NCERT Solutions For Class 10 Physics: Human Eye and Colorful World

Intext Questions (Page 164/165)

Question 1: What is meant by power of accommodation of the eye?

Answer-
The power of accommodation is the ability of the eye lens to adjust its focal length. This adjustment is achieved by the action of the ciliary muscles, which modify the curvature of the eye lens. This modification enables us to focus on both near and distant objects clearly.

Question 2: A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?

Answer-
The person is suffering from myopia (or near-sightedness). Myopia is corrected by using a concave lens (or diverging lens) of suitable power. This lens ensures the image of distant objects is formed back onto the retina.

Question 3: What is the far point and near point of the human eye with normal vision?

Answer-
The near point (or least distance of distinct vision) is the minimum distance at which objects can be seen most distinctly without strain. For a young adult with normal vision, the near point is about 25 cm. The far point of the eye is the farthest point up to which the eye can see objects clearly, and for a normal eye, it is 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-
The child is suffering from difficulty seeing distant objects distinctly, which is the defect known as Myopia (or near-sightedness). In a myopic eye, the image of a distant object is formed in front of the retina. This defect can be corrected by using an eye-glass fitted with a concave lens of suitable power.

Exercise Questions (Page 170)

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-
The ability of the eye lens to adjust its focal length to focus on objects at different distances is called (b) accommodation.

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

Answer-
The eye lens system forms an inverted real image on the light-sensitive screen called the (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-
The least distance of distinct vision (near point) for a young adult with normal vision is about (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-
The curvature and thus the focal length of the eye lens is modified by the action of the (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-
The focal length ($\text{f}$) is calculated using the formula $\text{f} = 1/\text{P}$ (where $\text{f}$ is in metres).
(i) Distant Vision Correction: Power $\text{P} = -5.5 \text{ D}$.
$\text{f} = \frac{1}{-5.5 \text{ D}} \approx \underline{-0.182 \text{ m}}$
(ii) Near Vision Correction: Power $\text{P} = +1.5 \text{ D}$.
$\text{f} = \frac{1}{+1.5 \text{ D}} \approx \underline{+0.67 \text{ m}}$

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-
1.Nature of the lens: The defect is myopia, which is corrected by using a concave lens.

2.Power of the lens: The corrective lens must bring a distant object (at infinity) to the far point of the eye ($v = -80 \text{ cm}$). Thus, $f = -80 \text{ cm} = -0.80 \text{ m}$.
$\text{P} = \frac{1}{\text{f}} = \frac{1}{-0.80 \text{ m}} = \underline{-1.25 \text{ D}}$

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-
Correction Diagram: Hypermetropia is corrected by using a convex lens (or converging lens). The diagram should show the convex lens providing additional focussing power so that light rays from a close object (25 cm) focus directly on the retina, instead of behind it.
Power of the lens: The lens must form a virtual image of an object placed at the normal near point ($u = -25 \text{ cm}$) at the hypermetropic near point ($v = -1 \text{ m} = -100 \text{ cm}$).
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-100} - \frac{1}{-25}$
$\frac{1}{f} = \frac{-1 + 4}{100} = \frac{3}{100} \text{ cm}^{-1}$
$\text{f} = \frac{100}{3} \text{ cm} \approx 0.33 \text{ m}$
$\text{P} = \frac{1}{f} \approx \underline{+3.0 \text{ D}}$

Question 8: Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Answer-
A normal eye cannot see objects clearly closer than 25 cm because the focal length of the eye lens cannot be decreased below a certain minimum limit, even with maximum effort from the ciliary muscles. If an object is held closer than 25 cm, the eye lens cannot converge the light rays sufficiently, and the image will be blurred or appear strained.

Question 9: What happens to the image distance in the eye when we increase the distance of an object from the eye?

Answer-
The image distance in the eye remains approximately the same. The image is always formed on the retina, which is the fixed, light-sensitive screen at the back of the eyeball. When the object distance increases, the ciliary muscles relax, causing the eye lens to become thinner and its focal length to increase. This adjustment ensures that the image is maintained sharply focused on the retina.

Question 10: Why do stars twinkle?

Answer-
Stars twinkle due to atmospheric refraction of starlight.
1.Stars are very distant and act as point-sized sources of light.

2.The atmosphere has a gradually changing refractive index, and its physical conditions are not stationary.

3.As starlight passes through these turbulent layers, the path of the light rays varies slightly. This variation causes the amount of starlight entering the eye to flicker, making the star appear sometimes brighter and sometimes fainter (twinkling).

Question 11: Explain why the planets do not twinkle.

Answer-
Planets do not twinkle because they are much closer to the earth than stars and are seen as extended sources of light. An extended source can be considered a collection of many point-sized sources. Since the path variations from these individual point sources are randomly averaged out, the total variation in the amount of light entering our eye averages out to zero, thereby nullifying the twinkling effect.

Question 12: Why does the sky appear dark instead of blue to an astronaut?

Answer-
The blue colour of the sky is caused by the scattering of light by the molecules of air and fine particles in the atmosphere. Astronauts orbit at very high altitudes where the atmosphere is not prominent. Since there are no air molecules or particles to scatter the sunlight effectively at that height, the sky appears dark to the astronaut.