This captivating chapter from NCERT Solutions for Class 10 Science Chapter 10,The Human Eye and the Colourful World explores the fascinating science behind human vision and the colorful natural phenomena we witness daily. From understanding how our eyes function as nature's perfect optical instrument to discovering why the sky is blue and sunsets are red, this chapter bridges the gap between biology and physics in the most visually stunning way. This chapter is part of the comprehensive NCERT Solutions Class 1o Science series, which covers all chapters in detail.
The chapter covers the detailed structure of the human eye, common vision defects like myopia and hypermetropia and their correction, and the beautiful optical phenomena created by atmospheric conditions—including rainbows, the twinkling of stars, and the scattering of light. Every solution has been designed keeping CBSE board exam patterns in mind, with clear ray diagrams, scientific explanations, and numerical problems that ensure students develop both conceptual clarity and problem-solving confidence for their examinations.
NCERT Solutions for Class 10 Science Chapter 10 - All Exercise Questions
Q.
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.
Q.
The human eye forms the image of an object at its
(a) cornea.
(b) iris.
(c) pupil.
(d) retina.
Q.
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.
Q.
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.
Q.
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?
Q.
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?
Q.
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.
Q.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Q.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Q.
Explain why the planets do not twinkle.
Q.
Why does the Sun appear reddish early in the morning?
Class 10 Chapter 10 Science Questions & Answers –The Human Eye and the Colourful World
Q.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.
Solution:
The correct option is (b).
Explanation: Power of accommodation is the ability of eye lens to change its focal length so that the image of an object can be focused on the retina.
Q2. The human eye forms the image of an object at its
(a) cornea.
(b) iris.
(c) pupil.
(d) retina.
Solution:
The correct option is (d).
Explanation: In the case of human eye, image is formed at retina.
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.
Solution:
The correct option is (c).
Explanation: Least distance of distinct vision is the smallest distance at which the human eye can see the objects clearly without any strain. For a normal eye, it is 25 cm.
Q4. 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.
Solution:
The correct option is (c).
Explanation: The curvature of the eye lens can be changed by the relaxation or contraction of ciliary muscles. The change in curvature of the eye lens changes the focal length of the eyes. Thus, the change in focal length of an eye lens is caused by the action of ciliary muscles.
Q5. 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 dioptres. What is the focal length of the lens required for correcting:
(i) distant vision, and
(ii) near vision?
Solution:
(i)Power of the lens used for correcting distant vision: \( P = -5.5\,\text{D} \)
The focal length \( f \) of the required lens is:
\[
f = \frac{1}{P} = \frac{1}{-5.5} = -0.181\,\text{m}
\]
Hence, the focal length of the lens for correcting distant vision is -0.181 m.
(ii)Power of the lens used for correcting near vision: \( P = +1.5\,\text{D} \)
The focal length \( f \) of the required lens is:
\[
f = \frac{1}{P} = \frac{1}{1.5} = +0.667\,\text{m}
\]
Hence, the focal length of the lens for correcting near vision is +0.667 m.
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 the problem?
Solution:
Given:
Object distance, \( u = \infty \) (since the far point is at infinity)
Image distance, \( v = -80\,\text{cm} \)
Using the lens formula:
\[
\frac{1}{v} - \frac{1}{u} = \frac{1}{f}
\]
Substituting the values:
\[
\frac{1}{-80} - \frac{1}{\infty} = \frac{1}{f}
\]
This simplifies to:
\[
\frac{1}{f} = \frac{-1}{80}
\]
So,
\[
f = -80\,\text{cm} = -0.8\,\text{m}
\]
Now, the power \( P \) of the lens is:
\[
P = \frac{1}{f} = \frac{1}{-0.8} = -1.25\,\text{D}
\]
Therefore, a concave lens of power -1.25 D is required to correct the defect.
Q7. A hypermetropic person has a near point at 1 meter. What is the power of the lens required to correct this defect? Assume that the near point of a normal eye is 25 cm.
Solution: It is a visual defect in which the eye cannot see nearby objects clearly. The image of the object is formed beyond the retina, causing the person to experience difficulty in seeing the object.
This defect can be corrected by using a convex lens. A convex lens of suitable power converges the incoming light in such a way that the image is formed on the retina. The convex lens actually creates a virtual image of a nearby object at the near point of vision (o) for the person suffering from hypermetropia.
A person can clearly see the object kept at 25 cm if the image of the object is formed at his near point, which is given as 1 m.
Given:
- \( u = -25 \, \text{cm} \)
- \( v = -1 \, \text{m} = -100 \, \text{cm} \)
Using the lens formula:
\[
\frac{1}{v} - \frac{1}{u} = \frac{1}{f}
\]
Substituting the values:
\[
\frac{1}{-100} - \frac{1}{-25} = \frac{1}{f}
\]
Simplifying:
\[
\frac{1}{f} = \frac{-1}{100} + \frac{1}{25} = \frac{3}{100}
\]
So,
\[
f = \frac{100}{3} = 33.3 \, \text{cm} = 0.33 \, \text{m}
\]
Now, the power \( P \) of the lens is:
\[
P = \frac{1}{f} = \frac{1}{0.33} = +3 \, \text{D}
\]
Hence, a convex lens of power +3 D is required to correct the defect.
Q8. Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Solution:
As the ciliary muscles of eyes are incapable to contract beyond a certain limit hence, a normal eye cannot see clearly the objects kept closer than 25 cm. If an object is kept at a distance less than 25 cm from the eye, then the object seems blurred and produces strain in the eyes.
Q9. What happens to the image distance in the eye when we increase the distance of an object from the eye?
Solution:
For an eye, the image distance always remains constant as the size of eyes cannot increase or decrease. The increase in the object distance is balanced by the change in the focal length of the eye lens. The eye changes its focal length focal length in such a way that the image is always formed at retina.
Q10. Why do stars twinkle?
Solution: Twinkling of stars is due to the atmospheric refraction of light. As stars are too far from the earth therefore, they can be considered as point sources of light. When the light coming from stars passes through the earth’s atmosphere, it gets refracted at different levels because of the variation in the air density at different levels of the atmosphere. When the star light refracted by the atmosphere, it comes more towards us. It appears brighter when it comes less towards us.
Q11. Explain why the planets do not twinkle.
Solution: Planets do not twinkle like stars because planets are much closer to the earth and are thus seen as extended objects. If we consider planet as a collection of large number of point-sized sources of light, sources of light the variation in amount of light entering our eye from all the individual point sized sources will average out to zero, thereby nullifying the twinkling effect.
Q12. Why does the Sun appear reddish early in the morning?
Solution: During sunrise, the light coming from the Sun has to travel a larger distance in the earth’s atmosphere before entering to our eyes. During this, the shorter wavelengths of lights are scattered out and only longer wavelengths are able to reach our eyes. As blue colour has a shorter wavelength and red colour has a longer wavelength, the red colour reaches to our eyes after the atmospheric scattering of light. Hence, the Sun seems reddish while sunrise.
More Resources of NCERT Solutions for Class 10 Science
NCERT Solutions for Cl0ass 10 Science Chapter 10 – FAQs
Q1. What are the common defects of vision and how are they corrected?
The most common vision defects are myopia (short-sightedness), where distant objects appear blurred because the image forms before the retina, corrected using concave lenses; hypermetropia (long-sightedness), where nearby objects appear blurred as the image forms behind the retina, corrected using convex lenses; and presbyopia, an age-related condition where the eye's lens loses flexibility, making it difficult to focus on both near and distant objects, corrected using bifocal lenses. Another defect is astigmatism, caused by irregular curvature of the cornea or lens, corrected using cylindrical lenses. These corrections work by adjusting the focal point of light entering the eye to fall precisely on the retina.
Q2. Why does the sky appear blue during the day and red during sunrise and sunset?
The blue color of the sky is due to the scattering of light by atmospheric particles. Sunlight consists of seven colors with different wavelengths. Blue light has a shorter wavelength and gets scattered more effectively by air molecules than other colors (following Rayleigh scattering). This scattered blue light reaches our eyes from all directions, making the sky appear blue. During sunrise and sunset, sunlight travels a longer distance through the atmosphere. Most of the blue light gets scattered away, and the longer wavelength red and orange light reaches our eyes with minimal scattering, making the sky appear reddish. This beautiful phenomenon demonstrates how the path length of light through the atmosphere affects the colors we perceive.
Q3. How does the human eye adjust to see objects at different distances (accommodation)?
The ability of the eye to focus on both near and distant objects by adjusting its focal length is called accommodation. This is achieved through the ciliary muscles that control the shape of the eye lens. When viewing distant objects, the ciliary muscles relax, making the lens thinner and increasing its focal length. When viewing nearby objects, the ciliary muscles contract, making the lens thicker and decreasing its focal length. This automatic adjustment ensures that images form sharply on the retina regardless of object distance. The nearest point at which the eye can see clearly is called the near point (about 25 cm for a normal eye), and the farthest point is called the far point (infinity for a normal eye). This remarkable ability allows us to seamlessly shift our focus between objects at varying distances.