AP 10th Class Physics 9th Lesson Questions and Answers Light Reflection and Refraction

AP State Board new syllabus AP Board Solutions Class 10 Physics 9th Lesson Light Reflection and Refraction Questions and Answers.

10th Class Physics 9th Lesson Light Reflection and Refraction Questions and Answers

10th Class Physics 9th Lesson Questions and Answers (Exercise)

Question 1.
Which one of the following materials cannot be used to make a lens?
a) Water
b) Glass
c) Plastic
d) Clay
Answer:
d) Clay

Question 2.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object ?
a) Between the principal focus and the centre of curvature
b) At the centre of curvature
c) Beyond the centre of curvature
d) Between the pole of the mirror and its principal focus
Answer:
d) Between the pole of the mirror and its principal focus.

Question 3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object ?
a) At the principal focus of the lens
b) At twice the focal length
c) At infinity
d) Between the optical centre of the lens and its principal focus,
Answer:
b) At twice the focal length

Question 4.
A spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and the lens are likely to be
a) both concave.
b) both convex.
c) the mirror is concave and the lens is convex.
d) the mirror is convex, but the lens is concave.
Answer:
a) both concave.

Question 5.
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
a) only plane.
b) only concave.
c) only convex.
d) either plane or convex.
Answer:
c) only convex.

Question 6.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
a) A convex lens of focal length 50 cm.
b) A concave lens of focal length 50 cm.
c) A convex lens of focal length 5 cm.
d) A concave lens of focal length 5 cm.
Answer:
c) A convex lens of focal length 5 cm.

Question 7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror ? What is the nature of the image ? Is the image larger or smaller than the object ? Draw a ray diagram to show the image formation in this case.
Answer:
Focal length of concave mirror, f = -15 cm.

Range of distance of the object from the mirror : For getting an erect image using a concave mirror, the object should be placed at a distance less than the focal length.
i. e., 15 cm from the pole.
Nature of the image : Image will be virtual, erect and the image is larger than the object (magnified).

Question 8.
Name the type of mirror used in the following situations,
(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace. Support your answer with reason.
Answer:
a) A concave mirror, this allows a powerful parallel beam of light.
b) A convex mirror, this allows a greater field of view.
c) A concave mirror, this allows concentration of light at the focus of the mirror.

Question 9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object ? Verify your answer experimentally. Explain your observations.
Answer:
Every part of a lens forms an image. Therefore, if the lower half of the lens is covered it will still form a complete image. However, the intensity of the image will be reduced. This can be verified experimentally by observing the image of a distance object like tree on a screen, when lower half of the lens is covered with a black paper.

Question 10.
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Answer:
Given that u = -25 cm, f = 10 cm and h_O = 5 cm using lens formula, we have

By lens formula, we have = \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) ⇒ \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\) = \(\frac{1}{10}\) + \(\frac{1}{-25}\) = \(\frac{3}{50}\)
Therefore, v = \(\frac{50}{3}\) = 16.67 cm
Also, magnification m = \(\frac{h^{\prime}}{\mathrm{h}}\) = \(\frac{\mathrm{v}}{\mathrm{u}}\)
Therefore, h’ = h × \(\frac{\mathrm{v}}{\mathrm{u}}\) = 5 × \(\frac{16.67}{-25}\) = -3.33 cm
Thus, the image is real and inverted and is formed at a distance of 16.67 cm on the other side of the lens.

Question 11.
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far
is the object placed from the lens ? Draw the ray diagram.
Answer:
Given, f = -15 cm, v = -10 cm, u = ?
By lens formula, we have = \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\(\frac{1}{u}=\frac{1}{v}-\frac{1}{f}\) = \(\frac{1}{-10}\) – \(\frac{1}{-15}\)
Therefore, u = \(\frac{15 \times 10}{10-15}\) = – 30 cm
The object is placed at a distance of 30 cm from the lens. The ray diagram is as shown.

Question 12.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
Given u = -10 cm, f = +15 cm, v = ?
Using the mirror formula \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}\)
We have \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\) = \(\frac{1}{15}\) – \(\frac{1}{-10}\) = \(\frac{1}{15}\) + \(\frac{1}{10}\)
Therefore, v = \(\frac{15 \times 10}{15+10}\) = \(\frac{150}{25}\) = 6 cm
The image is formed 6 cm behind the mirror. Thus, the image is virtual, erect and smaller in size than the object.

Question 13.
The magnification produced by a plane mirror is +1. What does this mean ?
Answer:
It means that the size of the image is equal to the size of the object. The positive sign ‘ indicates the image is virtual and erect.

Question 14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Given h = 5 cm, u = -20 cm, R = +30 cm or f = \(\frac{\mathrm{R}}{2}\) = + \(\frac{30}{2}\) = 15 cm, v = ?, h’ = ?
Using the mirror formula = \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}\)
We have \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\) = \(\frac{1}{15}\) – \(\frac{1}{-20}\) = \(\frac{1}{15}\) + \(\frac{1}{20}\)
Therefore, v = \(\frac{15 \times 20}{15+20}\) = \(\frac{300}{35}\) = 8.6 cm
The image is formed 8.6 cm behind the mirror. Thus, the image is virtual and erect.
Now, m = \(\frac{\mathrm{h}^{\prime}}{\mathrm{h}}\) = \(\frac{\mathrm{v}}{\mathrm{u}}\)
Therefore, we have h’ = \(\frac{\mathrm{vh}}{\mathrm{u}}\) = \(\frac{8.6 \times 5}{-20}\) = 2.15 cm
Thus, the size of the image is 2.15 cm or 2.2 cm. The image is reduced.

Question 15.
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained ? Find the size and the nature of the image.
Answer:
For concave mirror, h₀ = 7.0 cm, u = -27 cm, f = -18 cm
Using mirror formula \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\) we get
\(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\) = \(\frac{1}{-18}\) – \(\frac{1}{-27}\) = \(\frac{-1}{18}\) + \(\frac{1}{27}\) = \(\frac{- 3 + 2}{54}\) = \(\frac{-1}{54}\) ∴ v = – 54 cm
Hence the screen should be placed at a distance of 54 cm infront of the concave mirror to get the sharp focussed image of the object on it.
Magnification of spherical mirror is given by m = \(\frac{h_i}{h_o}\) = –\(\frac{\mathrm{v}}{\mathrm{u}}\)
⇒ h_i = – h_o \(\frac{\mathrm{v}}{\mathrm{u}}\) = -7 × \(\frac{-54}{-27}\) = -14 cm
∴ The height of image is 14 cm
Since h_i > h_o the image is enlarged
Since ‘v’ is -ve, the image is real and inverted.

Question 16.
Find the focal length of a lens of power – 2.0 D. What type of lens is this ?
Answer:
Given P = -2D
We know that P = \(\frac{1}{\mathrm{f}(\mathrm{~m})}\)
Therefore, f = \(\frac{1}{\mathrm{P}}\) = \(\frac{1}{-2}\) = -0.5 m
Since the power of the lens is negative, therefore, is must be a concave lens.

Question 17.
A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging ?
Answer:
Given P = + 1.5 D, f = ?
Using the relation P = \(\frac{1}{\mathrm{f}}\) or f = \(\frac{1}{\mathrm{P}}\) = \(\frac{1}{1.5}\) = \(\frac{100}{15}\) = 6.67 cm
The prescribed lens is converging or convex as its power is positive.

10th Class PS 9th Lesson Questions and Answers (InText)

Page No. 182

Question 1.
Define the principal focus of a concave mirror.
Answer:
It is a point on the principal axis where a beam of light coming parallel to the principal axis after reflection from the mirror, the rays of light converge.

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length ?
Answer:
We know f = \(\frac{\mathrm{R}}{2}\)
Therefore, f = \(\frac{20}{2}\) = 10 cm

Question 3.
Name a mirror that can give an erect and enlarged image of an object.
Answer:
Concave mirror.

Question 4.
Why do we prefer a convex mirror as a rear-view mirror in vehicles ?
Answer:
Convex mirror is used because

1. It always produces a virtual and erect image.
2. The size of image formed is smaller than the object. Therefore, it enables the driver to see a wide field view of the traffic behind the vehicle in a small mirror.

Page No. 186

Question 5.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
We know that f = \(\frac{\mathrm{R}}{2}\)
Therefore, f = \(\frac{32 \mathrm{~cm}}{2}\) = 16 cm

Question 6.
A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located ?
Answer:
Given, m = -3 (since image is real), u = -10 cm, v = ?
Using the (mirror formula m = \(\frac{-v}{u}\), we have
v = -mu = -(-3 × -10) = – 30 cm
Therefore, the image is formed 30 cm in front of the the mirror.

Page No. 194

Question 7.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
It bends towards the normal. This is because it is travelling from an optically rarer into an optically denser medium.

Question 8.
Light enters from air to glass having refractive index 1.50. What is the speed of light ‘ in the glass? The speed of light in vacuum is 3 × 108 ms^-1.
Answer:
Given, n = 1.50 and c = 3 × 108ms^-1
We know that the refractive index is given by the expression n = \(\frac{c}{v}\) , where ’c’ is the
velocity of light in vacuum and V is the velocity of light in the given medium. Hence, we have
v = \(\frac{c}{n}\) = \(\frac{3 \times 10^8}{1.50}\) 2 × 10⁸ ms^-1

Question 9.
Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density.
Answer:

1. The medium with highest optical density is diamond, as its refractive index is maximum i.e. 2.42.
2. The medium with lowest optical density is air, as its refractive index is minimum

Question 10.
You are given .kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 9.3.
Answer:

1. The refractive index of a medium is given by the expression n = \(\frac{c}{v}\) or v = \(\frac{c}{n}\).
2. This expression shows that light travels fastest in the medium whose refractive index is minimum. Water has the lowest refractive index among the three given media. Hence, the light will travel fastest in the water.

Question 11.
The refractive index of diamond is 2.42. What is the meaning of this statement ?
Answer:
This statement means that light travels 2.42 times faster in vacuum than in diamond.

Page No. 210

Question 12.
Define 1 dioptre of power of a lens.
Answer:
One dioptre is the power of a lens of focal length 1 m.

Question 13.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object ? Also, find the power of the lens.
Answer:
Given, v = +50 cm, m = 1, u = ?, P = ?
A convex lens forms a real and inverted image equal to the size of the object if the object is placed at the centre of curvature i.e., at 2f from the lens. In this position, u = v = 50 cm.
Now, 2f = 50 cm or f = 25 cm = 0.25 m
Also, P = \(\frac{1}{f}\) = \(\frac{1}{0.25}\) = 4 D

Question 14.
Find the power of a concave lens of focal length 2 m.
Answer:
Given f = – 2m, P = ?
We know that P = \(\frac{1}{f}\) = \(\frac{1}{-2}\) = -0.5 D

Example Problems [Textbook]

Question 1.
A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature and size of the image. (T.B. Page No. 184)
Solution:
Radius of curvature, R = +3.00 m; Object-distance, u = – 5.00 m;
Image-distance, v = ? Height of the image, h’ = ?
Focal length, f = \(\frac{\mathrm{R}}{2}\) = \(\frac{+3.00 \mathrm{~m}}{2}\) = +1.50 m (as the principal focus of a convex mirror is behind the mirror)

The image is 1.15 m at the back of the mirror.
Magnification, m = \(\frac{h^{\prime}}{h}\) – \(\frac{\mathrm{v}}{\mathrm{u}}\) = \(\frac{1.15 \mathrm{~m}}{-5.00 \mathrm{~m}}\) = +0.23
The image is virtual, erect and smaller in size by a factor of 0.23.

Question 2.
An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image ? Find the nature and the size of the image. (T.B. Page No. 186)
Solution:
Object-size, h = +4.0 cm ; Object-distance, u = – 25.0 cm ;
Focal length, f = -15.0 cm ; Image-distance, v = ? Image-size, h’ = ?

or, v = – 37.5 cm
The screen should be placed at 37.5 cm in front of the mirror. The image is real.
Also, magnification, m = \(\frac{h^{\prime}}{h}\) = –\(\frac{\mathrm{v}}{\mathrm{u}}\)
or, h’ = –\(\frac{v h}{\mathrm{u}}\) = \(\frac{(-37.5 \mathrm{~cm})(+4.0 \mathrm{~cm})}{(-25.0 \mathrm{~cm})}\) (-25.0cm)
Height of the image, h’ = -6.0 cm
The image is inverted and enlarged.

Question 3.
A concave lens has focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens ? Also, find the magnification produced by the lens. (T.B. Page No. 200)
Solution:
A concave lens always forms a virtual, erect image on the same side of the object. Image-distance v = -10 cm ; Focal length f – -15 cm ;
Object-distance u = ?

or, u = – 30 cm
Thus, the object-distance is 30 cm.
Magnification m = \(\frac{\mathrm{v}}{\mathrm{u}}\)
m = \(\frac{-10 \mathrm{~cm}}{-30 \mathrm{~cm}}\) = \(\frac{1}{3}\) = +0.33
The positive sign shows that the image is erect and virtual. The image is one-third of the size of the object.

Question 4.
A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. Also find its magnification. (T.B. Page No. 208)
Solution:
Height of the object h = + 2.0 cm ; Focal length f = + 10 cm
Object-distance u = – 15 cm ; Image-distance v = ?
Height of the image h’ = ?

or, v = + 30 cm
The positive sign of v shows that the image is formed at a distance of 30 cm on the
other side of the optical centre. The image is real and inverted.
Magnification m = \(\frac{h^{\prime}}{h}\) = –\(\frac{\mathrm{v}}{\mathrm{u}}\)
or, h’ = h(\(\frac{\mathrm{v}}{\mathrm{u}}\))
Height of the image, h’ = (2.0) (+ 30 / – 15) = -4.0 cm
Magnification m = \(\frac{\mathrm{v}}{\mathrm{u}}\)
or, m = \(\frac{+30 \mathrm{~cm}}{-15 \mathrm{~cm}}\) = -2
The negative signs of m and h’ show that the image is inverted and real. It is formed below the principal axis. Thus, a real, inverted image, 4 cm tall, is formed at a distance of 30 cm on the other side of the lens. The image is two times enlarged.

AP 10th Class Physical Science Chapter 9 Questions and Answers (Lab Activities)

Activity – 9.1 (Page. No. 168)

Question 1.
Write an activity to understand the formation of images by a concave surface and a convex surface.
Answer:
Aim : To study the formation of images by both sides of a large shining spoon. Apparatus required : A large shining spoon.
Procedure :

1. Take a large shining spoon. Try to view your face in its curved surface.
2. Do you get the image ? Is it smaller or larger ?
3. Move the spoon slowly away from your face. Observe the image. How does it change?
4. Reverse the spoon and repeat the activity. How does the image look like now ?
5. Compare the characteristics of the image on the two surfaces.

Observations and conclusions : First we view our face from the surface of the spoon that is curved inwards. It acts like a concave mirror. Initially the image formed is erect and enlarged and then gets inverted and begins to reduce in size as the face is moved away from the spoon.

Next we view our face from the surface that bulges outwards. It behaves as a convex mirror. It forms erect and smaller images of the face, irrespective of our distance from the spoon.

Activity – 9.2 (Page. No. 170)

Question 2.
Write paper burning activity to understand the focal length of a concave mirror.
Answer:
Aim : To find the approximate focal length of a concave mirror.
Apparatus required : Concave mirror, a sheet of paper and metre scale.
Procedure :

1. Hold a concave mirror in your hand and direct its reflecting surface towards the sun.
2. Direct the light reflected by the mirror on to a sheet of paper held close to the mirror.
3. Move the sheet of paper back and forth gradually until you find on the paper sheet a bright, sharp spot of light.
4. Hold the mirror and the paper in the same position for a few minutes. What do you observe? Why?

Observations and conclusions : At first the paper begins to burn producing smoke. Then; it may catch fire. This is because the light from the sun is converged at a point, as a sharp bright spot by the concave mirror. This spot of light is the image of the sun on the sheet of paper. This point is the focus of the concave mirror. The heat produced due to concentration of sun light is so large that it ignites the paper. The distance of image from the mirror gives the focal length of the concave mirror.

Activity – 9.3 (Page. No. 172)

Question 3.
Write an activity to understand the nature of images formed by a concave mirror for a different position of the object.
Answer:
Aim : To study the formation of images by a concave mirror by an object at different positions on its principle axis.
Apparatus required : Concave mirror, a metre scale, a chalk piece, a burning candle, a paper screen.
Procedure :

1. Take a concave mirror. Find out its approximate focal length as described in activity 9.2. Note down the value of focal length. (You can also find it out by obtaining image of a distant object on a sheet of paper.)
2. Mark a line on a table with a chalk. Place the concave mirror on a stand. Place the stand over the line such that its pole lies over the line.
3. Draw with a chalk two more lines parallel to the previous line such that the distance between any two successive lines is equal to the focal length of the mirror.
4. These lines will now correspond to the positions of the points P, F and C, respectively. Remember – For a spherical mirror of small aperture, the principal focus F lies mid¬way between the pole P and the centre of curvature C.
5. Keep a bright object, say a burning candle, at a position far beyond C.
6. Place a paper screen and move it in front of the mirror till you obtain a sharp bright image of the candle flame on it.
7. Observe the image carefully. Note down its nature, position and relative size with
   respect to the object size.
8. Repeat the experiment by placing the candle – (a) just beyond C, (b) at C, (c) between F and C, (d) at F, and (e) between P and F.
9. In one of the cases, you may not get the image on the screen. Identify the position of the object in such a case. Then, look for its virtual image in the mirror itself.
10. Note down and tabulate your observations.

Observations and conclusions : We see that nature, position and size of the image formed by a concave mirror depends on the position of the object in relation to points P, F and C. We note its nature (Whether real and inverted / virtual and erect) and also its size whether magnified, reduced or has the same size as the object. We tabulate our observations as in table.

Table : Image formation by a concave mirror for different positions of the object

Position of the object	Position of the image	Size of the image	(Nature of the image
At infinity	At the focus F	Highly diminished, point-sized	Real and inverted
Beyond C	Between F and C	Diminished	Real and inverted
At C	At C	Same size	Real and inverted
Between C and F	Beyond C	Enlarged	Real and inverted
At F	At infinity	Highly enlarged	Real and inverted
Between P and F	Behind the mirror	Enlarged	Virtual and erect

Activity – 9.4 (Page. No. 178)

Question 4.
Draw neat ray diagrams for different position of the object for a concave mirror. Describe the nature, position and relative size of the image formed in each case. Tabulate the results in a convenient format.
Answer:

Position of the object	Position of the image	Size of the image	Nature of the image
At infinity	At the focus F	Highly diminished, point-sized	Real and inverted
Beyond C	Between F and C	Diminished	Real and inverted
At C	AtC	Same size	Real and inverted
Between C and F	Beyond C	Enlarged	Real and inverted
At F	At infinity	Highly enlarged	Real and inverted
Between P and F	Behind the mirror	Enlarged	Virtual and erect

 

Activity – 9.5 (Page. No. 180)

Question 5.
What happens to the size of image formed by a convex mirror, when an object is gradually moved towards the mirror ?


Answer:
As shown in figure, when the object at position AB, its virtual image is at A’B’. When the object is at position. A₁, B₁, its virtual image is at A₁‘, B₁‘. So, when an object is gradually moved towards the pole of a convex mirror, its image also moves towards its pole and gradually increases in size till it has a size almost equal to the of the object. However, the image is always formed between F and P.

Activity – 9.6 (Page. No. 180)

Question 6.
Write an activity to understand the sizes of the images formed by a plane mirror, a concave mirror and a convex mirror.
(OR)
Observe the image of a distant object, say a distant tree, in a plane mirror. Could you see a full-length image ?
Try with plane mirrors of different sizes. Did you see the entire object in the image? Repeat this Activity with a concave mirror. Did the mirror show full length image of the object ?
Now try using a convex mirror. Did you succeed ?
Explain your observations with reason.
Answer:
1) When you observe the image of a distant object, such as a distant tree, in a plane mirror, you will see a full-length image. Plane mirrors produce images that are the same size as the actual object and are upright. Therefore, you can see the entire object in the image, just as you would if you were looking directly at the tree.

2) When you try this activity with plane mirrors of different sizes, you will continue to see a full-length image, regardless of the mirror’s size, because plane mirrors create virtual images that maintain the same size and orientation as the object.

3) Now, if you repeat this activity with a concave mirror, you will observe a different result. A concave mirror can either produce a full-length image or a partially distorted one, depending on where the object is located in relation to the focal point of the mirror.

4 ) When you use a convex mirror, you will consistently see a smaller, upright and full-length image of the object. Convex mirrors are designed to produce images that are diminished in size and maintain their upright orientation, but they do not provide magnification. The image will appear to be farther away from the mirror than the actual object.

Activity – 9.7 (Page. No. 188)

Question 7.
Write an activity to demonstrate the phenomenon of refraction.
Answer:
Aim To study the difficulty in picking a coin placed at the bottom of a bucket filled with water.
Apparatus required : A coin and bucket filled with water.
Procedure :

1. Place a coin at the bottom of a bucket filled with water.
2. With your eye to side above water, try to pick up the coin in one go. Did you succeed in picking up the coin ?
3. Repeat the Activity. Why did you not succeed in doing it in one go ?
4. Ask your friends to do this. Compare your experience with theirs.

Observations and conclusions : Due to refraction of light from water to air, the coin appears to be slightly raised above its actual position. So in one go it is difficult to pick the coin. After doing few trails, we can succeed in picking up the coin.

Activity – 9.8 (Page. No. 188)

Question 8.
Write activity to show how the refraction of light can change the apparent position of objects when they are viewed through different media.
Answer:
Aim : To look for the apparent image of a coin formed due to refraction in water. Apparatus required : A shallow bowl, a coin, a table and a beaker of water.
Procedure :

1. Place a large shallow bowl on a table and put a coin in it.
2. Move away slowly from the bowl. Stop when the coin just disappears from your sight.
3. Ask a friend to pour water gently into the bowl without disturbing the coin.
4. Keep looking for the coin from your position. Does the coin become visible again from your position ? How could this happen ?

Observations and conclusions : The coin becomes visible again on pouring water into the bowl. The coin appears slightly raised above its actual position due to refraction of light.

Activity – 9.9 (Page. No. 188)

Question 9.
Write an activity to demonstrate the phenomenon of apparent depth or apparent shift, which happens due to refraction.
Answer:
Aim : To study the refracted image of a straight line formed a glass slab.
Apparatus required : A white sheet of paper, a glass slab and an ink pen.
Procedure :

1. Draw a thick straight line in ink, over a sheet of white paper placed on a table.
2. Place a glass slab over the line in such a way that one of its edges makes an angle with the line.
3. Look at the portion of the line under the slab from the sides. What do you observe?
   Does the line under the glass slab appear to be bent at the edges?
4. Next, place the glass slab such that it is normal to the line. What do you observe now ? Does the part of the line under the glass slab appear bent ?
5. Look at the line from the top of the glass slab. Does the part of the line, beneath the slab, appear to be raised? Why does this happen?

Observations and conclusions:

1. The line under the glass slab appears to be bent at the edges due to the refraction of light from glass to air.
2. The part of the line under the glass slab does not appear bent. This is because light rays incident normally at the interface of glass and air do not suffer refraction or bending at the interface.
3. Yes, the part of the line beneath the glass slab appears to be raised. This is due to refraction of light rays as they travel from glass to air.

Activity – 9.10 (Page. No. 190)

Question 10.
Write an activity to demonstrate the refraction of light through a glass slab.
Answer:
Aim : To study the refraction of ray of light through a glass slab and to measure its lateral displacement.
Apparatus required : A white sheet of paper, drawing board, drawing pins, alpins, a sharp pencil, a scale.
Procedure :

1. Fix a sheet of white paper on a drawing board using drawing pins.
2. Place a rectangular glass slab over the sheet in the middle.
3. Draw the outline of the slab with a pencil. Let us name the outline as ABCD.
4. Take four identical pins.
5. Fix two pins, say E and F, vertically such that the line joining the pins is inclined to the edge AB.
6. Look for the images of the pins E and F through the opposite edge. Fix two other . pins, say G and H, such that these pins and the images of E and F lie on a straight line.
7. Remove the pins and the slab.
8. Join the positions of tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly, join the positions of tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O’.
9. Join O and O’. Also produce EF up to P, as shown by a dotted line in Figure.

Observations and conclusions :

1. Draw a perpendicular NN’ to AB at ‘O’ and another perpendicular MM’ to CD at O’.
2. The light ray at point O has entered from a rarer medium to a denser medium, that is, from air to glass.
3. Note that the light ray has bent towards the normal. At O’, the light ray has entered from glass to air, that is, from a denser medium to a rarer medium.
4. The light here has bent away from the normal. Compare the angle of incidence with the angle of refraction at both refracting surfaces AB and CD.
5. For refraction at face AB, we observe that angle of refraction r₁ is less than angle of incidence i₁.
6. Again, for refraction at face CD, the angle of refraction r₂ is greater than the angle of incidence i₂.
7. in the figure, EO is the incident ray, OO’ is the refracted ray and O’H is the emergent ray.
8. We observe that emergent ray O’H is parallel to the incident ray EO. This is because the extent of bending of the ray of light at the face AB is equal and opposite to that at face CD.
9. How ever, the emergent ray is shifted sideward slightly by a perpendicular distance OC with respect to the incident ray.
10. This lateral shift in the path of light on emerging from a medium with parallel faces is called lateral displacement.

Activity – 9.11 (Page. No. 198)

Question 11.
What happens when parallel rays of light are incident on a lens ? Describe through an activity.
Answer:
Aim : To find approximate focal length of a convex lens.
Apparatus required : A convex lens, a sheet of paper and a metre scale;
Procedure :

1. Hold a convex lens in your hand. Direct it towards the sun. ?
2. Focus the light from the sun on a sheet of paper. Obtain a sharp bright image of the Sun.
3. Hold the paper and the lens in the same position for a while. Keep observing the paper. What happenes ? Why? How can we find the focal length of convex lens ?

Observations and conclusions :

1. After a while, the paper begins to burn producing smoke. It may even catch fire. This is because the light from the sun travels in the form of parallel rays.
2. These rays are converged by the lens forming a bright spot on the paper. In fact, this bright spot is the real image of the sun. The concentrated sunlight produces heat. This causes the paper to burn.
3. We measurable the distance between the position of the lens and position of the image (bright spot on the paper) of the sun. This distance gives the approximate focal length of the lens.

Activity – 9.12 (Page. No. 198)

Question 12.
How do you find the nature, position and relative size of the image formed by a convex lens for various positions of the object ?
Answer:
Aim : To study the formation of images by a convex lens for various positions of an object placed on its principal axis.
Apparatus required : A convex lens, a metre scale, a burning candle, a screen of paper.
Procedure :

1. Take a convex lens.
2. Draw five parallel straight lines, using chalk, on a long Table such that the distance between the successive lines is equal to the focal length of the lens.
3. Place the lens on a lens stand. Place it on the central line such that the optical centre of the lens lies just over the line.
4. The two lines on either side of the lens correspond to F and 2F of the lens respectively. Mark them with appropriate letters such as 2F₁, F₁, F₂ and 2F₂, respectively.
5. Place a burning candle, far beyond 2F₁ to the left. Obtain a clear sharp image on a screen on the opposite side of the lens.
6. Note down the nature, position and relative size of the image.
7. Repeat this Activity by placing object just behind 2F₁, between F₁ and 2F₁ at F₁, between F₁ and O. Note down and tabulate your observations.

Table : Nature, position and relative size of the image formed by a convex lens for various positions of the object

Position of the object	Position of the image	Relative size of the image	Nature of the image
At infinity	At focus F2	Highly diminished, point-sized	Real and inverted
Beyond 2F1	Between F2 and 2F2	Diminished	Real and inverted
At 2F1	At 2F2	Same size	Real and inverted
Between F1 and 2F1	Beyond 2F2	Enlarged	Real and inverted
At focus F1	At infinity	Infinitely large or highly enlarged	Real and inverted
Between focus F1 and optical centre O	Behind the mirror of the lens as the object	Enlarged	Virtual and erect

Activity – 9.13 (Page No. 200)

Question 13.
How do you do an activity to study the nature, position and relative size of the image formed by a concave lens? Write a conclusion draw from the observations.
Answer:
Aim : To study the formation of image of an object by a concave lens.
Procedure:

1. Take a concave lens. Place it on a lens stand.
2. Place a burning candle on one side of the lens.
3. Look through the lens from the other side and observe the image. Try to get the image on a screen, if possible. If not, observe the image directly through the lens.
4. Note down the nature, relative size and approximate position of the image.
5. Move the candle away from the lens. Note the change in the size of the image. What happens to the size of the image when the candle is placed too far away from the lens.

Table : Nature, position and relative size of the image formed by a concave lens for
various positions of the object

Position of the object	Position of the image	Relative size of the image	Nature of the image
At infinity	At focus F1	Highly diminished, point-sized	Virtual and erect
Between infinity and optical centre 0 of the lens	Between focus F1 and optical centre O	Diminished	Virtual and erect