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NCERT Solutions

The Human Eye and the Colourful World — NCERT Solutions

CBSE · Class 10 · Science

NCERT Solutions for The Human Eye and the Colourful World, CBSE Class 10 Science: 12 textbook questions solved step by step. Covers Exercises.

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12 Questions Solved · 1 Section

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Exercises

1The 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.Show solution

Correct option: (b) accommodation.

The ability of the eye to adjust its focal length (by changing the curvature of the eye lens through ciliary muscles) to focus on objects at different distances is called accommodation. Presbyopia is the loss of this ability in old age; near-sightedness and far-sightedness are refractive defects, not the focusing mechanism itself.

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

Correct option: (d) retina.

The eye lens forms a real, inverted image of the object on the retina, which is the light-sensitive screen at the back of the eye. The retina contains photoreceptor cells (rods and cones) that convert the light signal into electrical signals sent to the brain via the optic nerve.

3The 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.Show solution

Correct option: (c) 25 cm.

The least distance of distinct vision (also called the near point) is the minimum distance at which the eye can see an object clearly without strain. For a young adult with normal vision, this distance is approximately 25 cm.

4The change in focal length of an eye lens is caused by the action of the (a) pupil. (b) retina. (c) ciliary muscles. (d) iris.Show solution

Correct option: (c) ciliary muscles.

The ciliary muscles hold the eye lens in position and can contract or relax to change the curvature (and hence the focal length) of the eye lens. When ciliary muscles contract, the lens becomes thicker (shorter focal length) for near objects; when they relax, the lens becomes thinner (longer focal length) for distant objects.

5A 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?Show solution

Given:

  • Power of lens for distant vision, P1=−5.5P_1 = -5.5 D
  • Power of lens for near vision, P2=+1.5P_2 = +1.5 D

Formula used:
f=1P(where f is in metres and P is in dioptres)f = \frac{1}{P} \quad (\text{where } f \text{ is in metres and } P \text{ is in dioptres})

(i) Focal length for correcting distant vision:
f1=1P1=1−5.5 mf_1 = \frac{1}{P_1} = \frac{1}{-5.5} \text{ m}
f1=−0.182 m≈−18.2 cmf_1 = -0.182 \text{ m} \approx -18.2 \text{ cm}

The negative sign indicates it is a concave (diverging) lens.

(ii) Focal length for correcting near vision:
f2=1P2=1+1.5 mf_2 = \frac{1}{P_2} = \frac{1}{+1.5} \text{ m}
f2=+0.667 m≈+66.7 cmf_2 = +0.667 \text{ m} \approx +66.7 \text{ cm}

The positive sign indicates it is a convex (converging) lens.

Answers:

  • Focal length for distant vision correction = –18.2 cm (concave lens)
  • Focal length for near vision correction = +66.7 cm (convex lens)
6The 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?Show solution

Given:

  • Far point of the myopic person = 80 cm = 0.80 m (in front of the eye)
  • A myopic person cannot see objects beyond 80 cm clearly.

Concept:
To correct myopia, we need a lens that forms the image of a distant object (at infinity) at the person's far point (80 cm in front of the eye).

So, object distance u=−∞u = -\infty and image distance v=−80v = -80 cm =−0.80= -0.80 m (image is on the same side as the object, hence negative by sign convention).

Using the lens formula:
1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u}
1f=1−80−1−∞\frac{1}{f} = \frac{1}{-80} - \frac{1}{-\infty}
1f=−180−0=−180 cm−1\frac{1}{f} = \frac{-1}{80} - 0 = \frac{-1}{80} \text{ cm}^{-1}
f=−80 cm=−0.80 mf = -80 \text{ cm} = -0.80 \text{ m}

Power of the lens:
P=1f(in metres)=1−0.80P = \frac{1}{f(\text{in metres})} = \frac{1}{-0.80}
P=−1.25 D\boxed{P = -1.25 \text{ D}}

Conclusion: The person requires a concave (diverging) lens of power –1.25 dioptres to correct myopia.

7Make 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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8Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

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

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10Why do stars twinkle?

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11Explain why the planets do not twinkle.

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

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Frequently Asked Questions

What are the important topics in The Human Eye and the Colourful World for CBSE Class 10 Science?
Key topics in The Human Eye and the Colourful World include Structure and Working of the Human Eye, Power of Accommodation and Vision Limits, Defects of Vision and Their Correction, Prism, Dispersion and Rainbow. Study these first, then practise questions on each for the CBSE Class 10 board exam.
Are these NCERT Solutions for The Human Eye and the Colourful World free?
The first 6 of the 12 solutions on this page are open to read. The other 6 are free with a Super Tutor account — signing up is free and needs no card.
How should I revise The Human Eye and the Colourful World for the CBSE Class 10 board exam?
Learn the core ideas first, then work through the 56 practice questions on The Human Eye and the Colourful World. Revise definitions regularly and use flashcards for quick recall before the exam.

Sources & Official References

Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.

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