E X E R C I S E S CHAPTER 10 The Human Eye and the Colourful World SCIENCE NCERT TEXTBOOK (CBSE) CLASS 10:

 E X E R C I S E S  

CHAPTER  10  The Human Eye and  the Colourful World


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

  • Answer: (d) accommodation.
    • Explanation: Accommodation refers to the ability of the eye to change the shape of the lens to focus on objects at different distances.

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

  • Answer: (d) retina.
    • Explanation: The retina is the light-sensitive layer at the back of the eye where the image is formed and transmitted to the brain.

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

  • Answer: (c) 25 cm.
    • Explanation: The least distance of distinct vision (near point) for a normal human eye is typically about 25 cm.

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

  • Answer: (c) ciliary muscles.
    • Explanation: The ciliary muscles control the shape of the eye lens, thus changing its focal length to focus on objects at different distances.

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:
    • Power (P) = -5.5 D
    • Focal length (f) = 1/P = 1/(-5.5) = -0.182 m = -18.2 cm.
    • Explanation: A negative focal length is required for myopia (nearsightedness), and the lens needed will be a diverging lens.
  • (ii) Near vision:
    • Power (P) = +1.5 D
    • Focal length (f) = 1/P = 1/(1.5) = 0.67 m = 67 cm.
    • Explanation: A positive focal length is required for hypermetropia (farsightedness), and the lens will be a converging lens.

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 of the lens: A concave lens is required to correct myopia (nearsightedness).
    • Power (P) = 1/f.
    • The far point is 80 cm, which is the object distance. The corrected image should form at infinity, so:
      • Using the lens formula: 1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u}where v=∞v = \infty (image at infinity), and u=−80 cm.1f=0−1−80=180 m\frac{1}{f} = 0 - \frac{1}{-80} = \frac{1}{80} \, \text{m} Thus, the focal length f=−80 cmf = -80 \, \text{cm}, and the power of the lens is: P=1f=1−0.80=−1.25 DP = \frac{1}{f} = \frac{1}{-0.80} = -1.25 \, \text{D}

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:
    • Explanation:
      • For hypermetropia (farsightedness), the eye’s near point is farther than the normal 25 cm. The corrective lens is a converging lens that brings the near point closer to the normal 25 cm.
      • The required power of the lens can be calculated using: P=1fP = \frac{1}{f} where ff is the focal length. The near point for the corrected eye should be at 25 cm. 1f=125−1100\frac{1}{f} = \frac{1}{25} - \frac{1}{100} P=125100−1100=124100=4.17 D.P = \frac{1}{\frac{25}{100} - \frac{1}{100}} = \frac{1}{\frac{24}{100}} = 4.17 \, \text{D}.
      • Conclusion: The power of the lens required is +4.17 dioptres (converging lens).

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

  • Answer:
    • The normal eye can focus on objects only within a certain range. The nearest point at which a normal eye can focus clearly is 25 cm, known as the least distance of distinct vision. Objects closer than 25 cm cannot be focused on clearly because the eye's lens cannot accommodate enough to bring them into focus.

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

  • Answer:
    • As the distance of an object increases, the image distance in the eye also changes. The eye's lens adjusts to focus the image on the retina. For distant objects, the image distance is greater (closer to the retina), and for near objects, the image distance is shorter (further from the retina).

10. Why do stars twinkle?

  • Answer:
    • Stars twinkle because their light passes through various layers of the Earth's atmosphere, which have different temperatures and densities. These variations cause the light to refract and change in intensity, making the star appear to twinkle.

11. Explain why the planets do not twinkle.

  • Answer:
    • Planets do not twinkle because they appear as extended sources of light (due to their larger size relative to stars). The light from a planet is not affected as much by atmospheric refraction, unlike the point sources of light (stars).

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

  • Answer:
    • The blue color of the sky is due to the scattering of sunlight by molecules in Earth's atmosphere. However, in space, there is no atmosphere to scatter sunlight, so the sky appears dark (black), even though the Sun is shining. This is why astronauts see a black sky while in space.

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