NCERT Solutions for Class 10 Science Chapter 10: The Human Eye and the Colourful World (NCERT 2026–27)

These Class 10 Science Chapter 10 solutions cover The Human Eye and the Colourful World with every in-text question and end-of-chapter exercise reproduced verbatim from the NCERT textbook and solved step by step. The chapter explains how the human eye works, its power of accommodation, the three common defects of vision and their correction, and the beautiful optical phenomena around us — refraction and dispersion through a prism, atmospheric refraction, and scattering of light. All numericals are worked out with the lens formula and the power relation, and answers are updated for session 2026–27.

Class: 10 Subject: Science Chapter: 10 Title: The Human Eye and the Colourful World Type: Physics (Optics) Session: 2026–27
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Class 10 Science Chapter 10 Solutions – Overview

Chapter 10, The Human Eye and the Colourful World, applies the ideas of refraction and lenses learnt in Chapter 9 to the human eye and to nature. The eye works like a camera: the cornea does most of the refraction, the eye lens makes the fine adjustment, and an inverted real image is formed on the light-sensitive retina. The ability of the eye lens to change its focal length is called the power of accommodation. The chapter then studies the three common refractive defects — myopia (near-sightedness, corrected by a concave lens), hypermetropia (far-sightedness, corrected by a convex lens) and presbyopia (age-related, often needing bi-focal lenses). Finally it explains the colourful world: refraction and dispersion of white light by a prism, the rainbow, atmospheric refraction (twinkling of stars, advance sunrise and delayed sunset) and scattering of light (Tyndall effect, the blue sky and the red of danger signals).

Key Concepts & Definitions

Power of accommodation: the ability of the eye lens to adjust its focal length so that both near and distant objects are focused sharply on the retina. Ciliary muscles change the curvature of the lens.

Near point (least distance of distinct vision): the closest distance at which the eye can see an object distinctly without strain — about 25 cm for a normal young adult.

Far point: the farthest point up to which the eye can see clearly — infinity for a normal eye.

Myopia (near-sightedness): distant objects are blurred because their image forms in front of the retina; corrected with a concave lens.

Hypermetropia (far-sightedness): nearby objects are blurred because their image forms behind the retina; corrected with a convex lens.

Presbyopia: loss of accommodation with age; near point recedes. May need bi-focal lenses (concave above, convex below).

Dispersion: the splitting of white light into its seven component colours (VIBGYOR) by a prism; violet bends most, red least.

Atmospheric refraction: bending of light by air layers of changing density — causes twinkling of stars and the apparent advance of sunrise / delay of sunset.

Scattering of light (Tyndall effect): spreading of light by fine particles; fine particles scatter shorter (blue) wavelengths more, making the clear sky look blue.

Important Formulas (Chapter 10)

Power of a lens: P = 1 / f  (f in metres, P in dioptre, D). Convex (converging) lens → P positive; concave (diverging) lens → P negative.

Lens formula: 1/v − 1/u = 1/f, with the sign convention (distances measured from the optical centre; real-is-positive only along incident-light direction as per NCERT cartesian convention).

Myopia correction: object at infinity must appear at the far point ⇒ f = −(far point distance).

Hypermetropia correction: object at normal near point (25 cm) must appear at the eye’s own near point ⇒ u = −25 cm, v = −(near point of defective eye).

VIBGYOR: Violet, Indigo, Blue, Green, Yellow, Orange, Red — sequence of the spectrum.

In-text Questions (Page 164) — Solutions

Reproduced verbatim from the NCERT textbook; answers are original and exam-ready.

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 (change) its focal length so that objects at different distances are focused sharply on the retina. The ciliary muscles do this: when they relax, the lens becomes thin and its focal length increases, letting us see distant objects; when they contract, the lens becomes thicker and its focal length decreases, letting us see nearby objects. However, the focal length cannot fall below a certain minimum, which is why an object cannot be seen distinctly when held closer than about 25 cm.

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 has myopia (near-sightedness) with the far point at 1.2 m. To restore normal vision, a concave (diverging) lens is required. The lens must form a virtual image of a very distant object (at infinity) at the person’s far point, 1.2 m, so its focal length equals −1.2 m. Power P = 1/f = 1/(−1.2) ≈ −0.83 D. So a concave lens of power about −0.83 D (focal length −1.2 m) is needed.

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

ANSWER For a normal eye, the near point (least distance of distinct vision) is about 25 cm from the eye. The far point is at infinity. Thus a normal eye can see objects clearly anywhere between 25 cm and infinity.

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 student cannot see distant objects (the blackboard) clearly, so the child is most likely suffering from myopia (near-sightedness). In this defect, the image of a distant object forms in front of the retina, due to either excessive curvature of the eye lens or elongation of the eyeball. It can be corrected by using spectacles fitted with a concave (diverging) lens of suitable power, which diverges the rays slightly so that the image is shifted back exactly onto the retina.

End-of-Chapter Exercises — Solutions

All questions reproduced verbatim from the NCERT “Exercises”; numericals verified with units.

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. The ciliary muscles change the curvature, and hence the focal length, of the eye lens to focus objects at different distances — this property is called the power of accommodation.

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

ANSWER (d) retina. The eye lens forms an inverted, real image on the light-sensitive retina, whose cells send signals to the brain through the optic nerve.

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. This is the near point of a normal young adult eye — the closest distance at which an object can be seen distinctly without strain.

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. By contracting or relaxing, the ciliary muscles change the curvature of the eye lens, which changes its focal length. (The pupil and iris only control the amount of light entering the eye.)

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 Given: formula P = 1/f, so f = 1/P (with f in metres). (i) Distant vision: P = −5.5 D ⇒ f = 1/(−5.5) = −0.182 m = −18.2 cm (a concave lens, since f is negative). (ii) Near vision: P = +1.5 D ⇒ f = 1/(+1.5) = +0.667 m = +66.7 cm (a convex lens, since f is positive).

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 Nature: the person has myopia, so a concave (diverging) lens is required. Working: the lens must form a virtual image of a very distant object (u = −∞) at the far point, 80 cm in front of the eye, so v = −80 cm = −0.8 m. Using 1/f = 1/v − 1/u = 1/(−0.8) − 1/(−∞) = −1.25 − 0 = −1.25, so f = −0.8 m. Power: P = 1/f = 1/(−0.8) = −1.25 D. A concave lens of focal length −80 cm (power −1.25 D) corrects the defect.

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 Diagram (described in words): Draw the eyeball with a convex correcting lens in front of it. Rays from a near object placed at 25 cm enter the convex lens; without correction these rays would converge to a point behind the retina (the defect). The convex lens converges them more so that, after passing through the eye lens, they meet exactly on the retina. Effectively, the convex lens takes the object at 25 cm and forms its virtual image at the eye’s own near point, 1 m, which the eye can then see clearly. Working: object distance u = −25 cm = −0.25 m; the lens must form a virtual image at the defective near point, so v = −100 cm = −1 m. 1/f = 1/v − 1/u = 1/(−1) − 1/(−0.25) = −1 + 4 = +3. So f = 1/3 m ≈ +0.333 m (+33.3 cm). Power: P = 1/f = +3 = +3.0 D. A convex lens of power +3.0 D (focal length about +33.3 cm) corrects this hypermetropia.

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

ANSWER To focus a nearer object, the eye lens must become thicker so that its focal length decreases. But the focal length of the eye lens cannot decrease below a certain minimum limit, set by the maximum curvature the ciliary muscles can produce. For a normal eye that minimum corresponds to an object distance of about 25 cm (the near point). For objects closer than this, the lens cannot reduce its focal length enough, so the image is not formed on the retina and appears blurred, with strain on the eye.

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 stays practically the same — the image is always formed on the retina, which is at a fixed distance from the eye lens. When the object moves farther away, the eye keeps the image on the retina by increasing the focal length of the eye lens (the ciliary muscles relax and the lens becomes thinner). So the focal length changes, but the image distance does not.

10. Why do stars twinkle?

ANSWER Stars twinkle because of atmospheric refraction of starlight. Stars are so far away that they act as point sources of light. As starlight enters the earth’s atmosphere, it is refracted continuously through layers of air of gradually changing refractive index, bending the light towards the normal, so the star appears slightly higher than its true position. Because the physical conditions of the atmosphere (temperature, density) keep changing, the apparent position of the star and the amount of starlight reaching the eye keep fluctuating. The star therefore appears bright at one moment and faint the next — this flickering is the twinkling effect.

11. Explain why the planets do not twinkle.

ANSWER Planets are much closer to the earth than stars, so they appear as extended sources of light rather than point sources. A planet can be thought of as a large collection of point-sized sources of light. The brightness from these many points fluctuates independently, so the total variation in the amount of light reaching the eye averages out to nearly zero. This cancels the twinkling effect, and so planets shine with a steady light instead of twinkling.

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 sunlight by the fine molecules and particles of the atmosphere, which scatter shorter (blue) wavelengths more strongly. At very high altitudes where an astronaut is, there is little or no atmosphere, so there are almost no particles to scatter sunlight. With no scattering, no scattered (blue) light reaches the eye, and the sky therefore appears dark (black) instead of blue.

Extra Practice Questions

Short Answer Type Questions

Q1. Name the part of the eye that does most of the refraction of incoming light.

ANSWERThe cornea — the transparent front bulge of the eye — carries out most of the refraction; the eye lens only makes the fine adjustment of focal length.

Q2. What is the function of the iris and the pupil?

ANSWERThe iris is a dark muscular diaphragm that controls the size of the pupil; the pupil regulates the amount of light entering the eye — it widens in dim light and narrows in bright light.

Q3. Why does the red light glow as the “danger” or “stop” signal?

ANSWERRed light has the longest wavelength among visible colours, so it is scattered the least by fog or smoke. It therefore travels the farthest with its colour unchanged and remains visible from a long distance, making it ideal for danger signals.

Q4. Why is the colour of the clear sky blue?

ANSWERThe fine molecules and particles of the atmosphere are smaller than the wavelength of visible light, and such fine particles scatter shorter (blue) wavelengths much more strongly than longer (red) wavelengths. This strongly scattered blue light reaches our eyes, so the clear sky appears blue.

Q5. What is the angle of deviation in a prism?

ANSWERWhen light passes through a prism, the emergent ray bends away from the direction of the incident ray. The angle between the direction of the incident ray (produced forward) and the emergent ray is called the angle of deviation (∠D).

Long Answer Type Questions

Q1. Explain the dispersion of white light by a glass prism. Why does a rainbow form in the sky?

ANSWERWhen a narrow beam of white light passes through a glass prism, it splits into a band of seven colours — Violet, Indigo, Blue, Green, Yellow, Orange, Red (VIBGYOR). This splitting is called dispersion. It happens because different colours bend through different angles in the prism: violet light bends the most and red light the least, so each colour emerges along a slightly different path and becomes distinct. This band of colours is the spectrum. Isaac Newton showed with a second, inverted prism that these colours recombine into white light, proving that sunlight is made up of seven colours. A rainbow is a natural spectrum: tiny water droplets in the atmosphere after a shower act like small prisms. Each droplet refracts and disperses the entering sunlight, reflects it internally, and refracts it again on leaving. Because of this dispersion and internal reflection, different colours emerge at slightly different angles and reach the observer’s eye, forming the coloured arc. A rainbow always appears in the direction opposite to the Sun.

Q2. Describe the three common refractive defects of the eye and how each is corrected.

ANSWER(i) Myopia (near-sightedness): the person can see nearby objects clearly but not distant ones; the image of a distant object forms in front of the retina, due to excessive curvature of the eye lens or elongation of the eyeball. It is corrected with a concave lens of suitable power, which diverges the rays so the image forms on the retina. (ii) Hypermetropia (far-sightedness): the person can see distant objects clearly but not nearby ones; the image of a near object forms behind the retina, because the focal length of the eye lens is too long or the eyeball too short. It is corrected with a convex lens of suitable power, which converges the rays so the image forms on the retina. (iii) Presbyopia: an age-related defect in which the power of accommodation decreases as the ciliary muscles weaken and the lens loses flexibility, so the near point recedes. A person suffering from both myopia and hypermetropia needs bi-focal lenses — a concave portion on top for distant vision and a convex portion below for near vision.

Q3. Explain atmospheric refraction and describe two phenomena caused by it.

ANSWERAtmospheric refraction is the refraction of light by the earth’s atmosphere, which consists of layers of air with gradually changing density and hence changing refractive index. As light passes through these layers it bends continuously. (1) Twinkling of stars: starlight is refracted continuously while passing through the atmosphere; since stars are point sources and atmospheric conditions keep changing, the apparent position of a star and the amount of light reaching the eye keep fluctuating, so the star appears to twinkle. (2) Advance sunrise and delayed sunset: due to atmospheric refraction, the Sun is visible about 2 minutes before the actual sunrise and about 2 minutes after the actual sunset, because the light from the Sun (below the horizon) is bent towards us by the atmosphere. The apparent flattening of the Sun’s disc at sunrise and sunset is also due to this refraction.

MCQs & Assertion–Reason

1. Most of the refraction of light entering the eye takes place at the:

(a) eye lens    (b) retina    (c) cornea    (d) iris

2. The far point of a normal human eye is at:

(a) 25 cm    (b) infinity    (c) 2.5 cm    (d) 1 m

3. Myopia is corrected by using a:

(a) convex lens    (b) concave lens    (c) bifocal lens    (d) cylindrical lens

4. In hypermetropia, the image of a nearby object is formed:

(a) in front of the retina    (b) on the retina    (c) behind the retina    (d) on the cornea

5. Which colour of white light is deviated the most by a prism?

(a) red    (b) green    (c) yellow    (d) violet

6. The splitting of white light into its component colours is called:

(a) refraction    (b) dispersion    (c) scattering    (d) reflection

7. The twinkling of stars is due to:

(a) scattering of light    (b) dispersion of light    (c) atmospheric refraction    (d) total internal reflection

8. The clear sky appears blue because air molecules scatter:

(a) red light most    (b) blue light most    (c) all colours equally    (d) no light at all

9. A person needing a lens of power +2 D to read has the defect:

(a) myopia    (b) hypermetropia    (c) cataract    (d) colour-blindness

10. Danger signal lights are red because red light is:

(a) scattered the most    (b) scattered the least    (c) absorbed by fog    (d) the brightest colour

Answer key: 1-(c), 2-(b), 3-(b), 4-(c), 5-(d), 6-(b), 7-(c), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: A normal eye cannot see objects clearly when held closer than 25 cm.

Reason: The focal length of the eye lens cannot be decreased below a certain minimum limit.

A-R 2. Assertion: Myopia is corrected by a convex lens.

Reason: In myopia the image of a distant object is formed in front of the retina.

A-R 3. Assertion: Planets do not twinkle but stars do.

Reason: Planets are extended sources of light, so the variation in light reaching the eye averages out to nearly zero.

A-R 4. Assertion: The sky appears dark to an astronaut at high altitude.

Reason: At high altitudes there is little atmosphere, so there is almost no scattering of sunlight.

A-R 5. Assertion: Violet light bends more than red light while passing through a prism.

Reason: Different colours of light deviate through different angles in a prism.

Answer key: 1-(A), 2-(D), 3-(A), 4-(A), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Mixing up the corrections: myopia → concave lens (negative power), hypermetropia → convex lens (positive power). Do not swap them.
  • Forgetting sign conventions in numericals — distances measured against the incident-light direction are negative; a negative focal length means a concave lens.
  • Writing that the image distance in the eye changes when an object moves — it is the focal length that changes; the image always stays on the retina.
  • Confusing dispersion (splitting of white light by a prism) with scattering (spreading of light by fine particles, which gives the blue sky).
  • Saying stars do not twinkle — it is the planets that do not twinkle.
  • Stating the near point as 25 m or 2.5 cm — it is 25 cm for a normal eye.

How to score full marks in this chapter

For every numerical, write the formula P = 1/f, convert distances to metres, show the sign of f, and finish with the correct unit (D for power, m or cm for focal length). State the nature of the lens (concave/convex) along with its power. For theory answers on twinkling, blue sky and the rainbow, always name the exact phenomenon (atmospheric refraction, scattering, or dispersion + internal reflection) and explain it in one or two clear sentences. Remember VIBGYOR and the “violet bends most, red bends least” rule.

Frequently Asked Questions

What is Class 10 Science Chapter 10 about?

Chapter 10, The Human Eye and the Colourful World, explains the structure and working of the human eye, its power of accommodation, the three common refractive defects (myopia, hypermetropia and presbyopia) and their correction, and optical phenomena in nature — refraction and dispersion of light by a prism, the rainbow, atmospheric refraction, and the scattering of light that makes the sky blue.

How is myopia different from hypermetropia?

In myopia (near-sightedness), a person sees nearby objects clearly but not distant ones because the image forms in front of the retina; it is corrected with a concave lens. In hypermetropia (far-sightedness), a person sees distant objects clearly but not nearby ones because the image forms behind the retina; it is corrected with a convex lens.

Why does the sky appear blue?

The fine molecules and particles of the atmosphere scatter shorter (blue) wavelengths of sunlight much more strongly than longer (red) wavelengths. This strongly scattered blue light reaches our eyes from all directions, so the clear sky looks blue. Where there is no atmosphere, the sky looks dark.

Are these Class 10 Science Chapter 10 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 10 Science are free and follow the official NCERT textbook for session 2026–27, with every exercise and in-text question solved and numericals verified.

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