---
title: "NCERT Solutions Class 10 Science The Human Eye and the Colourful World"
url: https://www.swavid.com/science/class/10/chapter/the-human-eye-and-the-colourful-world/ncert-solutions
dateModified: 2026-10-07T16:45:46+00:00
---

# NCERT Solutions Class 10 Science The Human Eye and the Colourful World

This chapter's questions cover concepts related to the human eye, its defects of vision, power of accommodation, and various atmospheric and optical phenomena such as refraction, dispersion, and scattering of light.

Free PDF (7 pages): https://www.swavid.com/api/seo/pdf/ncert/science/class-10/swavid-ncert-solutions-class-10-science-chapter-10-the-human-eye-and-the-colourful-world-9f3b5fff55.pdf

## QUESTIONS

### Question 1

*2 marks · Very short answer*

What is meant by power of accommodation of the eye?

**Solution**

1. The ability of the eye lens to adjust its focal length is called accommodation.
2. This ability allows the eye to focus on both near and distant objects clearly.

**Answer:** The power of accommodation is the ability of the eye lens to adjust its focal length to see objects at different distances clearly.

> Common mistake: Confusing power of accommodation with the least distance of distinct vision.

### Question 2

*3 marks · Numerical*

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?

**Solution**

1. Given: Far point of the myopic person, $d = 1.2\text{ m} = 120\text{ cm}$.
2. Formula: To correct myopia, a concave lens is used whose focal length is equal to the far point of the given myopic eye, so $f = -1.2\text{ m}$.
3. Substitution: The corrective lens should be a concave lens.

**Answer:** Concave lens

> Common mistake: Forgetting to specify the sign or nature of the lens.

### Question 3

*2 marks · Very short answer*

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

**Solution**

1. The near point of the human eye with normal vision is about $25\text{ cm}$.
2. The far point of the human eye with normal vision is at infinity.

**Answer:** The near point is $25\text{ cm}$ and the far point is infinity.

> Common mistake: Writing incorrect units for the far point.

### Question 4

*3 marks · Short answer*

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?

**Solution**

1. The student is suffering from myopia or near-sightedness, where a person cannot see distant objects distinctly.
2. This defect arises because the image of a distant object is formed in front of the retina.
3. It can be corrected by using a concave lens of suitable power.

**Answer:** The child is suffering from myopia and it can be corrected by using a concave lens.

> Common mistake: Confusing myopia with hypermetropia.

## EXERCISES

### Question 1

*1 mark · MCQ*

The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to

- presbyopia.
- accommodation.
- near-sightedness.
- far-sightedness.

**Solution**

1. The ability of the eye lens to adjust its focal length is called accommodation.

**Answer:** (b) accommodation.

> Common mistake: Confusing accommodation with refractive defects like presbyopia.

### Question 2

*1 mark · MCQ*

The human eye forms the image of an object at its

- cornea.
- iris.
- pupil.
- retina.

**Solution**

1. The eye lens forms an inverted real image of the object on a light-sensitive screen called the retina.

**Answer:** (d) retina.

> Common mistake: Choosing cornea where most refraction occurs instead of retina where the image is formed.

### Question 3

*1 mark · MCQ*

The least distance of distinct vision for a young adult with normal vision is about

- 25 m.
- 2.5 cm.
- 25 cm.
- 2.5 m.

**Solution**

1. For a young adult with normal vision, the near point or least distance of distinct vision is about 25 cm.

**Answer:** (c) 25 cm.

> Common mistake: Confusing centimeters with meters.

### Question 4

*1 mark · MCQ*

The change in focal length of an eye lens is caused by the action of the

- pupil.
- retina.
- ciliary muscles.
- iris.

**Solution**

1. The curvature of the eye lens can be modified to some extent by the ciliary muscles, changing its focal length.

**Answer:** (c) ciliary muscles.

> Common mistake: Confusing ciliary muscles with iris, which controls the size of the pupil.

### Question 5

*3 marks · Numerical*

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?

**Part (i)**

1. Given: Power for distant vision $P_1 = -5.5\text{ D}$
2. Formula: $f = \frac{1}{P}$
3. Substitution: $f_1 = \frac{1}{-5.5} = -0.182\text{ m}$

Answer (i): $-0.182\text{ m}$ (or $-18.2\text{ cm}$)

**Part (ii)**

1. Given: Power for near vision $P_2 = +1.5\text{ D}$
2. Formula: $f = \frac{1}{P}$
3. Substitution: $f_2 = \frac{1}{+1.5} = +0.667\text{ m}$

Answer (ii): $+0.667\text{ m}$ (or $+66.7\text{ cm}$)

**Answer:** Focal length for distant vision is $-0.182\text{ m}$ and for near vision is $+0.667\text{ m}$.

> Common mistake: Forgetting the negative sign for concave lens focal length.

### Question 6

*3 marks · Numerical*

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?

**Solution**

1. Given: Far point of the myopic person $d = 80\text{ cm} = -0.8\text{ m}$
2. The corrective concave lens must virtual-image an object at infinity to the person's far point, so object distance $u = -\infty$ and image distance $v = -80\text{ cm} = -0.8\text{ m}$.
3. Formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
4. Substitution: $\frac{1}{f} = \frac{1}{-0.8} - \frac{1}{-\infty} = \frac{1}{-0.8}$
5. Result: $f = -0.8\text{ m}$, Power $P = \frac{1}{f} = \frac{1}{-0.8\text{ m}} = -1.25\text{ D}$

**Answer:** Nature of lens is concave and power is $-1.25\text{ D}$.

> Common mistake: Taking positive sign for the focal length of a myopic correction lens.

### Question 7

*5 marks · Long answer*

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.

**Solution**

1. Diagram: Draw Fig. 10.3(c) from the textbook showing correction of hypermetropia using a convex lens, with the object placed at the normal near point $N$ ($25\text{ cm}$), its virtual image formed at the defective near point $N'$ ($1\text{ m}$), and the ray diagram correctly labelled.
2. Given: Object distance $u = -25\text{ cm}$ (near point of a normal eye), Image distance $v = -100\text{ cm}$ (near point of the hypermetropic eye).
3. Formula: Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
4. Substitution: $\frac{1}{f} = \frac{1}{-100} - \frac{1}{-25} = -\frac{1}{100} + \frac{1}{25} = \frac{-1 + 4}{100} = \frac{3}{100}\text{ cm}^{-1}$.
5. Focal length: $f = \frac{100}{3}\text{ cm} = +\frac{1}{3}\text{ m} = +0.33\text{ m}$.
6. Power of the lens: $P = \frac{1}{f(\text{in metres})} = \frac{1}{1/3} = +3.0\text{ D}$.

**Answer:** $+3.0\text{ D}$

> Common mistake: Taking positive signs for object and image distances in the lens formula or forgetting to convert cm into meters while calculating power.

### Question 8

*3 marks · Short answer*

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

**Solution**

1. The ciliary muscles of a normal eye cannot contract beyond a certain limit to increase the curvature and thickness of the eye lens further.
2. Consequently, the focal length of the eye lens cannot be decreased below a certain minimum limit.
3. Therefore, objects placed closer than $25\text{ cm}$ (the near point) cannot be focused clearly on the retina and appear blurred, causing eye strain.

**Answer:** A normal eye cannot see objects closer than $25\text{ cm}$ clearly because the ciliary muscles cannot increase the curvature of the eye lens beyond a certain limit to decrease its focal length further.

> Common mistake: Stating that the retina cannot receive light instead of explaining the limitation of the ciliary muscles and focal length.

### Question 9

*3 marks · Short answer*

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

**Solution**

1. The distance between the eye lens and the retina (the image distance) is fixed and does not change regardless of the position of the object.
2. When we increase the distance of an object from the eye, the ciliary muscles relax to make the eye lens thinner.
3. This increases the focal length of the eye lens such that the image is still focused sharply on the retina.

**Answer:** The image distance in the eye remains constant because the distance between the eye lens and the retina is fixed; the eye adjusts its focal length using ciliary muscles to keep the image on the retina.

> Common mistake: Assuming that image distance changes like in a normal camera with a movable lens.

### Question 10

*3 marks · Short answer*

Why do stars twinkle?

**Solution**

1. Stars are very distant and act as point-sized sources of light.
2. As starlight enters the Earth's atmosphere, it undergoes continuous refraction in a medium of gradually changing refractive index.
3. Due to changing physical conditions of the earth's atmosphere, the apparent position and amount of starlight reaching the eye fluctuate, causing the twinkling effect.

**Answer:** Stars twinkle due to the atmospheric refraction of point-sized starlight passing through layers of varying refractive indices and densities.

> Common mistake: Writing that stars themselves produce flickering light instead of attributing it to atmospheric refraction.

### Question 11

*3 marks · Short answer*

Explain why the planets do not twinkle.

**Solution**

1. Planets are much closer to the Earth and are seen as extended sources of light rather than point-sized sources.
2. A planet can be considered as a collection of a large number of point-sized light sources.
3. The total variation in the amount of light entering our eye from all these individual point-sized sources averages out to zero, nullifying the twinkling effect.

**Answer:** Planets do not twinkle because they are closer, acting as extended sources of light where fluctuations from individual points average out to zero.

> Common mistake: Stating that planets do not reflect sunlight.

### Question 12

*3 marks · Short answer*

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

**Solution**

1. The blue colour of the sky is due to the scattering of sunlight by fine dust particles and air molecules in the atmosphere.
2. An astronaut flies at very high altitudes where the atmosphere is absent or extremely thin.
3. In the absence of sufficient atmosphere and scattering particles, no scattered light reaches the astronaut's eyes, making the sky appear dark.

**Answer:** The sky appears dark to an astronaut because there is no atmosphere at high altitudes to scatter sunlight.

> Common mistake: Saying that there is no sunlight in space instead of explaining the lack of atmospheric scattering.

## Frequently asked questions

### How many questions are there in the NCERT solutions for Class 10 Science Chapter 10?

This chapter contains 4 questions in the in-text exercises and 12 questions in the main exercises. SwaVid provides free PDF and step-by-step solutions for all of them on this page only.

### Which topics do the questions in this chapter cover?

The questions cover important concepts like far point and near point, myopia and its correction, power of accommodation, atmospheric refraction, ciliary muscles, and least distance of distinct vision. You can find detailed explanations for these topics in SwaVid's free PDF and step-by-step solutions available on this page only.

### What are the hardest question types in this chapter and how should I approach them?

Numerical problems involving lens formula, focal length, power of lens calculation, and defect correction are usually considered the hardest. To approach them, first identify the given values for image and object distances, apply the correct sign convention, and use KaTeX formulas like $P = \frac{1}{f}$ carefully.

### How can I write answers to score full marks in Class 10 board exams for this chapter?

To secure full marks, write clear steps for numericals, include proper SI units, and draw neat ray diagrams for eye defects like myopia and hypermetropia. SwaVid's free PDF and step-by-step solutions on this page only demonstrate the exact answering style required by examiners.

### Is the free PDF for these NCERT solutions available for download?

Yes, you can easily access and download the complete study material for this chapter. SwaVid's free PDF and step-by-step solutions are on this page only to help you revise offline.

## Related pages

- [The Human Eye and the Colourful World: CBSE previous year questions](https://www.swavid.com/cbse/class-10/science/pyq/the-human-eye-and-the-colourful-world)
- [Class 10 Science chapters](https://www.swavid.com/science/class/10)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student