Ch 10: The Human Eye and the Colourful World (NCERT Solutions)

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NCERT Solutions for Class 10 Science Ch 10: The Human Eye and the Colourful World, is given below.

You will get solutions for;

  • In-text questions as well as,
  • Exercises questions.

So let’s dive into it!

In-text Questions Set-1

1) What is meant by power of accommodation of eye?

Solution:
The power of accommodation is the eye’s ability to adjust its lens so it can focus clearly on both distant and nearby objects on the retina.

2) A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of corrective lens used to restore proper vision?

Solution:
A person with myopia should use a concave lens with a focal length of 1.2 meters to correct their vision.

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

Solution:
The near point of the eye is the closest distance at which an object can be seen clearly without straining. For a normal eye, this distance is 25 cm.

4) 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:
The student has short-sightedness (myopia). This can be corrected with the use of a concave (diverging) lens of the right strength.

Exercise Questions

1) The human eye can focus 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:
(b) accommodation

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

Solution:
(d) retina

3) 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:
(c) 25 cm

4) 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:
(c) ciliary muscles

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, and (ii) near vision?

Solution:
The power (P) of a lens of focal length f is given by the relation
Power (P) = 1/f

(i) Power of the lens (used for correcting distant vision) = – 5.5 D
Focal length of the lens (f) = 1/P
f = 1/-5.5 = -0.181 m
The focal length of the lens (for correcting distant vision) is – 0.181 m  or 18.1 cm.

(ii) Power of the lens (used for correcting near vision) = +1.5 D
Focal length of the required lens (f) = 1/P
f = 1/1.5 = +0.667 m
The focal length of the lens (for correcting near vision) is 0.667 m or 66.7 cm.

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?

Solution:

To correct the myopia the person concerned should use a concave lens of focal length (f) = -80 cm = -0.80 m.

Power of lens (P) = 1/f(m)
= 1/-0.80
= 100/-80
= -1.25 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.

Solution:

Diagram for the correction of hypermetropia is given below:


Near point of defective eye is 1 m and that of normal eye is 25 cm.

Here, u = -25 cm, v = -1 m = 100 cm.

Using lens formula,
1/f = 1/v – 1/u

1/f = 1/-100 + 1/25 = 3/100

f = 100/3 cm = 1/3m.

P = 1/f(m) = 1/0.33 = +3.0 D.

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

Solution:
A normal eye cannot see objects clearly if they are closer than 25 cm because the ciliary muscles cannot contract enough to focus on such close objects.

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

Solution:
The image forms on the retina even when an object is moved farther away from the eye. As the object gets farther, the eye lens becomes thinner and its focal length increases.

10) Why do stars twinkle?

Solution:
Stars twinkle because of the way Earth’s atmosphere bends their light. Since stars are so far away, they appear as tiny points of light. As their light passes through the atmosphere, it gets bent slightly, causing the star’s position to shift and its brightness to change. This makes the star look like it’s flickering or twinkling.

11) Explain why the planets do not twinkle.

Solution:
Planets are much closer to Earth and appear as extended sources of light. They can be thought of as many tiny light sources combined. While individual points of light might flicker, the overall light from all these sources averages out, making planets look steadily bright without twinkling.

12) Why does the Sun appear reddish early in the morning?

Solution:
White light from the sun travels through the atmosphere before reaching us. As it does, most colors scatter, but red light scatters less and reaches us more directly. This is why the sun looks red at sunrise and sunset.

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

Solution:
The blue color of the sky is due to the scattering of shorter wavelength light by particles in Earth’s atmosphere. Without an atmosphere, there would be no scattering, and the sky would look dark. When astronauts travel above the atmosphere, the sky appears dark to them because there’s no scattering of light.

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