Class 10 Science · Chapter 10

The Human Eye and the Colourful World

163 Class 10 Science questions on The Human Eye and the Colourful World from the NCERT textbook, NCERT Exemplar and CBSE board papers 2023–2026, each with its answer and a step-by-step solution.

NCERT exercises

Exercises (12) Intext Questions (Page 164) (4)

NCERT Exemplar exercises

Multiple Choice Questions (14) Short Answer Questions (10) Long Answer Questions (6)

CBSE Class 10 Science previous year papers, with answers →

Questions · page 1 of 9

Q1 NCERT · Exercises
The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to
1. (a)presbyopia.
2. (b)accommodation.
3. (c)near-sightedness.
4. (d)far-sightedness.
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(b) accommodation.

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Q2 NCERT · Intext Questions (Page 164)

What is meant by power of accommodation of the eye?

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It is the ability of the eye lens to change its focal length so that objects at different distances focus on the retina.
The power of accommodation of the eye is the ability of the eye lens to adjust its own focal length, so that objects lying at different distances are all focused sharply on the retina. The eye lens is made of a fibrous, jelly-like material whose curvature is changed by the ciliary muscles: when these muscles relax the lens becomes thin and its focal length increases, which lets distant objects be seen clearly, and when they contract the lens becomes thicker and its focal length decreases, which lets nearby objects be seen clearly. This adjustment is necessary because the retina is fixed about $2.5$ cm behind the eye lens, so the image distance cannot change and only the focal length can. The ability is limited, because the focal length cannot be reduced below a certain minimum; that is why a normal eye can see clearly only from about $25$ cm, its near point, out to infinity, its far point.

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Q3 NCERT · Exercises
The human eye forms the image of an object at its
1. (a)cornea.
2. (b)iris.
3. (c)pupil.
4. (d)retina.
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(d) retina.

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Q4 NCERT · Intext Questions (Page 164)

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?

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A concave (diverging) lens of focal length $- 1.2$ m and power $- 0.83$ D is needed.
A concave lens, that is a diverging lens, should be used. Since the person cannot see beyond $1.2$ m, the far point of the eye is $1.2$ m in front of it instead of infinity. The corrective lens must form the image of an object at infinity exactly at that far point, so $u = - \infty$ , giving $\frac{1}{u} = 0$ , and $v = - 1.2$ m. The lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ then gives $\frac{1}{f} = - \frac{1}{1.2}$ , so $f = - 1.2$ m. The focal length is negative, which is the mark of a concave lens, and its power is $P = \frac{1}{f} = \frac{1}{- 1.2} = - 0.83$ D.

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Q5 NCERT · Exercises
The least distance of distinct vision for a young adult with normal vision is about
1. (a) $25$ m.
2. (b) $2.5$ cm.
3. (c) $25$ cm.
4. (d) $2.5$ m.
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(c) $25$ cm.

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Q6 NCERT · Intext Questions (Page 164)

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

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For a normal eye the far point is at infinity and the near point is about 25 cm.
For a human eye with normal vision the far point is at infinity and the near point is about $25$ cm from the eye. The far point is the farthest point up to which the eye can see objects clearly, and a normal eye has no such limit, so it lies at infinity. The near point is the minimum distance at which an object can be seen most distinctly and without strain, also called the least distance of distinct vision; for a young adult with normal vision it is about $25$ cm. A normal eye can therefore see clearly everything lying between $25$ cm and infinity.

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Q7 NCERT · Exercises
The change in focal length of an eye lens is caused by the action of the
1. (a)pupil.
2. (b)retina.
3. (c)ciliary muscles.
4. (d)iris.
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(c) ciliary muscles.

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Q8 NCERT · Intext Questions (Page 164)

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?

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The defect is myopia, corrected with spectacles having a concave lens of suitable power.
The child is most likely suffering from myopia, that is near-sightedness or short-sightedness. The blackboard seen from the last row is a distant object, and a myopic eye can see nearby objects clearly but cannot see distant objects distinctly, because its far point has come nearer than infinity and may be only a few metres away. In such an eye the image of a distant object is formed in front of the retina instead of on it, either because the eye lens is too strongly curved or because the eyeball has become too long, and the blackboard therefore looks blurred. It can be corrected by wearing spectacles fitted with a concave, that is diverging, lens of suitable power. The concave lens spreads the parallel rays coming from the blackboard so that the eye's own lens now brings them to a focus on the retina; equivalently, it forms the image of the distant blackboard at the child's far point, which is a distance the eye can still focus.

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Q9 NCERT · Exercises
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
1. (i)distant vision, and
2. (ii)near vision?
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The focal lengths are $- 0.18$ m for distant vision (concave lens) and $+ 0.67$ m for near vision (convex lens).
(i) For distant vision, $f_{1} = \frac{1}{P _{1}} = \frac{1}{- 5.5} = - 0.18$ m (about $- 18$ cm), a concave lens; (ii) For near vision, $f_{2} = \frac{1}{P _{2}} = \frac{1}{+ 1.5} = + 0.67$ m (about $+ 67$ cm), a convex lens. Both values are rounded to two decimal places, and the focal lengths are in metres because power is measured in dioptres, where $1$ D $= 1$ m $^{- 1}$ .

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Q10 NCERT · Exercises

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?

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A concave (diverging) lens of power $- 1.25$ D is required.
A concave (diverging) lens of power $- 1.25$ D is required. The lens must form, at the eye's far point $80$ cm away, the image of an object at infinity, so $u = - \infty$ and $v = - 80$ cm. The lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ then gives $\frac{1}{f} = - \frac{1}{80}$ , that is $f = - 80$ cm $= - 0.8$ m, and $P = \frac{1}{f} = \frac{1}{- 0.8} = - 1.25$ D. The negative sign shows that the lens is concave.

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Q11 NCERT · Exercises

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.

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A convex lens of power $+ 3.0$ D corrects the defect.
A convex (converging) lens of power $+ 3.0$ D is required. The lens must form, at the eye's near point of $100$ cm, a virtual image of an object placed at the normal near point of $25$ cm, so $u = - 25$ cm and $v = - 100$ cm. The lens formula gives $\frac{1}{f} = \frac{1}{- 100} - \frac{1}{- 25} = \frac{- 1 + 4}{100} = \frac{3}{100}$ , so $f = + \frac{100}{3} = + 33.3$ cm $= + 0.333$ m and $P = \frac{1}{f} = + 3.0$ D. The image is erect and four times as tall as the object, and the diagram above shows the construction: the ray parallel to the axis refracted through $F^{'}$ , the ray through the optical centre undeviated, and their backward extensions meeting at the $100$ cm mark.

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Q12 NCERT · Exercises

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

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The lens cannot shorten its focal length below $2.27$ cm, so for objects nearer than $25$ cm the image falls behind the retina.
Because the eye has run out of accommodation. The retina is fixed about $2.5$ cm behind the eye lens, so the image distance cannot change; the only way to focus a nearer object is to shorten the focal length of the lens, and the ciliary muscles do that by making the lens more curved. From $\frac{1}{f} = \frac{1}{2.5} + \frac{1}{d}$ , an object at $25$ cm needs $f = 2.27$ cm, which is the shortest focal length a normal eye lens can reach. An object at $10$ cm would need $f = 2.00$ cm, which the lens cannot supply, so with $f$ stuck at $2.27$ cm the image is formed $2.94$ cm behind the lens — behind the retina. The rays are still converging as they cross the retina, so a blurred patch falls on it and the object cannot be seen clearly. That is why $25$ cm is called the least distance of distinct vision.

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Q13 NCERT · Exercises

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

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The image distance stays constant at about 2.5 cm because the retina is fixed; the eye lens changes its focal length instead.
The image distance in the eye does not change: it stays at about $2.5$ cm, the fixed distance from the eye lens to the retina. Because the retina cannot move, the image must always be formed on it, so in $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ the quantity $v$ is held constant and the **focal length** of the eye lens changes instead. As the object is moved further away, the ciliary muscles relax, the eye lens becomes thinner and less curved, and its focal length increases — from about $2.27$ cm for an object at $25$ cm to about $2.5$ cm for a very distant one. This adjustment is called accommodation. What does get smaller as the object recedes is the **size** of the image on the retina, not its distance.

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Q14 NCERT · Exercises

Why do stars twinkle?

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Atmospheric refraction by moving air layers keeps changing the light from a point-like star, so it seems brighter and fainter by turns.
Stars twinkle because of atmospheric refraction. The earth's atmosphere is made of layers of air whose density increases towards the ground, so light from a star is refracted continuously as it comes down, bending towards the normal and following a curved path; this is also why a star seen near the horizon appears a little higher in the sky than it really is. The layers of air are constantly moving and changing density, so the amount of bending keeps changing, and with it the path the starlight takes and the amount of it that enters the eye. Because a star is so far away that it acts as a single point source, all its light shifts together, so the star seems brighter at one moment and fainter the next. This continual variation in brightness is the twinkling of stars.

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Q15 NCERT · Exercises

Explain why the planets do not twinkle.

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A planet is a disc of many point sources whose independent flickers average out, so its light stays steady.
Planets do not twinkle because they are extended sources of light, not point sources. A planet is millions of times nearer to us than any star, so it subtends a measurable angle at the eye and is seen as a small disc, which can be treated as a very large number of point sources lying side by side. The atmosphere refracts the light from each of those points along a path that keeps changing, exactly as it does for a star, so each point does flicker. But the flickers are independent of one another, so at any instant the points that are momentarily brighter are balanced by those that are momentarily dimmer. The total light the eye receives from the whole disc therefore stays very nearly constant, and the planet shines steadily. A star, being effectively a single point source, has nothing to average its fluctuations against, so it twinkles.

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Q16 NCERT · Exercises

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

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There is no atmosphere above the astronaut to scatter sunlight, so no scattered blue light reaches the eye and the sky looks dark.
Because there is no atmosphere above the astronaut to scatter sunlight. The sky looks blue from the ground only because sunlight passing through the air is scattered by its molecules, and since light of shorter wavelength is scattered much more strongly than light of longer wavelength, it is mainly blue light that is thrown sideways and reaches the eye from every direction of the sky. An astronaut is above practically all the air, so there are no molecules to scatter the sunlight and no scattered light arrives from the empty directions. Light reaches the astronaut only directly from the Sun and from lit bodies such as the earth, the moon and the stars, so the rest of the sky appears completely dark. There is no shortage of sunlight up there — it is stronger than at the ground — only a shortage of anything to scatter it.

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Q17 Exemplar · Multiple Choice Questions
A person cannot see distinctly objects kept beyond $2$ m. This defect can be corrected by using a lens of power
1. (a) $+ 0.5$ D
2. (b) $- 0.5$ D
3. (c) $+ 0.2$ D
4. (d) $- 0.2$ D
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(b) $- 0.5$ D

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Q18 Exemplar · Multiple Choice Questions
A student sitting on the last bench can read the letters written on the blackboard but is not able to read the letters written in his text book. Which of the following statements is correct?
1. (a)The near point of his eyes has receded away
2. (b)The near point of his eyes has come closer to him
3. (c)The far point of his eyes has come closer to him
4. (d)The far point of his eyes has receded away
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(a) The near point of his eyes has receded away

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Q19 Exemplar · Multiple Choice Questions
A prism ABC (with BC as base) is placed in different orientations. A narrow beam of white light is incident on the prism as shown in Figure 11.1. In which of the following cases, after dispersion, the third colour from the top corresponds to the colour of the sky?
(i)
(ii)
(iii)
(iv)
1. (a)(i)
2. (b)(ii)
3. (c)(iii)
4. (d)(iv)
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(b) (ii)

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Q20 Exemplar · Multiple Choice Questions
At noon the sun appears white as
1. (a)light is least scattered
2. (b)all the colours of the white light are scattered away
3. (c)blue colour is scattered the most
4. (d)red colour is scattered the most
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(a) light is least scattered

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