది ఒక నివేదిక. దీనిని సరుకుతో పాటుగా కన్సైనీకి పంపుతాడు. ఈ నివేదికలో సరుకు వివరాలు అనగా వర్ణన, బరువు, పరిమాణము, ధర, ప్యాకింగ్ మొదలగునవి ఉంటాయి.

→ అకౌంట్ సేల్స్న కనైన్ కన్సైనార్కు పంపుతాడు. దీనిలో అమ్మిన సరుకు విలువ, అయిన ఖర్చులు, రావలసిన కమీషన్, పంపిన బయానా మినహాయించి ఎంత పంపవలెనో చూపుతాడు.

→ కన్సానార్ పంపిన సరుకు అంతా అమ్ముడు కాకపోతే కొంత సరుకు మిగిలితే, దీనిని అసలు ధర లేదా మార్కెట్ ధర ఏది తక్కువైతే దానికి విలువ కట్టవలెను. ఖరీదుతోపాటు సరుకు విలువను పెంపొందించే పునరావృత ఖర్చులను కలుపవలెను.

→ కన్నానార్ సరుకు పంపినపుడు మార్గమధ్యములో కాని, కన్సైన్ వద్ద కొంత సరుకు నష్టము కావచ్చు. సహజసిద్ధ కారణాల వలన సరుకు నష్టము జరిగితే దానిని సాధారణ నష్టము అని, అనుకోని సంఘటనల వల్ల నష్టం జరిగితే దానిని అసాధారణ నష్టం అంటారు.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 2nd Lesson Ray Optics and Optical Instruments Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 2nd Lesson Ray Optics and Optical Instruments

Very Short Answer Questions

Question 1.
Define focal length and radius of curvature of a concave lens.
Answer:
Focal length (f) : The distance of principal focus from the optical centre of the lens is called focal length of the lens. The focal length (f) = CF

Radius of curvature : Radius of curvature is the radius of the sphere from which the curved surface is taken a part.

Question 2.
What do you understand by the terms ‘focus’ and ‘principal focus’ in the context of lenses ?
Answer:
Focus : The point where image of an object placed at infinity is formed is called the focus of the lens.
Principal focus: A narrow beam of light incident on a lens in a direction parallel to its principal axis, after refraction through the lens, the rays converge to a point on the principal axis. This point is called principal focus.

Question 3.
What is optical density and how is it different from mass density ? (T.S. Mar. ’16)
Answer:
Optical density: Optical density is defined as the ratio of the speed of light in media.
Mass density: Mass per unit volume is defined as mass density.
Mass density of an optically denser medium less than that of optically rarer medium.

Question 4.
What are the laws of reflection through curved mirrors ?
Answer:

1. “The angle of reflection equals to the angle of incidence”.
2. “The incident ray, reflected ray and the normal to the reflecting surface at the point of incidence lie in the same plane”.

Question 5.
Define ‘power’ of a convex lens. What is its unit ?
Answer:
Power of a lens : Power of a lens is defined as its bending ability and is measured as reciprocal of focal length in metre.
∴ Power of a lens (P) = \(\frac{1}{\text { f(in metres) }}\) = \(\frac{100}{\mathrm{f}(\text { in } \mathrm{cms})}\)
Unit → Dioptre (D)

Question 6.
A concave mirror of focal length 10 cm is placed at a distance 35 cm from a wall. How far from the wall should an object be placed so that its real image is formed on the wall ?
Answer:
f = 10 cm, υ = 35 cm
\(\frac{1}{f}\) = \(\frac{1}{v}\) + \(\frac{1}{-u}\) (using sign convention)
\(\frac{1}{\mathrm{u}}\) = \(\frac{1}{v}\) – \(\frac{1}{f}\) = \(\frac{1}{35}\) – \(\frac{1}{10}\)
\(\frac{1}{\mathrm{u}}\) = \(\frac{10-35}{35 \times 10}\) = \(\frac{-1}{14}\)
U = – 14 cm.
Distance of the object from the wall = 35 – 14 = 21cm.

Question 7.
A concave mirror produces an image of a long vertical pin, placed 40cm from the mirror, at the position of the object. Find the focal length of the mirror.
Answer:
Give u = υ = 40cm]
\(\frac{1}{f}\) = \(\frac{1}{v}\) + \(\frac{1}{u}\)
\(\frac{1}{f}\) = \(\frac{1}{40}\) + \(\frac{1}{40}\)
\(\frac{1}{f}\) = \(\frac{2}{40}\)
f = 20 cm.

Question 8.
A small angled prism of 40 deviates a ray through 2.48°. Find the refractive index of the prism. (A.P. Mar. ’19)
Answer:
A = 4°, D_m = 2.48°
D_m = A(μ – 1)
μ – 1 = \(\frac{D_m}{A}\) = \(\frac{2.48}{4}\) = 0.62
μ = 1 + 0.62
μ = 1.62

Question 9.
What is dispersion ? Which colour gets relatively more dispersed?
Answer:
Dispersion : The phenomenon of splitting of white light into its constituent colours, on passing through a prism is called dispersion of light.
The deviation is maximum for violet colour.

Question 10.
The focal length of a concave lens is 30 cm. Where should an object be placed so that its image is 1/10 of its size? (T.S. Mar. ‘19)
Answer:
f = 30 cm, h₁ = h, h₂ = \(\frac{\mathrm{h}}{10}\)

Question 11.
What is myopia ? How can it be corrected ? (T.S. Mar. ’15)
Answer:
Myopia (or) Near sightedness :
The light from a distant object arriving at the eye-lens may get converged at a point infront of the retina. This type of defect is called myopia.
To correct this, we interpose a concave lens between the eye and the object.

Question 12.
What is hypermetropia ? How can it be corrected ? (A.P. Mar. ’16)
Answer:
Hypermetropia (or) Farsightedness :
The light from a distant object arriving at the eye-lens may get converged at a point behind the retina. This type of defect is called Hypermetropia.
To correct this, we interpose a convex lens (Convergent lens) between the eye and the object.

Short Answer Questions

Question 1.
A light ray passes through a prism of angle A in a position of minimum deviation. Obtain an expression for (a) the angle of incidence in terms of the angle of the prism and the angle of minimum deviation (b) the angle of refraction in terms of the refractive index of the prišm.
Answer:
In the quadrilateral AQNR
∠A + ∠QNR = 180° ……… (1)
From Δ^le QNR, r₁ + r₂ + ∠QNR = 180° ………. (2)
r₁ + r₂ = A …….. (3)
Total deviation δ = (i – r₁) + (e – r₂)
δ = i + e – A ……. (4)

a) At minimum deviation position M
δ = D_m, i = e and r₁ = r₂ = r
∴ From eq(4), D_m = 2i – A
i = \(\frac{\mathrm{A}+\mathrm{D}_{\mathrm{m}}}{2}\) …… (5)

b) From eq (3), r + r = A
r = A/2 ……. (6)
Refractive index of the prism μ = \(\frac{\sin i}{\sin r}\) …… (8)

Question 2.
Define focal length of a concave mirror. Prove that the radius of curvature of a concave mirror is double its focal length. (A.P. Mar. ’19)
Answer:
Focal length of concave mirror:
The distance between the focus F and the pole P of the mirror is called the focal length of the concave mirror.

Consider a ray AB parallel to principal axis incident on a concave mirror at B and is reflected along BF. The line CB is normal to the mirror.

Let θ be the angle of incidence, ∠ABC = ∠BCP = θ

Draw BD ⊥ CP
In right angled Δ^le BCD
Tan θ = \(\frac{\mathrm{BD}}{\mathrm{CD}}\) ….. (1)
From Δ^le BFD, Tan2θ = \(\frac{\mathrm{BD}}{\mathrm{FD}}\) ……. (2)
Dividing eq (2) by eq (1)
…….. (3)

If θ is very small, them tan θ ≈ θ and tan 2θ ≈ 2θ since the aperture of the lens is small
∴ The point B lies very close to p.
CD ≈ CP and FD ≈ FP
From eq (3), \(\frac{2 \theta}{\theta}\) = \(\frac{C P}{F P}\) = \(\frac{R}{f}\) ⇒ 2 = \(\frac{R}{f}\)
R = 2f

Question 3.
A mobile phone lies along the principal axis of a concave mirror longitudinally. Explain why the magnification is not uniform.
Answer:
The ray diagram for the formation of image of the phone is shown in figure. The image of part which is on the plane perpendicular to principal axis will be on the same plane. It will be the same size i.e, B’C = BC.

Question 4.
Explain the cartesian sign convention for mirrors.
Answer:
According to cartesian sign convention:

1. All distances are measured from the pole of the mirror (or) the optical centre of the lens.
2. The distances measured in the same direction as incident light are taken as positive.
   
3. The distances measured in the direction opposite to the incident light are taken as negative.
4. The heights measured upwards with respect to x-axis and normal to the x-axis are taken as positive.
5. The heights measured downwards are taken as negative.

Question 5.
Define critical angle. Explain total internal reflection using a neat diagram.(T.S. Mar. ’15)
Answer:
Critical angle:
When light ray travelling from denser medium to rarer medium, then the angle of inci-dence for which angle of refraction in air is 90° is called critical angle.
C = sin^-1 \(\left(\frac{1}{\mu}\right)\)

Total internal reflection:
When a light ray travels from denser to rarer medium, the angle of incidence is greater than the critical angle, then it reflects into the same medium is called total internal reflection.

Explanation:
Consider an object in the denser medium. A ray OA incident on XY bends away from the normal. As the angle of incidence is increased, the angle of refraction goes on increasing. For certain angle of incidence, the refracted ray parallel to XY surface (r = 90°)

When the angle of incidence is further increased, the ray is not refracted but is totally reflected back in the denser medium. This phenomenon is called total internal reflection.

Question 6.
Explain the formation of a mirage. (T.S. Mar. 19 & A.P. Mar. ’16)
Answer:
In a desert, the sand becomes very hot during the day time and it rapidly heats the layer of air which is in its contact. So density of air decreases. As a result the successive upward layers are denser than lower layers.

When a beam of light travelling from the top of a tree enters a rarer layer, it is refracted away from the normal. As a result at the surface of layers of air, each time the angle of incidence increases and ultimately a stage is reached, when the angle of incidence becomes greater than the critical angle between the two layers, the incident ray suffers total internal reflection.

So it appears as inverted image of the tree is formed and the same looks like a pool of water to the observer.

Question 7.
Explain the formation of a rainbow.
Answer:

Figure shows how sun light is broken into its segments in the process and a rainbow appears. The dispersion of the violet and the red rays after internal reflection in the drop is shown in figure.

The red rays emerge from the drops of water at one angle (43°) and the violet rays emerge at another angle (41°). The large number of water drops in the sky makes a rainbow. The rainbow appears semicircular for an observer on earth.

Question 8.
Why does the setting sun appear red ? (Mar. ’14)
Answer:
As sunlight travels through the earths atmosphere, gets scattered by the large number of molecules present. This scattering of sun light is responsible for the colour of the sky, during sunrise and sunset etc.

The light of shorter wave length is scattered much more than light of larger wavelength. Scattering ∝ \(\frac{1}{\lambda^4}\).

Most of blue light is scattered, hence the bluish colour of sky predominates.

At sunset (or) sunrise, sun rays must pass through a larger atmospheric distance. More of the blue colour is scattered away only red colour which is least scattered appears to come from sun. Hence it appears red.

Question 9.
With a neat labelled diagram explain the formation of image in a simple microscope. (T.S. Mar. 16 & A.P. Mar. 15)
Answer:
Simple microscope : It consists a single short focus convex lens. It increases the visual angle to see an object clearly. It is also called magnifying glass (or) reading glass.

Working : The object is adjusted within the principal focus of the convex lens to form the image at the near point. The image is formed on same side of the object and it is virtual, erect and magnified as shown in fig.

Magnifying power: The ratio of the angle subtended by the image at the eye to the angle subtended by the object at the eye is called magnifying power of a simple microscope.
It is denoted by ‘m’

Question 10.
What is the position of the object for a simple microscope ? What is the maximum magnification of a simple microscope for a realistic focal length?
Answer:
When an object is placed between principal focus and optical centre of a convex lens, a virtual and erect image will be formed on the same side of the object.

Magnifying power: It is defined as the ratio of the angle subtended at the eye by the image to the angle subtended by the object at the eye.

This shows that smaller the focal length of the lens, greater will be the magnifying power of microscope.

Long Answer Questions

Question 1.
a) What is the cartesian sign convention ? Applying this convention and using a neat diagram, derive an expression for finding the image distance using the mirror equation.
b) An object of 5 cm height is placed at a distance of 15 cm from a concave mirror of radius of curvature 20cm. Find the size of the image.
Answer:
According to cartesian sign convention:

1. All distances are measured from the pole of the mirror (or) the optical centre of the lens.
2. The distances measured in the same direction as incident light are taken as positive.
3. The distances measured in the direction opposite to the incident light are taken as negative.
4. The heights measured upwards with respect to x-axis and normal to the x-axis are taken as positive.
5. The heights measured downwards are taken as negative.

Image distance using mirror equation :

Consider an object AB is placed beyond centre of curvature of a concave mirror, on its principal axis.

A ray AD parallel to principal axis incident on the mirror at point D and is reflected to pass through F. Another ray AE passing through centre of curvature C is reflected along the same path. Two rays of light intersect at point A’. Thus A’B’ is real, inverted and diminished image of AB formed between C and F.
Δ^le DPF and Δ^le A’B’ F are similar
\(\frac{\mathrm{B}^{\prime} \mathrm{A}^{\prime}}{\mathrm{PD}}\) = \(\frac{\mathrm{B}^{\prime} \mathrm{F}}{\mathrm{FP}}\)
(or) \(\frac{\mathrm{B}^{\prime} \mathrm{A}^{\prime}}{\mathrm{BA}}\) = \(\frac{\mathrm{B}^{\prime} \mathrm{F}}{\mathrm{FP}}\) …… (1) (∴ PD = AB)

Since ∠APB = ∠A’P’V’
The right angle triangles A’B’P and ABP are similar
\(\frac{\mathrm{B}^{\prime} \mathrm{A}^{\prime}}{\mathrm{BA}}\) = \(\frac{B^{\prime} P}{B P}\) …….. (2)
From equations (1) and (2), \(\frac{\mathrm{B}^{\prime} \mathrm{F}}{\mathrm{FP}}\) = \(\frac{\mathrm{B}^{\prime} \mathrm{P}}{\mathrm{BP}}\) = \(\frac{\mathrm{B}^{\prime} \mathrm{P}-\mathrm{FP}}{\mathrm{FP}}\) ….. (3)

Now applying the sign convention
B’P = -v, FP = -f, BP = -u
\(\frac{-v+f}{-f}\) = \(\frac{-\mathrm{V}}{-\mathrm{u}}\) ⇒ \(\frac{v-f}{f}\) = \(\frac{\mathrm{v}}{\mathrm{u}}\)
\(\frac{v}{\mathrm{f}}\) – 1 = \(\frac{v}{\mathrm{u}}\) ⇒ \(\frac{1}{\mathrm{f}}\) = \(\frac{1}{\mathrm{v}}\) + \(\frac{v}{\mathrm{u}}\)

b) Given that h₁ = 5 cm
u = -15cm
R = 20cm
f = \(\frac{-\mathrm{R}}{2}\) = \(\frac{-20}{2}\) = -10cm

Question 2.
a) Using a neat labelled diagram derive the mirror equation. Define linear magnification.
b) An object is placed at 5cm from a convex lens of focal length 15cm. What is the position and nature of the image ?
Answer:
a) Derivation of mirror equation: Consider an object AB is placed beyond centre of curvature of a concave mirror, on its principal axis.

A ray AD parallel to principal axis incident on the mirror at point D and is reflected to pass through F.
Another ray AE passing through centre of curvature C is reflected along the same path. Two rays of light intersect at point A’. Thus A’ B’ is real, inverted and diminished image of AB formed between C and F.

Linear magnification:
Linear magnification is the ratio of the size of the image formed by the mirror to the size of the object.

Nature of the image is virtual.

Question 3.
a) Derive an expression for a thin double convex lens. Can you apply the same to a double concave lens too?
b) An object is placed at a distance of 20cm from a thin double convex lens of focal length 15cm. Find the position and magnification of the image.
Answer:
a)

1. A convex lens is made up of two spherical refracting surfaces of radii of curvatures, R₁ and R₂ and μ is the refractive index of the lens.
2. P₁, P₂ are the poles, C₁, C₂ are the centres of curvatures of two surfaces with optical centre C.
3. Consider a point object O lying on the principal axis of the lens and I₁ is the real image of the object.
   If CI₁ ≈ P₁ I₁ = v₁
   and CC₁ ≈ PC₁ = R₁
   CO ≈ P₁ = u
4. As refraction is taking place from rarer to denser medium
   \(\frac{\mu_1}{-u}\) + \(\frac{\mu_2}{v_1}\) = \(\frac{\mu_2-\mu_1}{R_1}\) ……. (1)
5. The refracted ray suffers further refraction
   Therefore I is the final real image of O.
6. For refraction at second surface, I₁ as virtual object, whose real image is formed at I.
   ∴ u ≈ CI₁ ≈ P₂I₁ = V₁
   Let CI ≈ P₂ I = V
7. Now refraction taking place from denser to rarer medium

When the object on the left of the lens is at infinity (∝), image is formed at principal focus of the lens.
∴ u = ∝, υ = f = focal length of the lens

This is the lens maker’s formula
Yes, same formula applies to double concave lens too.

b) Given that u = 20 cm, f = 15 cm
\(\frac{1}{f}\) = \(\frac{1}{v}\) + \(\frac{1}{u}\)
\(\frac{1}{v}\) = \(\frac{1}{f}\) – \(\frac{1}{u}\) = \(\frac{1}{15}\) – \(\frac{1}{20}\)
\(\frac{1}{v}\) = \(\frac{20-15}{15 \times 20}\) + \(\frac{5}{15 \times 20}\)
\(\frac{1}{v}\) = \(\frac{1}{60}\)
v = 60 cm.
Magnification (m) = \(\frac{-\mathbf{v}}{\mathbf{u}}\)
m = \(\frac{-60}{-20}\) (u = -20cm)
m = 3

Question 4.
Obtain an expression for the combined focal length for two thin convex lenses kept in contact and hence obtain an expression for the combined power of the combination of the lenses.
Answer:

1. Consider two lenses A and B of focal lengths f₁ and f₂ placed in contact with each other.
2. Let the object be placed at a point O, the first lens forms the image at I₁, it is real. It serves as the virtual object of lens B, Producing final image at I.
3. For the image formed by lens A, we get

Question 5.
a) Define Snell’s Law. Using a neat labelled diagram derive an expression for the refractive index of the material of an equilateral prism.
b) A ray of light, after passing through a medium, meets the surface separating the medium from air at an angle of 45° and is just not refracted. What is the refractive index of the medium ?
Answer:
a) Snell’s law:
The ratio of the sine of the angle of incidence to the sine of angle of refraction is constant, called the refractive index of the medium.
\(\frac{\sin i}{\sin r}\) = μ (constant).
Let ABC be the glass prism. Its angle of prism is A. The refractive index of the material of the prism is μ. Let AB and AC be the two refracting surfaces PQ = incident ray, RS = emergent ray.
Let angle of incidence = i₁
angle of emergence = i₂
angle of refraction = r₁
angle of refraction at R = r₂
After travelling through the prism it falls on AC and emerges as RS.
The D = angle of deviation.

From the Δ QRT
r₁ + r₂ + ∠T = 180° —— (1)
From the quadrilateral AQTR
∠A + ∠T = 180°
∠T = 180° – A ……. (2)
From the equations (1) and (2)
r₁ + r₂ + ∠T = 180° we get
r₁ + r₂ + 180° – A = 180°
r₁ + r₂ = A …… (3)
from the Δ QUR
i₁ – r₁ + i₂ – r₂ + 180° – D = 180°
i₁ + i₂ – (r₁ + r₂) = D
i₁ + i₂ – A = D [∵ r₁ + r₂ = A]
i₁ + i₂ = A + D …. (4)

Minimum deviation: Experimentally it is found that as the angle of incidence increased the angle of deviation decreases till it reaches a minimum value and then it increases. This least value of deviation is called angle of minimum deviation ‘δ’ as shown in the fig.

When D decreases the two angles i₁ and i₂ become closer to each other at the angle of minimum deviation, the two angles of incidence are same i.e, i₁ = i₂
As i₁ = i₂, r₁ = r₂
∴ i₁ = i₂ = r₁ = r₂ = r

substituting this in (1) and (2) we get
2r = A ⇒ r = A/2
i + i = A + δ ⇒ i = \(\frac{\mathrm{A}+\delta}{2}\)

Note :The minimum deviation depends on the refractive index of the prism material and the angle of the prism.

b) Given that i = C = 45°
μ = \(\frac{1}{\sin c}\) ⇒ μ = \(\frac{1}{\sin 45^{\circ}}\)
μ = \(\frac{1}{1 / \sqrt{2}}\) = \(\sqrt{2}\)
μ = 1.414

Question 6.
Draw a neat labelled diagram of a compound microscope and explain its working. Derive an expression for its magnification.
Answer:
Description: It consists of two convex lenses separated by a distance. The lens near the object is called objective and the lens near the eye is called eye piece. The objective lens has small focal length and eye piece has of Idrger focal length. The distance of the object can be adjusted by means of a rack and pinion arrangement.

Working: The object OJ is placed outside the principal focus of the objective and the real image is formed on the other side of it. The image I₁ G₁ is real, inverted and magnified.
This image acts as the object for the eyepiece. The position of the eyepiece is so adjusted that the image due to the objective is between the optic centre and principal focus to form the final image at the near point. The final image IG is virtual, inverted and magnified.

Magnifying Power : It is defined as the ratio of the angle subtended by the final image at the eye when formed at near point to the angle subtended by the object at the eye when imagined to be at near point.

Imagining that the eye is at the optic centre, the angle subtended by the final image is α. When the object is imagined to be taken at near point it is represented by IJ’ and OJ = IJ’. The angle made by I J’ at the eye is β. Then by the definition of magnifying power

Magnifying power of the objective (me) = I₁ G₁ / OJ = Height of the image due to the objective / Height of its object.
Magnifying power of the eye piece (me) = IG/I₁G₁ = Height of the final image / Height of the object for the eyepiece.
∴ m = m_o × m_e ….. (1)

To find m_o : In figure OJ O’ and I₁ G₁ O’ are similar triangles. \(\left(\frac{\mathrm{I}_1 \cdot \mathrm{G}_1}{\mathrm{OJ}}\right)\) = \(\frac{O^{\prime} I_1}{O^{\prime} O}\)
Using sign convention, we find that O’I₁ = + v₀ and O’O = -u where v₀ is the image distance due to the objective and u is the object distance for the objective or the compound microscope.
I₁G₁ is negative and OJ is positive.

To find m_e : The eyepiece behaves like a simple microscope. So the magnifying power of the eye piece.
∴ m_e = \(\left(1+\frac{D}{f_e}\right)\)
Where f_e is the focal length of the eyepiece.
Substituting m₀ and m_e in equation (1),
m = \(+\frac{\mathrm{v}_0}{\mathrm{u}}\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right)\)
When the object is very close to the principal focus F₀ of the objective, the image due to the objective becomes very close to the eyepiece.
u ≈ f₀ and v₀ ≈ L
Where L is the length of the microscope. Then
m = \(-\frac{L}{f_0}\left(1+\frac{D}{f_e}\right)\)

Problems

Question 1.
A light wave of frequency 4 × 10¹⁴ Hz and a wavelength of 5 × 10^-7 m passes through a medium. Estimaté the refractive index of the medium.
Answer:
υ = 4 × 10¹⁴ Hz
A = 5 × 10^-7 m
v = υλ = 4 × 10¹⁴ × 5 × 10^-7 = 20 × 10⁷
= 2 × 10⁸ m fs
we know that C = 3 × 10⁸ m/s
µ = \(\frac{C}{v}\)
µ = \(\frac{3 \times 10^8}{2 \times 10^8}\)
µ = \(\frac{3}{2}\) = 1.5

Question 2.
A ray of light is incident at an angle of 60° on the face of a prism of angle 30°. The emergent ray makes an angle of 30° with the incident ray. Calculate the refractive index of the material of the prism.
Answer:
i₁ = 60°, r = 30°, i² = 30° .
µ = \(\frac{\sin i}{\sin r}\)
µ = \(\frac{\sin 60^{\circ}}{\sin 30^{\circ}}\)
µ = \(\frac{\sqrt{3}}{2 \times \frac{1}{2}}\)
µ = \(\sqrt{3}\)
µ = 1.732

Question 3.
Two lenses of power – 1.75D and +2.25D respectivetly, are placed in contact. Calculate the focal length of the combination.
Answer:
P₁ = – 1.75 D, P₂ = + 2.25 D.
P = P₁ + P₂
P = -1.75 + 2.25
P = 0.5
\(\frac{1}{\mathrm{~F}}\) = P
F = \(\frac{1}{p}\) = \(\frac{1}{0.5}\) = 2m
F = 200cm

Question 4.
Some rays falling on a converging lens are focussed 20cm from the lens. When a diverging lens is placed in contad with the converging lens, the rays are focussed 30cm from the combination. What is the focal length of the diverging lens?
Answer:
u = -20 cm
υ = 30cm
\(\frac{1}{f}\) = \(\frac{1}{v}\) + \(\frac{1}{\mathrm{u}}\) ⇒ \(\frac{1}{\mathrm{f}}\) = \(\frac{1}{\mathrm{30}}\) – \(\frac{1}{\mathrm{20}}\)
\(\frac{1}{\mathrm{f}}\) = \(\frac{20-30}{30 \times 20}\) = \(\frac{-10}{30 \times 20}\)
\(\frac{1}{\mathrm{f}}\) = \(-\frac{1}{60}\)
f = -60cm.

Question 5.
A double convex lens of focal length 15cm is used as a magnifying glass in order to produce an erect image which
is 3 times magnified. What is the distance between the object and the lens?
Answer:
f = 15cm
m = 3
Magnifying power (m) = \(\frac{-v}{\mathbf{u}}\) = \(\frac{f}{f-u}\)
3 = \(\frac{15}{15-u}\)
45 – 3u = 15
3u = 45 – 15
3u = 30
u = \(\frac{30}{3}\) = 10cm.

Question 6.
A compound microscope consists of an object lens of focal length 2cm and an eyepiece of focal length 5cm. When an object is placed at 2.2cm from the object lens, the final Image is at 25cm from the eye lens. What is the distance between the lenses ? What is the total linear magnification?
Answer:
Given that f₀ = 2, f_e = u₀ = 2.2, D = 25cm
\(\frac{1}{\mathrm{f}_0}\) = \(\frac{1}{\mathrm{u}_0}\) + \(\frac{1}{\mathrm{v}_0}\)
\(\frac{1}{\mathrm{v}_0}\) = \(\frac{1}{\mathrm{f}_0}\) – \(\frac{1}{\mathrm{u}_0}\) = \(\frac{1}{2}\) – \(\frac{1}{2.2}\) ⇒ \(\frac{1}{v_0}\) = \(\frac{2.2-2}{2 \times 2.2}\)
v₀ = 22 cm
For the eye-piece, the distance of the image V_e = 25cm
\(\frac{1}{f_e}\) = \(\frac{1}{\mathrm{u}_{\mathrm{e}}}\) – \(\frac{1}{v_e}\) (For virtual image)
\(\frac{1}{u_e}\) = \(\frac{1}{f_e}\) + \(\frac{1}{v_e}\) = \(\frac{1}{5}\) + \(\frac{1}{25}\)
⇒ \(\frac{1}{u_e}\) = \(\frac{25+5}{5 \times 25}\)
u_e = 4.166
i) The distance between the two lenses
L = v₀ + u_e
L = 22 + 4.166
L = 26.166

Question 7.
The distance between two point sources of light is 24cm. Where should you place a converging lens, of focal length 9 cm, so that the images of both sources are formed at the same point ?
Answer:
Distance between two point sources of light = 24cm Focal length (f) = 9cm Radius of curvature (R) = 2f
R = 2 × 9 = 18cm
∴ Converging lens is placed at 18 cm (or) Second position of converging
lens = 24 – 18 = 6cm.
∴ position of converging lens = 18 cm (or) 6cm.

Question 8.
Find two positions of an object, placed in front of a concave mirror of focal length 15cm, so that the image formed is 3 times the size of the object.
Answer:
f = 15cm
m = 3
i) m = \(\frac{-v}{u}\) = \(\frac{f}{f-u}\)
3 = \(\frac{15}{15-u}\)
45 – 3u = 15
3u = 30
u = 10 cm

ii) m = \(\frac{\mathrm{f}}{\mathrm{u}-\mathrm{f}}\)
3 = \(\frac{15}{\mathrm{u}-15}\)
3u – 45 = 15
3u = 60
u = 20

Question 9.
When using a concave mirror, the magnification is found to be 4 times as much when the object is 25cm from the mirror as it is with the object at 40cm from the mirror, the image being real in each case. What is the focal length of the mirror?
Answer:
Given that m = 4
u = 25cm
m = \(\frac{f}{u-f}\)
4 = \(\frac{f}{25-f}\)
100 – 4f = f
100 = 5f
f = \(\frac{100}{5}\) = 20cm.

Question 10.
The focal length of the objective and eyepiece of a compound microscope are 4cm and 6cm respectively. If an object is placed at a distance of 6cm from the objective, what is the magnification produced by the microscope ?
Answer:
Given that f₀ = -4cm, f_e = 6cm, u₀ = 6
\(\frac{1}{\mathrm{f}_0}\) = \(\frac{1}{\mathrm{v}_0}+\frac{1}{\mathrm{u}_0}\)
\(\frac{1}{v_0}\) = \(\frac{1}{\mathrm{f}_0}\) – \(\frac{1}{\mathrm{u}_0}\) = \(\frac{1}{4}-\frac{1}{6}\)
\(\frac{1}{v_0}\) = \(\frac{2}{24}\)
υ₀ = 12 cm.
Magnifying power m = \(\frac{\mathrm{v}_0}{\mathrm{u}_0}\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right)\)
= \(\frac{12}{6}\left[1+\frac{25}{6}\right]\) = \(2\left[\frac{31}{6}\right]\) = \(\frac{62}{6}\)
m = 10.33

Textual Exercises

Question 1.
A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved ?
Answer:
u = – 27cm, R = – 36cm, f = – 18cm
\(\frac{1}{\mathrm{u}}\) + \(\frac{1}{v}\) = \(\frac{1}{\mathrm{f}}\) ⇒ \(\frac{-1}{27}+\frac{1}{v}\) = \(\frac{-1}{18}\)
\(\frac{1}{v}\) = \(\frac{1}{27}-\frac{1}{18}\) ⇒ \(\frac{1}{\mathrm{v}}\) = \(\frac{-1}{54}\)
v = – 54 cm.
The Screen should be placed 54cm from the mirror
m = \(\frac{\mathrm{I}}{\mathrm{O}}\) = \(\frac{-\mathrm{V}}{\mathrm{u}}\) ⇒ \(\frac{\mathrm{I}}{2.5}\) = \(\frac{-54}{27}\)
U = – 5cm
∴ The image is real, inverted and magnified. If the candle is moved closer, the screen would have moved farther and farther. Closer than 18cm from the mirror, the image gets virtual and cannot be collected on the screen.

Question 2.
A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification.
Answer:
O = 4.5 cm, u = -12cm, f = 15cm

Image is virtual and erect and is formed behind the mirror.

As the needle is moved further from the mirror, the image moves towards the focus and gets progressively diminished in size.

Question 3.
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 2.4 cm. What is the refractive index of water ? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again ?
Answer:

Distance by which image becomes raised = 9.4 – 7.67 = 1.73 = 1.7cm the microscope will be moved up by 1.7cm to focus on the needle again.

Question 4.

The above 3 Figures (a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water- air figures interface respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with, the normal to a water-glass interface [Fig. (c)]
Answer:
1st Case:
Angle of incidence i = 60°
Angle of refraction r = 35°

3rd Case :
Angle of incidence i = 45°
Angle of refraction = ?
w_rg = \(\frac{\sin i}{\sin r}\) = \(\frac{\sin i}{\sin r}\) = 1.28
Sin r = \(\frac{\sin 45^{\circ}}{1.28}\) = 0.5525
sin r = sin 33°54′
r = 33°44′

Question 5.
A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out ? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Answer:
If r is the radius of the large circle from which light comes out, C is the critical angle for water – air interface, then
tan C = \(\frac{\mathrm{DB}}{\mathrm{DO}}\) = \(\frac{\mathbf{r}}{\mathbf{d}}\)
r = d tan C
Area of circle,
A = πr²
A = π(d tan C)²
A = πd².\(\frac{\sin ^2 C}{\cos ^2 C}\)
A = πd².\(\frac{\sin ^2 C}{1-\sin ^2 C}\)
But Sin C = \(\frac{1}{\mu}\) = \(\frac{1}{1.33}\) ≈ 0.75
A = \(\frac{\pi(0.8)^2(0.75)^2}{1-(0.75)^2}\) = 2.6m²

Question 6.
A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism ? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
Answer:

Question 7.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm ?
Answer:
\(\frac{\mu_2}{\mu_1}\) = μ = 1.55
R₁ = R, R₂ = -R
f = 20

Question 8.
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm ?
Answer:
Here the object is virtual and the image is real u = + 12 cm (object on right and virtual)
a) f = + 20cm
Lens formula is \(\frac{-1}{\mathrm{u}}\) + \(\frac{1}{v}[latex] = \)\frac{1}{f}[/latex]

i. e., u = 7.5 cm (image on right and real) It is located 7.5 cm from the lens,

b) f = – 16cm

Image will be located 48cm from the lens.

Question 9.
An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens ?
Answer:
‘O’ = 3.0cm
u = – 14cm, f = -21cm
\(\frac{1}{v}+\frac{1}{u}\) = \(\frac{1}{f}\) ⇒ \(\frac{1}{v}+\frac{1}{14}\) = \(\frac{-1}{21}\)
\(\frac{1}{v}\) = \(\frac{-1}{21}-\frac{1}{14}\)
υ = \(\frac{-42}{5}\) = – 8.4 cm
Image is erect, virtual and located 8.4cm from the lens on the same side as the object. Using the relation,
\(\frac{I}{O}\) = \(\frac{\mathrm{v}}{\mathrm{u}}\)
υ = \(\frac{8.4}{15}\) × 5 = 1.8cm

As the object is moved away from the lens, the virtual image moves towards the focus of the lens and progressively diminishes in size. (When u = 21 cm, v = -10.5 cm and when u = ∞, v = -21 cm)

Question 10.
What is the focal length of a convex lens of focal length 30cm in contact with a concave lens of focal length 20cm ? Is the system a converging or a diverging lens ? Ignore thickness of the lenses.
Answer:
Given f₁ = 30 cm, f₂ = -20 cm, f = ?
f = \(\frac{\mathrm{f}_1 \mathrm{f}_2}{\mathrm{f}_1+\mathrm{f}_2}\)
f = \(\frac{30 \times(-20)}{30-20}\) = -60cm
Thus the system is a diverging lens of focal length 60cm.

Question 11.
A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.2 5cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (2 5cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
Answer:
a) V_e = 25 cm
f_e = 6.25 cm.
Using lens formula,

Question 12.
A person with a normal near point (25cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 2.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
Answer:
Angular magnification of the eye piece for image at 25 cm

u = –\(\frac{25}{11}\) = -2.27 cm
∴ Distance between objective and eye piece
= v + |u| = 7.2 + 2.27 = 9.47cm
Magnifying power of microscope
= \(\frac{7.2}{0.9} \times \frac{-25}{\frac{-25}{11}}\) = 88

Question 13.
A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope ? What is the separation between the objective and the eyepiece?
Answer:
a) For normal adjustment
M.P of telescope = \(\frac{f_0}{f_e}\) = \(\frac{144}{6}\) = 24
b) The length of the telescope in normal adjustment
L = f₀ + f_e = 144 + 6
= 150 cm.

Question 14.
a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of
focal length 1.0 cm is used, what is the angular magnification of the telescope?
b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens ? The diameter of the moon is 3.48 × 10⁶m, and radius of lunar orbit is 3.8 × 10⁸m.
Answer:
a) Anglular magnification
= \(\frac{f_0}{f_e}\) = \(\frac{15}{0.01}\) = 1500
b) if d is the diameter of the image (in cm)
\(\frac{\mathrm{d}}{1500}\) = \(\frac{3.48 \times 10^6}{3.8 \times 10^8}\)
d = 13.7

Question 15.
Use the mirror equation to deduce that:
a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
b) a convex mirror always produces a virtual image independent of the location of the object.
c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

[Note : This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
Answer:

Since for concave mirror, f is negative, υ becomes negative.
It means image produced is real and beyond 2f.

b) For mirror formula,
υ = \(\frac{\mathbf{u f}}{\mathbf{u}-\mathbf{f}}\)
since for a convex mirror, fis positive and u is always negative, u will be always positive image and will always be formed behind the mirror and will be virtual.

c) For relation, m = \(\frac{v-f}{f}\) positive for convex mirror, m will always be negative and less than one. Hence virtual image formed will always be diminished.
For relation, m = \(\frac{v-f}{f}\) and m being negative, υ will always be less than f. Hence image will be formed between pole and focus.

d) When u > 0 < f we get m = \(\frac{f}{u-f}\) = \(\frac{f}{(>0<f)-f}\) = \(\frac{f}{(>-f<0)}\) = > -1
(∵ m is negative, image is virtual and enlarged because is numerically >1).

Question 16.
A small pin fixed on a table top is viewed from above from a distance of 50 cm. By What distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table ? Refractive index of glass = 1.5. Does the answer depend on the location of the slab ?
Answer:
μ = 1.5; real thickness = 15 cm

∴ Apparent Depth = \(\frac{15}{1.5}\) = 10 cm
∴ Pin appears raised by 15 – 10 = 5 cm.
The result is independent of the location of the slab.

Question 17.
a) Figure shows a cross-section of a light pipe made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.

b) What is the answer if there is no outer covering of the pipe ?

Answer:
a)
μ = \(\frac{1.68}{1.44}\) = \(\frac{1}{\sin C}\)
Sin C = \(\frac{1.44}{1.68}\) = 0.8571
C = 59°
Total internal reflection takes place when i > 59° or angle r may have value between 0 to 31°
r_max = 31°
Now \(\frac{\sin \mathrm{i}_{\max }}{\sin \mathrm{r}_{\max }}\) = 1.68
\(\frac{\sin i_{\max }}{\sin 31^{\circ}}\) = 1.68
sin i_max = 0.8562, i<sub.max = 60°
Thus all incident rays of angles in the range 0 < i < 60° will suffer total internal reflections in the pipe.

b) If there is no outer covering of the pipe
Sin C = \(\frac{1}{\mu}\)
= \(\frac{1}{1.68}\) = 0.5962
sin C = sin 36.5°
C = 36.5°

Question 18.
Answer the following questions :
a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce, real images under some circumstances ? Explain.
b) A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction ?
c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is ?
d) Does the apparent depth of a tank of water change if viewed obliquely ? If so, does the apparent depth increase or decrease ?
e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter ?
Answer:
a) Rays converging to a point (behind) a plane or convex mirror are reflected to a point infront of the mirror on the screen. In other words a plane or convex mirror can produce a real image if the object is virtual

b) When the reflected or refracted rays are divergent, the image is virtual. The divergent rays can be converged on to a screen by means of an appropriate converging lens. The convex lens of the eye does just that the virtual image here serves as a object for the lens to produce a real image. The screen here is not located at the position of the virtual image. There is no contradiction.

c) The rays starting from the head of the fisherman and incident on water become bent towards normal and appear to come from a higher point.
AF is real height of fisherman. Rays starting from A, bend towards normal. For diver they appear to come from A₁, A₁ F becomes apparent height of fisherman, which is more than real height.

d) The apparent depth for oblique viewing decreases from its value for near-normal viewing.

e) Refractive index of diamond is about 2.42, much larger than that of ordinary glass. The critical angle for diamond is above 24°, much less than that of glass. A skilled diamond cutter exploits the large range of angles of incidence (in the diamond) 24° to 90° to ensure that light entering the diamond is totally reflected from many faces before getting out thus producing a sparkling effect.

Question 19.
The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3m away by means of a large convex lens: What is the maximum possible focal length of the lens required for the purpose ?
Answer:
Let,
υ = + υ
∴ u = -(3 – υ)
f_max = ?
Now, \(\frac{1}{f}\) = \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{v}-\frac{1}{-(3-v)}\)
\(\frac{1}{f}\) = \(\frac{1}{v}+\frac{1}{3-v}\) ⇒ \(\frac{1}{f}\) = \(\frac{3-v+v}{(3-v) v}\)
3υ – υ² = 3f
For f to be maximum d(f) = 0
i.e d(3υ – υ²) = 0
3 – 2 υ = 0
υ = 3/2 = 1.5 m
Hence, u = -(3 – 1.5)
= -1.5m
and
\(\frac{1}{f}\) = \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{1.5}\) – \(\frac{1}{-1.5}\) = \(\frac{1+1}{1.5}\)
= \(\frac{2}{1.5}\) = 0.75m

Question 20.
A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
Answer:
a) Distance between object and image D = 90 cm = u + υ
Distance between two’ positions of lens d = 20 = u = υ
u = 55 cm and υ = 35 cm.
For lens formula,
\(\frac{1}{\mathrm{f}}\) = \(\frac{1}{55}\) + \(\frac{1}{35}\) = \(\frac{18}{385}\)
f = \(\frac{385}{18}\) = 21.4

Question 21.
a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 10, if they are placed 8.0cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident ? Is the notion of effective focal length of this system useful at all ?
b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40cm. Determine the magni-fication produced by the two-lens is 40cm.
Answer:
a) Here f₁ = 30cm, f₂ = -20cm,
d = 8.0cm, f = ?

i) Let a parallel beam be incident on the convex lens first. If 2nd lens were absent, then
u₁ = ∞ and f₁ = 30cm
As \(\frac{1}{v_1}-\frac{1}{\mathrm{u}_1}\) = \(\frac{1}{f_1}\)
\(\frac{1}{v_1}\) = \(\frac{1}{\infty}\) = \(\frac{1}{30}\)
υ₁ = 30 cm
The image would now act as a virtual object for 2nd lens.
υ₂ = + (30 – 8) = + 22 cm
υ₂ = ? f₂ = – 20 cm
As \(\frac{1}{v_2}\) = \(\frac{1}{\mathrm{f}_2}+\frac{1}{\mathrm{u}_2}\)
\(\frac{1}{v_2}\) = \(\frac{1}{-20}+\frac{1}{22}\) = \(\frac{-11+10}{220}\) = \(\frac{-1}{220}\)
υ₂ = -220 cm
∴ Parallel incident beam would appear to diverge from a point 220 – 4 = 216 cm from the centre of the two lens system,

ii) Suppose a parallel beam of light from the left is incident first on the concave lens.

This image acts as a real object for the 2nd lens.

∴ The parallel beam appears to diverge from a point 420 – 4 = 416 cm, on the left of the centre of the two lens system From the above discussion, we observe that the answer depends on which side of the lens system the parallel beam is incident. Therefore, the notion of effective focal length does not seem to be useful here.

b) Here, h₁ = 1.5 cm, u₁ = 40 cm, m = ?, h₂ = ? for the 1st lens, \(\frac{1}{v_1}-\frac{1}{u_1}\) = \(\frac{1}{f_1}\)
\(\frac{1}{v_1}\) = \(\frac{1}{p_1}+\frac{1}{u_1}\) = \(\frac{1}{30}-\frac{1}{40}\) = \(\frac{1}{120}\)
υ₁ = 120 cm
Magnitude of magnification produced by first lens,
m₁ = \(\frac{v_1}{u_1}=\frac{120}{40}\) = 3
The image formed by 1st lens as virtual object for the 2nd lens.
υ₂ = 120 – 8 = 112 cm, f₂ = – 20cm
υ₂ = ?

Magnitude of Magnification produced by second lens
m₂ = \(\frac{\mathrm{v}_2}{\mathrm{u}}\) = \(\frac{112 \times 20}{92 \times 112}\) = \(\frac{20}{92}\)
Net magnification produced by the combination
m = m₁ × m₂
= 3 × \(\frac{20}{92}\) = \(\frac{60}{92}\) = 0.652
∴ size of image h₂ = mh₁
= 0.652 × 1.5
= 0.98 cm

Question 22.
At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face ? The refractive index of the material of the prism is 1.524.
Answer:
i₁ = ?, A = 60°, μ = 1.524
μ = \(\frac{1}{\sin C}\)
C = r₂
sin C = sin r₂ = \(\frac{1}{\mu}=\frac{1}{1.524}\) = 0.6561
r₂ = 41°
As r₁ + r₂ = A
r₁ = A – r₂ = 60° – 41° = 19°
μ = \(\frac{\sin i_1}{\sin r_1}\)
sin i₁ = 1.524 sin 19°
= 1.524 × 0.3256
= 0.4962
i₁ = 29°45¹

Question 23.
You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will
a) deviate a pencil of white light without much dispersion.
b) disperse (and displace) a pencil of white light without much deviation.
Answer:
i) For no dispersion, angular dispersion produced by two prisms should be zero
i.e. (μ_b – μ) A + (μ_b – μ’_r) A’ = 0
As (μ’_b – μ’_r) for flint glass is more than that for grown glass, therefore A’ < A i.e ., flint glass prism of smaller angle has to be suitably combined with crown glass prism of larger angle.

ii) For almost no deviation (μ_v – 1) A + (μ’_y – 1) A’ = 0
Taking crown glass prism of certain angle, we go on-increasing angle of flint glass prism till this condition is met. In the final combination however, angle of flint glass prism will be smaller than the angle of crown glass prism as μ’_y for flint glass is more than μ_y for crown glass.

Question 24.
For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm in front of the eye. The cornea of the eye provides a converging power of about 40 diopters, and the least converging power of the eye- lens behind the cornea is about 20 diopters. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye. ”
Answer:
To observe objects at infinity, the eye uses its least converging power = 40 + 20 = 60D
∴ Distance between cornea eye lens and retina
focal length of eye lens = \(\frac{100}{p}\) = \(\frac{100}{60}\) = \(\frac{5}{3}\) cm
To focus an object at the near point u = -25cm, v = 5/3 cm f = ? –
\(\frac{1}{f}\) = \(-\frac{1}{\mathrm{u}}+\frac{1}{v}\) ⇒ \(\frac{1}{f}\) = \(\frac{1}{25}+\frac{3}{5}\) = \(\frac{1+15}{25}\)
= 16/25
f = 25/16 cm
Power = \(\frac{100}{f}\) = \(\frac{100}{25 / 16}\) = 64D
Power of eye lens = 64 – 40 = 24D
Hence range of accommodation of eye lens is roughly 20 to 24 dioptre.

Question 25.
Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?
Answer:
No, a person may have normal ability of accommodation and yet he may be myopic or hyper metropic. Infact, myopia arises when length of eye ball gets shortened.
However, when eye ball has normal length, but the eye lens loses partially its power of accommodation, the defect is called Presbiopia.

Question 26.
A myopic person has been using spectacles of power – 1.0 dioptre for distant vision. During old age he also needs to use separate reading glass of power + 2.0 dioptres. Explain whàt may have happened.
Answer:
Here u = -25cm, y = -50cm, f = ?
\(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{\mathrm{f}}\) ⇒ –\(\frac{1}{50}\) + \(\frac{1}{25}\) = \(\frac{1}{f}\)
\(\frac{-1+2}{50}\) = \(\frac{1}{\mathrm{f}}\) or f = 50 cm
As P = \(\frac{100}{f}\) = \(\frac{100}{50}\) = + 2 dioptres.

Question 27.
A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to ? How is such a defect of vision corrected ?
Answer:
This defect is called Astigmatism. It arises because curvature of cornea plus eye lens refracting system is-not the same in diffreht planes. As vertical lines are seen distinctly, the curvature in the vertical plane is enough, but in the horizontal , plane, curvature is insufficient.
This defect is removed by using a cylindrical lens with its axis along the vertical.

Question 28.
A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.
a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass ?
b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope ?
Answer:
a) Here, f = 5 cm, u = ?
For the closest distance; v = – 25cm
As \(\frac{1}{f}\) = \(\frac{1}{v}-\frac{1}{u}\)
\(\frac{1}{u}\) = \(\frac{1}{v}-\frac{1}{f}\) = \(\frac{1}{-25}-\frac{1}{5}\) = \(\frac{-1-5}{25}\)
u’ = \(\frac{25}{-6}\) – 4.2 cm
This is the closest distance at which he can read the book.
For the farthest distance, v’ = ∞, u’ = ?
As \(\frac{1}{\mathrm{f}}\) = \(\frac{1}{\mathrm{v}^{\prime}}-\frac{1}{\mathrm{u}^{\prime}}\)
\(\frac{1}{v^{\prime}}\) = \(\frac{1}{v^i}-\frac{1}{f}\) = \(\frac{1}{\infty}-\frac{1}{5}\) = \(\frac{-1}{5}\)
u’ = -5cm
This is the farthest distance at which he can real the book.

b) Max. Angular magnification
m = \(\frac{\mathrm{d}}{\mathrm{u}}\) = \(\frac{25}{25 / 6}\) = 6
Min. Angular magnification
m’ = \(\frac{\mathrm{d}}{\mathrm{u}^{\prime}}\) = \(\frac{25}{5}\) = 5

Question 29.
A card sheet divided into squares each of size 1 mm² is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9cm) held close to the eye.
a) What is the magnification in produced by the lens ? How much is the area of each square in the virtual image ?
b) What is the angular magnification (magnifying power) of the lens ?
c) Is the magnification in (a) equal to the magnifying power in (b) ? Explain.
Answer:
a) Here, area of each (object) square 1mm²,
u – 9cm, f = 10cm
\(\frac{1}{v}\) = \(\frac{1}{f}+\frac{1}{u}\) = \(\frac{1}{10} \frac{-1}{9}\) = \(\frac{-1}{90}\)
v = -90cm
Magnification, m = \(\frac{v}{|u|}=\frac{90}{9}=10\)
∴ Area of each square in virtual image = (10)² × 1 = 100 sq,mm

b) Magnifying power = \(\frac{\mathrm{d}}{\mathrm{u}}\) = 25/9 = 2.8
c) No, Magnification in (a) which is (υ/u) cannot be equal to magnifying power in (b) which is (dlu) unless v = d ie., image is located at the least distance of distinct vision.

Question 30.
a) At what distance should the lens be held from the figure in Exercise 2.29 in order to view the squares distinctly with the maximum possible magnifying power ?
b) What is the magnification in this case ?
c) Is the magnification equal to the magnifying power in this case ? Explain.
Answer:
i) Here, υ = -25cm, f = 10cm, u = ?

Yes, the magnification and magnify-ing power in this case are equal, because image is formed at the least distance of distinct vision.

Question 31.
What should be the distance between the object in Exercise 30 and the magnifying glass if the virtual image of each square in the figure is to have an , area of 6.25 mm². Would you be able to see the squares distinctly with your eyes very close to the magnifier ?
[Note : Exercises 29 to31 will help you clearly understand the difference between magnification in absolute size ‘ and the angular magnification (or magnifying power) of an instrument.]
Answer:
Here, magnification in area = 6.25
linear magnification m = \(\sqrt{6.25}\) = 2.5
As m = \(\frac{v}{u}\) or v = mu = 2.5u
As \(\frac{1}{v}\) – \(\frac{1}{\mathrm{u}}\) = \(\frac{1}{\mathrm{f}}\)
\(\frac{1}{2.5 \mathrm{u}}\) – \(\frac{1}{\mathrm{u}}\) = \(\frac{1}{10}\)
\(\frac{1-2.5}{2.5 u}\) = \(\frac{1}{10}\)
u = -6cm
y = 2.5u = 2.5 (-6) = -15cm
as the virtual image is at 15cm; where as distance of distinct vision is 25cm, therefore, the image cannot be seen distinctly by the eye.

Question 32.
Answer the following questions :
a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification ?
b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back ?
c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power ?
d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths ?
e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why ? How much should be that short distance between the eye and eyepiece ?
Answer:
a) It is true that angular size of image is equal to the angular size of the object. By using magnifying glass, we keep the object far more closer to the eye than at 25cm, its normal position without use of glass. The closer object has larger angular size than the same object at 25cm. It is in this sense that angular magnification is achieved.

b) yes, the angluar magnification changes, if the eye is moved back. This is because angle subtended at the eye would be slightly less than the angle subtended at the lens. The effect is negligible when image is at much larger distance.

c) Theoretically, it is true, However, when we decrease focal length, observations both spherical and chromatic become more pronounced. Further, it is difficult to grind lenses of very small focal lengths.

d) Angular magnification of eye piece is \(\left(1+\frac{d}{f_e}\right)\). This increase as f_e decreases. Further, magnification if objective lens is \(\frac{v}{u}\). As object lies close to focus of objective lens u ≈ f₀. To increase this magnification (υ/f₀), f₀ should be smaller.

e) The image of objective lens in eye pie is called ‘eye ring’ All the rays from the object refracted by the objective go through the eye ring. Therefore, ideal position for our eyes for viewing is this eye ring only.

When eye is too close to the eye piece, field of view reduces and eyes do not collect much of the light. The precise, location of the eye ring would depend upon the separation between the objective and eye piece, and also on focal length of the eye piece.

Question 33.
An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5cm. How will you set up the compound microscope ?
Answer:
In normal ajustment, image is formed at least distance of distinet vision, d = 25cm Angular Magnification of eye piece = \(\left(1+\frac{d}{f_e}\right)=\left(1+\frac{25}{5}\right)\) = 6 As total Magnification is 30, Magnification of objective lens.

i.e. object should be held at 1.5cm in-front of objective lens

∴ Seperation between the objective lens and eye piece
= |u_e| + |v₀|
= 4.17 + 7.5.
= 11.67cm

Question 34.
A small telescope has an objective lens of focal length 140cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
a) the telescope is in normal adjustment (i.e., when the final image is at infinity) ?
b) the final image is formed at the least distance of distinct vision (25 cm) ?
Answer:
Here, f₀ = 140cm, f_e = 5.0cm
Magnifying power = ?
a) In normal adjustment,
Magnifying power
= \(\frac{f_0}{-f_e}\) = \(\frac{140}{-5}\) = -28

b) When final image is at the least distance of distinct vision, Magnifying power
= \(\frac{-\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\left(1+\frac{\mathrm{f}_{\mathrm{e}}}{\mathrm{d}}\right)\) = \(\frac{-140}{5}\left(1+\frac{5}{25}\right)\) = -33.6

Question 35.
a) For the telescope described in Exercise 34 (a), what is the separation between the objective lens and the eyepiece ?
b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens ?
c) What is the height of the final image of the tower if it is formed at 25 cm ?
Answer:
a) Here, in normal adjustment, seperation between objective lens and eye lens = f₀ + f_e = 140 + 5 = 145cm

b) Angle subtended by tower 100m tall at 3km
α = \(\frac{100}{3 \times 1000}\) = \(\frac{1}{30}\) radian
if his the height of image formed by the objective, then
α = \(\frac{\mathrm{h}}{\mathrm{f}_0}\) = \(\frac{h}{140}\)
∴ \(\frac{\mathrm{h}}{140}\) = \(\frac{\mathrm{1}}{30}\)
h = \(\frac{\mathrm{140}}{30}\) = 4.7 cm

c) Magnifying produced by eye piece
= (1 + \(\frac{d}{f e}\)) = 1 + \(\frac{25}{5}\) = 6
∴ Height of final image = 4.7 × 6 = 28.2cm

Question 36.
A cassegrain telescope is built with the mirrors 20mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140mm. where will the final image of an object at infinity be ?
Answer:
Here, radius of curvature of objective mirror R₁ = 220 mm radius of curvature of secondary mirror R₂ = 140mm;
f₂ = \(\frac{\mathrm{R}_2}{2}=\frac{140}{2}\) = 70mm
Distance between the two mirrors, d = 20mm. When object is at infinity, parallel rays falling on objective mirror, on reflection, would collect at its focus at
f₁ = \(\frac{\mathrm{R}_1}{2}=\frac{220}{2}\) = 110mm
Instead, they fall on secondary mirror at 20mm from objective mirror.
∴ For secondary mirror, u = f₁ – d = 110 – 20 = 90 mm
From \(\frac{1}{v}+\frac{1}{u}\) = \(\frac{1}{\mathrm{f}_2}\)
υ = \(\frac{1}{\mathrm{f}_2} \frac{-1}{\mathrm{u}}\) = \(\frac{1}{70}-\frac{1}{90}\) = \(\frac{9-7}{630}\) = \(\frac{2}{630}\)
υ = \(\frac{630}{2}\) = 315 mm = 31.5 cm to the right of secondary mirror

Question 37.
Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

Answer:
Here, θ = 3.5°
x = 1.5m, d = ?
When the mirror turns through an angle θ, the reflected ray turns through double the angle.
2θ = 2 × 3.5° = 7° = \(\frac{7 \pi}{180}\) rad
from figure,
tan 2θ = \(\frac{\text { SA }}{\mathrm{OS}}\) = \(\frac{\mathrm{d}}{1.5} \times \mathrm{d}\)
= 1.5 × \(\frac{7 \pi}{180} \mathrm{~m}\) = 0.18m
d = 1.5 tan 2θ
≈ 1.5(2θ)
= 1.5 × \(\frac{7 \pi}{180} \mathrm{~m}\) = 0.18m

Question 38.
Figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A
small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid?

Answer:
Let focal length of convex lens of glass = f₁ = 30cm focal length of plano concave lens of liquid = f₂ combined focal length, F = 45.0cm

For liquid lens

Students get through AP Inter 2nd Year Physics Important Questions 9th Lesson Electromagnetic Induction which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 9th Lesson Electromagnetic Induction

Very Short Answer Questions

Question 1.
What did the experiments of Faraday and Henry show ?
Answer:
The discovery and understanding of electromagnetic induction are based on a long series of experiments carried out by Faraday and Henry.

Question 2.
Define magnetic flux.
Answer:
Magnetic flux is defined as the number of magnetic lines of force crossing through the surface.
Φ_B – \(\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}\) = BA cos θ
C.G.S unit → Maxwell .
S.I. unit → Weber (wb)
Magnetic flux is a scalar.

Question 3.
State Faraday’s law of electromagnetic induction.
Answer:
“Magnitude of induced e.m.f is directly proportional to the rate of change of magnetic flux”
ε ∝ \(-\frac{\mathrm{d} \phi}{\mathrm{dt}}\)

Question 4.
State Lenz’s law.
Answer:
The direction of induced e.m.f (or) current is such that it opposes the cause which produces it. Lenz’s law is in accordance with law of conservation of energy.

Question 5.
What happens to the mechanical energy (of motion) when a conductor is moved in a uniform magnetic field ?
Answer:
Motional e.m.f is produced to the motion of the conductor in a magnetic field.
Motion e.m.f (ε) B/υ.

Question 6.
What are Eddy currents ? [A.P. Mar. 15]
Answer:
Eddy currents (or) Focault currents : The induced circulating currents produced in a conductor itself due to change in magnetic flux linked with the conductor are called Eddy currents.
Due to Eddy currents, the energy is dissipated in the form of heat energy.

Question 7.
Define ‘inductance’.
Answer:
Inductance is a coefficient of electromagnetic induction and is an intrinsic property of a material just like capacitance.
Inductance is an important scalar quantity which depends upon the geometry (i.e, dimensions) of a coil.

Question 8.
What do you understand by ‘self inductance’ ?
Answer:
Self inductance of a coil is defined as the induced e.m.f produced in the coil through which the rate of change of current is unity.
ε = -L \(\frac{\mathrm{dI}}{\mathrm{dt}}\); ε = -L, If \(\frac{\mathrm{dI}}{\mathrm{dt}}\) = 1 A/s.

Short Answer Questions

Question 1.
Obtain an expression for the emf induced across a conductor which is moved in a uniform magnetic field which is perpendicular to the plane of motion.
Answer:
Consider a conductor PQ of length l moving freely in a uniform magnetic field \(\overrightarrow{\mathrm{B}}\) with uniform veiority υ on a rectangular conductor ABCD. Let any arbitrary charge q in the conductor also move in the field with same velocity.

Magnitude of Lorentz force on this charge
(F) = Bqυ ……………….. (1)
Workdone in moving the charge from P to Q is given by ‘
W = Force × displacement
W = Bqυ × l ………………. (2) (∵ Direction of force on the charge as per Flemings left hand rule)
Electromagnetic force (ε) = \(\frac{W}{Q}\)
ε = \(\frac{\mathrm{Bqυl}}{\mathrm{q}}\) ⇒ ε = Blυ ……………. (3)

Question 2.
Describe the ways in which Eddy currents are used to advantage. [A.P. Mar. 17, 16; A.P. & T.S. Mar. 15]
Answer:
Eddy currents are used to advantage in
i) Magnetic braking in trains : A strong magnetic field is applied across the metallic drum rotating with the axle of the electric train. Thus large eddy currents are produced in the metallic drum. These currents oppose the motion of the drum and hence the axle of the train which ultimately makes the train come to rest.

ii) Induction Motor: Eddy currents are used to rotate the short circuited motor of an induction motor. Ceiling fans are also induction motors which run on single phase alternating current.

iii) Electromagnetic damping : Certain galvanometers have a fixed core made of non magnetic metallic material. When the coil oscillates, the eddy currents generated in the core oppose the motion and bring the coil to rest quickly.

iv) Induction furnace : Induction furnace can be used to produce high temperatures and can be utilised to prepare alloys, by melting the constituent metals. A high frequency alternating current is passed through a coil. The eddy currents generated in the metals produce high temperatures sufficient to melt it.

v) Analogue energy meters : Concept of eddy currents is used in energy meters to record the consumption of electricity. Aluminium disc used in these meters get induced due to varying magnetic field. It rotates due to eddy currents produced in it.

Question 3.
Obtain an expression for the mutual inductance of two long co-axial solenoids.
Answer:
Consider two solenoids as shown in figure. The length of primary coil be l and area of cross section A. Let N₁ and N₂ are the total number of turns in the primary and secondary solenoids. Let n₁ and n₂ be the number of turns per unit length
(n₁ = \(\frac{\mathrm{N}_1}{l}\) and n₂ = \(\frac{\mathrm{N}_2}{l}\)). Current in the primary coil is i.

∴ Magnetic field inside the primary (B) = μ₀n₁ I = μ₀ \(\frac{\mathrm{N}_1}{l}\) I ……….. (1)
Magnetic flux through each turn of primary
Φ_B = \(\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}\) = μ₀ \(\frac{\mathrm{N}_1}{l}\) I × A ……………. (2)
The same magnetic flux is linked with the secondary coil.
∴ Total magnetic flux linked with secondary = μ₀\(\frac{\mathrm{N}_1 \mathrm{i}}{l}\) × A × N₂ ……………. (3)
If M be mutual inductance of the two coils, the total flux linked with the secondary is Mi.
∴ Mi = \(\frac{\mu_0 \mathrm{~N}_1 \mathrm{~N}_2 \mathrm{iA}}{l}\) ……………… (4)
M = \(\frac{\mu_0 \mathrm{~N}_1 \mathrm{~N}_2 \mathrm{~A}}{l}\) …………….. (5) (∵ A = πr²)
(or) M = \(\frac{\mu_0 \mathrm{~N}_1 \mathrm{~N}_2\left(\pi \mathrm{r}^2\right)}{l}\) ………….. (6) (∵μ_r = \(\frac{\mu}{\mu_0}\))

Problems

Question 1.
A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth’s magnetic field H_E at a place. If H_E = 0.4 G at the place, what is the induced emf between the axle and the rim of the wheel ? (Note that 1 G = 10^-4 T.)
Solution:
Induced emf = (1/2) ω B R²
= (1/2) × 4π × 0.4 × 10^-4 × (0.5)²
= 6.28 × 10^-5 V

Question 2.
Number of turns in a coil are 100. When a current of 5A is flowing through the coil, the magnetic flux is 10^-6Wb. Find the self induction. [Board Model Paper]
Solution:
Self inductance, L = \(\frac{n \phi}{\mathrm{i}}\)
number of turns, n = 100; magnetic flux, Φ = 10^-6Wb; Current, i = 5A
L = \(\frac{100 \times 10^{-6}}{5}\) = 20 × 10^-6 = 20 µH
∴ Self inductance, L = 20 µH

Question 3.
Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced, give an estimate of the self-inductance of the circuit. [Mar. 16 (T.S.) Mar. 14]
Solution:
Change in current, dI = 5 – 0 = 5A,
Time taken in current change dt = 0.1 s
Induced average emf e_av = 200 V
Induced emf in the circuit e = L . \(\frac{\mathrm{dI}}{\mathrm{dt}}\) ⇒ 200 = L\(\left(\frac{5}{0.1}\right)\) or L = \(\frac{200}{50}\) = 4 H.

Question 4.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil ? [T.S. Mar. 17]
Solution:
Given, mutual inductance of coil M = 1.5 H
Current change in coil dI = 20 – 0 = 20 A
Time taken in change dt = 0.5s, Induced emf in the coil e M = \(M \frac{d I}{d t}=\frac{d \phi}{d t}\)
dΦ = M.dI = 1.5 × 20, dΦ = 30 Wb,
Thus the change of flux linkage is 30 Wb.

Question 5.
A jet plane is travelling towards west at a speed of 1800 km/K What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10^-4 T and the dip angle is 30°.
Solution:
Speed of jet plane V = 1800 km/h = 1800 × \(\frac{5}{18}\) = 500 m/s
l = Distance between the ends of wings = 25 m
The magnitude of magnetic field B = 5 × 10^-4 T
Angle of dip γ = 30°.
Use the formula of motional emf
e = B_vVl, e = B sin γ Vl (B_v = B sin γ),
e = 5 × 10^-4 × sin 30° × 500 × 25, e = 3.1V
Thus, the voltage difference developed between the ends is 3.1 V.

Textual Examples

Question 1.
(a) What would you do to obtain a large deflection of the galvanometer ?
(b) How would you demonstrate the presence of an induced current in the absence of a gal-vanometer ?
Solution:
a) To obtain a large deflection, one or more of the following steps can be taken :

1. Use a rod made of soft iron inside the coil C₂.
2. Connect the coil to a powerful battery, and
3. Move the arrangement rapidly towards the test coil C₁.

b) Replace the galvanometer by a small bulb, the kind one finds in a small torch light. The relative motion between the two coils will cause the bulb to glow and thus demonstrate the presence of an induced current.

Question 2.
A square loop of side 10 cm and resistance 0.5 Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Determine the magnitudes of induced emf and current during this time interval.
Solution:
The angle θ made by the area vector of the loop with the magnetic field is 45°.
From eq. Φ_B = B.A. = BA cosθ the initial magnetic flux is Φ = BA cosθ
= \(\frac{0.1 \times 10^{-2}}{\sqrt{2}}\) Wb
Final flux, Φ_min = 0
The change in flux is brought about in 0.70 s. From Eq. ε = \(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\), the magnitude of the induced emf is given by
ε = \(\frac{\left|\Delta \phi_{\mathrm{B}}\right|}{\Delta \mathrm{t}}=\frac{|(\phi-0)|}{\Delta \mathrm{t}}=\frac{10^{-3}}{\sqrt{2} \times 0.7}\) = 1.0 mV
And the magnitude of the current is
I = \(\frac{\varepsilon}{\mathrm{R}}=\frac{10^{-3} \cdot \mathrm{V}}{0.5 \Omega}\) = 2 mA.
Note that the earth’s magnetic field also produces a flux through the loop.

Question 3.
A circular coil of radius 10 cm, 1500 turns and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. Estimate the magnitudes of the emf and current induced in the coil. .Horizontal component of the earth’s magnetic field at the place is 3.0 × 10^-5 T.
Solution:
Initial flux through the coil,
Φ_B(initial) = BA C0S θ
= 3.0 × 10^-5 × (π × 10^-2) × cos 0°
= 3π × 10^-7 Wb.
Final flux after the rotation,
Φ_B(final) = 3.0 × 10^-5 × (π × 10^-2) × cos 180°
= -3π × 10^-7 Wb.
Therefore, estimated value of the induced emf is,
ε = N\(\frac{\Delta \phi}{\Delta t}\) = 500 × (6π × 10^-7)/0.25
= 3.8 × 10^-3 V
I = ε/R = 1.9. × 10^-3 A.
Note that the magnitudes of ε and I are the estimated values.

Question 4.
The following figure shows planar loops of different shapes moving out of or into a region of a magnetic field which is directed normal to the plane of the loop away from the reader. Determine the direction of induced current in each loop using Lenz’s law.

Solution:

1. The magnetic flux through the rectangular loop abed increases, due to the motion of the loop into the region of magnetic field. The induced current must flow along the path bcdab so that it opposes the increasing flux.
2. Due to the outward motion, magnetic flux through the triangular loop abc decreases due to which the induced current flows along bacd, so as to oppose the change in flux.
3. As the magnetic flux decreases due to motion of the irregular shaped loop abed out of the region of magnetic field, the induced current flows along edabe, so as to oppose change in flux.

Question 5.
a) A closed loop is held stationary in the magnetic field between the north and south poles of two permanent magnets held fixed. Can we hope to generate current in the loop by using very strong magnets ?
b) A closed loop moves normal to the constant electric field between the plates of a large capacitor. Is a current induced in the loop (i) when it is wholly inside the region between the capacitor plates (ii) When it is partially outside the plates of the capacitor ? The electric field is normal to the plane of the loop.
c) A rectangular loop and a circular loop are moving out of a uniform magnetic field region as in the figure, to a field-free region with a constant velocity v. In which loop do you expect the induced emf to be constant during the passage out of the field region ? The field is normal to the loops.

d) Predict the polarity of the capacitor in the situation described by the following figure.

Solution:
a) No. However strong the magnet may be, current can be induced only by changing the magnetic flux through the loop.

b) No current is induced in either case. Current can not be induced by changing the electric flux.

c) The induced emf is expected to be constant only in the case of the rectangular loop. In the case of circular loop, the rate of change of area of the loop during its passage out of the field region is not constant, hence induced emf will vary accordingly.

d) The polarity of plate ‘A’ will be positive with respect to plate ‘B’ in the capacitor.

Question 6.
A metallic rod of 1 m length is rotated with a frequency of 50 rev/s, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular to the plane of the ring as in figure. A constant and uniform magnetic field of 1 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring ?

Solution:
Method I : As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached.
Using equation (ε = – Bl \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = BlV), the magnitude of the emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by
dε = Bυ dr. Hence,
ε = \(\int \mathrm{d} \varepsilon=\int_0^{\mathrm{R}} \mathrm{B} v \mathrm{dr}=\int_0^{\mathrm{R}} \mathrm{B} \omega \mathrm{rdr}=\frac{\mathrm{B} \omega \mathrm{R}^2}{2}\)
Note that we have used υ = ωr. This gives
ε = \(\frac{1}{2}\) × 1.0 × 2π × 50 × (1²) = 157 V.

Method II: To calculate the emf, we can imagine a closed loop OPQ in which point O and P are connected with a resistor R and OQ is the rotating rod. The potential difference across the resistor is then equal to the induced emf and equals B × (rate of change of area of loop). If θ is the angle between the rod and the radius of the circle at P at time t, the area of the sector OPQ is given by
πR² × \(\frac{\theta}{2 \pi}=\frac{1}{2}\), R²θ
where R is the radius of the circle. Hence, the induced emf is
e = B × \(\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{1}{2} \mathrm{R}^2 \theta\right]=\frac{1}{2} \mathrm{BR}^2 \frac{\mathrm{d} \theta}{\mathrm{dt}}=\frac{\mathrm{B} \omega \mathrm{R}^2}{2}\)
[Note: \(\frac{\mathrm{d} \theta}{\mathrm{dt}}\) = ω = 2πv]
This expression is identical to the – expression obtained by Method I and we get the same value of ε.

Question 7.
A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth’s magnetic field H_E at a place. If H_E = 0.4 G at the place, what is the induced emf between the axle and the rim of the wheel ? Note that 1 G = 10^-4 T.
Solution:
Induced emf = (1/2) ω B R²
= (1/2) × 4π × 0.4 × 10^-4 × (0.5)²
= 6.28 × 10^-5 V

Question 8.
Refer to fig. The arm PQ of the rectangular conductor is moved from x = 0, outwards. The uniform magnetic field is perpendicular to the plane and extends from x = 0 to x = b and is zero for x > b. Only the arm PQ possesses substantial resistance r. Consider the situation when the arm PQ is pulled outwards from x = 0 to x = 2b, and is then moved back to x = 0 with constant speed o. Obtain expressions for the flux, the induced emf, the force necessary to pull the arm and the power dissipated as Joule heat. Sketch the variation of these quantities with distance.

Solution:
Let us first consider the forward motion from x = 0 to x = 2b. The flux Φ_B linked with the circuit SPQR is
Φ_B = Blx 0 ≤ x < b
= Blυ b ≤ x < 2b
The induced emf is,
ε = – \(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\)
= -Blυ 0 ≤ x < b
= 0 b ≤ x < 2b.
When the induced emf is non-zero, the current I is (in magnitude)
I = \(\frac{\mathrm{B} l v}{\mathrm{r}}\)

The force required to keep the arm PQ in constant motion is I lB. Its direction is to the left. In magnitude
F = \(\frac{\mathrm{B}^2 l^2 v}{\mathrm{r}}\) 0 ≤ x < b
= 0 b ≤ x < 2b
The Joule heating loss is
P_j = I²r
= \(\frac{\mathrm{B}^2 l^2 v^2}{\mathrm{r}}\)   0 ≤ x < b
= 0   b ≤ x < 2b
One obtains similar expressions for the inward motion from x = 2b to x = 0.

Question 9.
Two concentric circular coils, one of small radius r₁ and the other of large radius r₂, such that r₁ << r₂, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement.
Solution:
Let a current I₂ flow through the outer circular coil. The field at the centre of the coil is B₂ = μ₀I₂/2r₂. Since the other co-axially placed coil has a very small radius. B₂ may be considered constant over its cross-sectional area. Hence,
Φ₂ = \(\pi \mathrm{r}_1^2 \mathrm{~B}_2\)
= \(\frac{\mu_0 \pi \mathrm{r}_1^2}{2 \mathrm{r}_2} \mathrm{I}_2\)
= M₁₂I₂
Thus,
M₁₂ = \(\frac{\mu_0 \pi r_1^2}{2 r_2}\)
From Equation M₁₂ = M₂₁ = M
M₁₂ = M₂₁ = \(\frac{\mu_0 \pi r_1^2}{2 r_2}\)
Note that we calculated M₁₂ from an approximate value of Φ₁ assuming the magnetic field B₂ to be uniform over the area \(\pi r_1^2\). However, we can accept this value because r₁ << r₂.

Question 10.
(a) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid.
(b) How does this magnetic energy compare with the electrostatic energy stored in a capacitor ?
Solution:
a) From Equation ε = – L \(\frac{\mathrm{dI}}{\mathrm{dT}}\),
the magnetic energy is
U_B = \(\frac{1}{2}\) LI²
= \(\frac{1}{2} \mathrm{~L}\left(\frac{\mathrm{B}}{\mu_0 \mathrm{n}}\right)^2\)
(since B = μ₀nI, for a solenoid)
= \(\frac{1}{2}\) (μ₀n²Al) \(\left(\frac{\mathrm{B}}{\mu_0 \mathrm{n}}\right)^2\)
[from Equation L = μ₀n²Al]
= \(\frac{1}{2 \mu_0}\) B²Al

b) The magnetic energy per unit volume is,
u_B = \(\frac{\mathrm{U}_{\mathrm{B}}}{\mathrm{V}}\)
[where Vis volume that contains flux]
= \(\frac{\mathrm{U}_{\mathrm{B}}}{\mathrm{Al}}=\frac{\mathrm{B}}{2 \mu_0}\) ……………….. (1)
We have already obtained the relation for the electrostatic energy stored per unit volume in a parallel plate capacitor.
u_E = \(\frac{1}{2}\) ε₀E² ………………. (2)
In both the cases energy is proportional to the square of the field strength.

Question 11.
Kamla peddles a stationary bicycle, the pedals of the bicycle are attached to a 100 turn coil of area 0.10 m². The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil ?
Solution:
Here f = 0.5 Hz; N = 100, A = 0.1 m² and B = 0.01 T.
Employing Equation ε = NBA ω sin ωt
ε₀ = NBA (2πv)
= 100 × 0.01 × 0.1 × 2 × 3.14 × 0.5
= 0.314 V
The maximum voltage is 0.314 V