NCERT Solutions for CBSE Class 10 Science Chapter 11 – Human Eye and Colourful World

Human Eye and Colourful World is one of the most important physics chapters for CBSE Class 10 board exams. It covers eye structure, vision defects, and light phenomena like dispersion, scattering, and atmospheric refraction. Below are complete, step-by-step NCERT solutions to help you revise and score better.

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In-Text Questions

Q1. What is meant by power of accommodation of the eye?

Answer: The eye lens can change its curvature (and hence its focal length) with the help of the ciliary muscles, allowing it to focus objects at different distances clearly on the retina. This ability is called the power of accommodation.

Q2. A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What type of corrective lens is required to restore proper vision?

Answer: This is a case of myopia (short-sightedness), where the far point of the eye has shifted from infinity to 1.2 m. To correct this, a concave lens is needed whose focal length equals the person’s far point, so that light rays from infinity appear to diverge from that near point after passing through the lens.

• Far point = 1.2 m, so focal length of the required lens, f = –1.2 m
• Power, P = 1/f = 1/(–1.2) = –0.83 D

A concave lens of power approximately –0.83 D is required.

Q3. What is the far point and near point of the human eye with normal vision?

Answer:

• Near point: The closest distance at which an object can be seen clearly without eye strain. For a normal eye, this is 25 cm.
• Far point: The farthest distance up to which the eye can see clearly. For a normal eye, this is at infinity.

Q4. A student has difficulty reading the blackboard while sitting in the last row of the classroom. What could be the defect, and how can it be corrected?

Answer: The student is likely suffering from myopia (short-sightedness) — they can see nearby objects clearly but struggle with distant ones. This happens because the image of a distant object forms in front of the retina rather than on it. It can be corrected using a concave lens of suitable power, which diverges the incoming light rays so the image shifts back onto the retina.

NCERT Exercise Questions (Page 197–198)

Q1. 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 adjust the curvature of the eye lens, changing its focal length so that both near and distant objects can be focused sharply on the retina.

Q2. The human eye forms the image of an object at its:

(a) cornea (b) iris (c) pupil (d) retina

Answer: (d) retina. The retina is the light-sensitive layer at the back of the eye. It contains rod and cone cells that convert incoming light into electrical signals, which the optic nerve carries to the brain to form the image we perceive.

Q3. 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. Objects closer than this distance cannot be seen clearly without straining the eye.

Q4. The change in focal length of the eye lens is caused by the action of the:

(a) pupil (b) retina (c) ciliary muscles (d) iris

Answer: (c) ciliary muscles. These muscles contract or relax to change the curvature of the lens, adjusting its focal length so the eye can focus on objects at varying distances.

Q5. A person needs a lens of power –5.5 D for correcting distant vision and a lens of power +1.5 D for correcting near vision. Find the focal length required for (i) distant vision and (ii) near vision.

Answer:
Using Power (P) = 1/f (f in metres):

(i) Distant vision:
P = –5.5 D
f = 1/(–5.5) = –0.181 m (approx.)

(ii) Near vision:
P = +1.5 D
f = 1/1.5 = +0.667 m (approx.)

Q6. 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 this defect?

Answer: In myopia, distant objects appear blurred because their image forms in front of the retina. A concave lens is used to correct this — it diverges incoming parallel rays so the image forms exactly at the person’s far point instead of at infinity.

Given: Object distance, u = ∞; Image distance, v = –80 cm = –0.8 m

Using the lens formula:
1/f = 1/v – 1/u = 1/(–0.8) – 0 = –1.25

f = –0.8 m, so Power P = 1/f = –1.25 D

A concave lens of power –1.25 D is required.

Q7. Draw a diagram showing how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What power of lens is required to correct this defect, assuming the normal near point is 25 cm?

Answer: In hypermetropia (far-sightedness), a person can see distant objects clearly but nearby objects appear blurred, because the image of a close object forms behind the retina. A convex lens is used to correct this — it converges the incoming rays before they enter the eye, moving the image forward onto the retina.

Given: Object distance, u = –25 cm = –0.25 m; Image distance (at the person’s near point), v = –1 m

Using the lens formula:
1/f = 1/v – 1/u = 1/(–1) – 1/(–0.25) = –1 + 4 = 3

f = 1/3 m, so Power P = +3 D

A convex lens of power +3 D is required. (Diagram: rays from a nearby object converge through a convex lens before entering the eye, forming a sharp image on the retina instead of behind it.)

Additional Important Questions

Q8. Why do we have two eyes for vision instead of one?

Answer: Having two eyes gives us binocular vision — a wider field of view (nearly 180°) and, more importantly, the ability to judge distance and depth accurately (3D vision), since each eye sees a slightly different image and the brain combines them.

Q9. Why do stars twinkle, but planets do not?

Answer: Stars are extremely far away and act as point sources of light. As starlight passes through the constantly shifting layers of Earth’s atmosphere, it undergoes continuous refraction, causing its apparent position and brightness to fluctuate — this is seen as twinkling. Planets, being much closer, appear as extended sources (a collection of many point sources), so the fluctuations from different points average out, and no twinkling is observed.

Q10. Why does the Sun appear reddish at sunrise and sunset?

Answer: At sunrise and sunset, sunlight has to pass through a much thicker layer of the atmosphere to reach an observer. Along this longer path, shorter wavelengths (blue, violet) are scattered away almost completely, leaving mostly the longer wavelength red and orange light to reach our eyes — making the Sun appear reddish.

Q11. Why does the sky appear dark instead of blue to an astronaut in space?

Answer: The blue colour of the sky is caused by the scattering of sunlight by air molecules in the atmosphere. In space, there is no atmosphere and hence no medium to scatter sunlight, so there is no scattered light reaching the eyes — this is why astronauts see a dark or black sky even during “daytime.”

Quick Revision Table

Concept	Key Value
Near point (normal eye)	25 cm
Far point (normal eye)	Infinity
Myopia correction	Concave lens
Hypermetropia correction	Convex lens
Presbyopia correction	Bifocal lens

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