AP 10th Class Physics 10th Lesson Questions and Answers The Human Eye and the Colourful World

AP State Board new syllabus AP Board Solutions Class 10 Physics 10th Lesson The Human Eye and the Colourful World Questions and Answers.

10th Class Physics 10th Lesson The Human Eye and the Colourful World Questions and Answers

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

Question 1.
The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to
a) presbyopia.
b) accommodation.
c) near-sightedness.
d) far-sightedness.
Answer:
b) accommodation.

Question 2.
The human eye forms the image of an object at its
a) cornea.
b) iris.
c) pupil.
d) retina.
Answer:
d) retina.

Question 3.
The least distance of distinct vision for a young adult with normal vision is about
a) 25 m.
b) 2.5 cm.
c) 25 cm.
d) 2.5 m.
Answer:
c) 25 cm.

Question 4.
The change in focal length of an eye lens is caused by’the action of the
a) pupil.
b) retina.
c) ciliary muscles.
d) iris.
Answer:
c) ciliary muscles.

Question 5.
A person needs a lens of power -5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?
Answer:
i) Focal length of the lens for distant vision = \(\frac{1}{\text { Power }}\) = \(\frac{100}{-5.5}\) = cm = -18 cm (approx)
ii) Focal length of the lens for near vision = \(\frac{100}{1.5}\) cm = 66.66 cm

Question 6.
The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem ?
Answer:
The far point of a normal eye is infinity. Since the far point of the defective eye is given as 80 cm, the eye is short-sighted. To correct it, the lens should be such that an object at infinity must form its image at the far point of defective eye.
∴ u = -∝, v = -80 cm.
Using lens formula \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
∴ \(\frac{1}{\mathrm{f}}\) = \(\frac{1}{-80}\) – \(\frac{1}{(-\infty)}\) = \(\frac{1}{-80}\)
∴ Focal length of lens is – 80 cm
The correction is done by using a concave lens of focal length 80 cm.
Power of the lens = \(\frac{100}{\mathrm{f}(\text { in } \mathrm{cm})}\) = –\(\frac{100}{80}\) = -1.25 D is needed.

Question 7.
Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect ? Assume that the near point of the normal eye is 25 cm.
Answer:

N = Near point of a hypermetropic eye
N’ = Near point of a normal eye
To correct the defect, the image of an object at 25 cm should be brought at 100 cm.
∴ \(\) = \(\frac{1}{-100}\) – \(\frac{1}{-25}\)
\(\frac{1}{\mathrm{f}}\) = \(\frac{-1}{100}\) + \(\frac{1}{25}\) = \(\frac{- 1 + 4}{100}\) = \(\frac{3}{100}\)
∴ f = + \(\frac{100}{3}\) = +33.3 cm
So, a convex lens of focal length 33.3 cm is required power, P = \(\frac{100}{33.3}\) = 3.0 D

Question 8.
Why is the normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
The focal length of the eye lens cannot be reduced below a certain limit.

Question 9.
What happens to the image distance in the eye when we increase the distance of an object from the eye ?
Answer:
In eye, the image is always formed on the retina. The image distance is the distance between the eye lens and the retina. When we increase the distance of the object from the eye, the focal length of the eye lens increases due to the action of ciliary muscless so that the image of object is formed on the retina and therefore, the image distance remains the same.

Question 10.
Why do stars twinkle ?
Answer:

1. Stars appear to twinkle due to atmospheric refraction.
2. The light of stars passes through the Earth’s atmosphere, which contains air layers of varying temperatures and densities.
3. These variations cause the star’s light to refract and create small, rapidly changing differences in brightness and position, leading to the twinkling effect.

Question 11.
Explain why the planets do not twinkle.
Answer:
The planets are much closer to the Earth, and are thus seen as extended sources. If we consider a planet as a collection of a large number of point- sized sources of light, the total variation in the amount of light entering our eye from’all the individual, point – sized sources will average out to zero, thereby nullifying the twinkling effect.

Question 12.
Why does the sky appear dark instead of blue is an astronaut ?
Answer:
At such huge heights due to absence of atmosphere, no scattering out the light takes place. Therefore, sky appears dark.

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

Page No. 222

Question 1.
What is meant by the power of accommodation of the eye ?
Answer:
Power of accommodation The ability of eye lens to adjust its focal length to form a sharp image of the object at varying distances on the retina is called power of accommodation.

Question 2.
A person with a myopic eye cannot see objects beyond 12m distinctly. What should be the type of the corrective lens used to restore proper vision ?
Answer:
A person with myopic eye defect should use concave lens of focal length 1.2 m to restore his proper vision.

Question 3.
What is the far point and near point of the human eye with normal vision ?
Answer:
For normal vision, the near point is about 25 cm and far point is infinity. Thus, a normal eye can see objects clearly that are between 25 cm and infinity.

Question 4.
A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected ?
Answer:
Child is suffering from myopia or short sightedness. The defect is corrected by using a concave lens of suitable power placed infront of eye defective.

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

Activity – 10.1 (Page. No. 224)

Question 1.
Describe an activity to show refraction of light through a prism. Draw relevant figure, marking angle of incidence, refraction, emergence and deviation.
Answer:
Aim : To study the refraction of a ray of light through a glass prism. Measure the angle of incidence, angle of prism, angle of deviation and angle of emergence and hence establish the relation between these angles.

Apparatus required : A glass prism, a drawing sheet, drawing pins, alpins, a drawing board and geometry box.

Procedure :

1. Fix a sheet of white paper on a drawing board using drawing pins.
2. Place a glass prism on it in such a way that it rests on its triangular base. Trace the outline of the prism using a pencil.
3. Draw a straight line PE inclined to one of the refracting surfaces, say AB, of the
   prism.
4. Fix two pins, say at points P and Q, on the line PE as shownn in Figure.
5. Look for the images of the pins, fixed at P and Q, through the other face AC.
6. Fix two more pins, at points R and S, such that the pins at R and S and the images of the pins at P and Q line on the same straight line.
7. Remove the PE meets the boundary of the prism at point E (see Figure). Similarly, join and produce the points R and S. Let these lines meet the boundary of the prism at E and F, respectively. Join E and F.
8. Draw perpendiculars to the refracting surfaces AB and AC of the prism at points E and F, respectively.
9. Mark the angle of incidence (∠i), the angle of refraction (∠r) and the angle of emergence (∠e) as shown in Figure.

Observations and Conclusion : As this incident ray PE suffers refraction from air to glass at face AB, it bends towards the normal NN’ along the path EF. Again, as the ray EF suffers refraction from glass to air at face AC, it bends away from the normal MM’. Thus the emergent ray FS bends towards the base of the prism. We measure the angle of incidence i, angle of prism A, angle of emergence e and the angle of deviation D. It will be seen that
i + e = A + D

Activity – 10.2 (Page. No. 226)

Question 2.
Explain dispersion of white light by a glass prism.
Answer:
Apparatus required : Aim To study the dispersion of white light by a glass prism
A glass prism, a cardboard having a narrow rectangular slit, a white screen.
Procedure :

1. Take a thick sheet of cardboard and make a small hole or narrow slit in its middle.
2. Allow sunlight to fall on the narrow slit. This gives a narrow beam of white light.
3. Now take a glass prism and allow the light that comes out of its appears on a nearly screen.
4. Turn the prism slowly until the light that comes out of it appears on a nearby screen.
5. What do you observe? You will find a beautiful band of colours. Why does this happen ? What is the sequence of colours that you see on the screen ?

Observations and conclusions : The glass prism splits the incident white light into a band of colours. The various colours are seen in the sequence.

Violet, Indigo, Blue, Green, Yellow, Orange and Red (VIBGYOR). This band of the coloured components of light beam is called its spectrum. The splitting of white light into its component colours is called dispersion.

Different colours of light bend through different angles with respect of the incident ray, as they pass through the prism. The red light bends the least while the violet bends the most. Thus the rays of different colours emerge along different paths and become distinct. Thus we see bands of different colours on the screen in the form of a spectrum.