NCERT Solutions for Class 10 Science Chapter 10 Human Eye and Colourful World

NCERT
SOLUTIONS
CLASS - 10th

aglase .co

Class : 10th
Subject : Science
Chapter : 11
Chapter Name : The Human Eye And The Colourful World

Q1 What is meant by power of accommodation of the eye?

Answer. When the ciliary muscles are relaxed, the eye lens becomes thin, the focal length
increases, and the distant objects are clearly visible to the eyes. To see the nearby objects
clearly, the ciliary muscles contract making the eye lens thicker. Thus, the focal length of the
eye lens decreases and the nearby objects become visible to the eyes. Hence, the human eye
lens is able to adjust its focal length to view both distant and nearby objects on the retina.
This ability is called the power of accommodation of the eyes.

Page : 190 , Block Name : Questions

Q2 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?

Answer. The person is able to see nearby objects clearly, but he is unable to see objects beyond
1.2 m. This happens because the image of an object beyond 1.2 m is formed in front of the
retina and not at the retina, as shown in the given gure.

To correct this defect of vision, he must use a concave lens. The concave lens will bring
the image back to the retina as shown in the given gure.

Page : 190 , Block Name : Questions

Q3 What is the far point and near point of the human eye with normal vision?

Answer. The near point of the eye is the minimum distance of the object from the eye, which
can be seen distinctly without strain. For a normal human eye, this distance is 25 cm. The far
point of the eye is the maximum distance to which the eye can see the objects clearly. The far
point of the normal human eye is in nity.

Page : 190 , Block Name : Questions

Q4 A student has dif culty reading the blackboard while sitting in the last row. What could be
the defect the child is suffering from? How can it be corrected?

Answer. A student has dif culty in reading the blackboard while sitting in the last row. It
shows that he is unable to see distant objects clearly. He is suffering from myopia. This defect
can be corrected by using a concave lens.

Page : 190 , Block Name : Questions

Q1 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) nearsightedness.
(d) farsightedness.

Answer. (b) Human eye can change the focal length of the eye lens to see the objects situated
at various distances from the eye. This is possible due to the power of accommodation of the
eye lens.

Page : 197 , Block Name : Exercise

Q2 The human eye forms the image of an object at its
(a) cornea.
(b) iris.
(c) pupil.
(d) retina.

Answer. (d) The human eye forms the image of an object at its 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.

Answer. (c) The least distance of distinct vision is the minimum distance of an object to see
clear and distinct image. It is 25 cm for a young adult with normal visions.

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

Answer. (c) The relaxation or contraction of ciliary muscles changes the curvature of the eye
lens. The change in curvature of the eye lens changes the focal length of the eyes. Hence, 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 dioptre. What is the focal length of the
lens required for correcting (i) distant vision, and (ii) near vision?

Answer. For distant vision = -0.181 m, for near vision = 0.667 m
The power P of a lens of focal length f is given by the relation
1
P =
f ( in metres )

(i) Power of the lens used for correcting distant vision = − 5.5D

1
Focal length of the required lens, f =
P

1
f = = −0.181m
−5.5

(ii) Power of the lens used for correcting near vision = + 1.5D

1
Focal length of the required lens, f =
P

1
f = = +0.667m
1.5

The focal length of the lens for correcting near vision is 0.667m.

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?

Answer. The person is suffering from an eye defect called myopia. In this defect, the image is
formed in front of the retina. Hence, a concave lens is used to correct this defect of vision.
Object distance, u = in nity =
Image distance, v = —80 cm
Focal length = f

According to the lens formula,
1 1 1
− =
v u f

1 1 1
− − =
80 ∞ f

1 1
= −
f 80

f = −80cm = −0.8m

1
Power, P =
f ( in metres )

1
Power, P =
f ( in metres )

1
P = = −1.25D
−0.8

A concave lens of power —1.25 D is required by the person to correct his defect.

Q7 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. A person suffering from hypermetropia can see distinct objects clearly but faces
dif culty in seeing nearby objects clearly. It happens because the eye lens focuses the
incoming divergent rays beyond the retina. This defect of vision is 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, as shown in the following gure.

The convex lens actually creates a virtual image of a nearby object (N' in the gure) at the near

point of vision (N) of the person suffering from hypermetropia. The given person will be able
to
clearly see the object kept at 25 cm (near point of the normal eye), if the image of the object is
formed at his near point, which is given as 1 m.
Object distance, u = -25 cm
Image distance, v = —1 m = -100 m
Focal length, f
Using the lens formula,

1 1 1
− =
v u f

1 1 1
− =
−100 −25 f

1 1 1
= −
f 25 100

1 4−1
=
f 100

100
f = = 33.3cm = 0.33m
3

1 1
Power, P = = = +3.0D
f ( in metres ) 0.33m

A convex lens of power + 3.0D is required to correct the defect.

Q8 Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Answer. Because the focal length of Normal eye is 25cm.
When any object placed closer than 25 cm then the image is FORMED behind the retina. Then
object seem blurred.

Q9 What happens to the image distance in the eye when we increase the distance of an object
from the eye?

Answer. Since the size of eyes cannot increase or decrease, the image distance remains
constant. When we increase the distance of an object from the eye, the image distance in the
eye does not change. The increase in the object distance is compensated by the change in the
focal length of the eye lens. The focal length of the eyes changes in such a way that the image
is always formed at the retina of the eye.

Q10 Why do stars twinkle?

Answer. Stars emit their own light and they twinkle due to the atmospheric refraction of light.
Stars are very far away from the earth. Hence, they are considered as point sources of light.
When the light coming from stars enters 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 comes more towards us, it appears brighter than
when it comes less towards us. Therefore, it appears as if the stars are twinkling at night.

Q11 Explain why the planets do not twinkle?

Answer. Planets do not twinkle because they appear larger in size than the stars as they are
relatively closer to earth. Planets can be considered as a collection of a large number of point-
size sources of light. The different parts of these planets produce either brighter or dimmer
effect in such a way that the average of brighter and dimmer effect is zero. Hence, the
twinkling effects of the planets are nulli ed and they do not twinkle.

Q12 Why does the Sun appear reddish early in the morning?

Answer. During sunrise, the light rays coming from the Sun have to travel a greater distance in
the earth's atmosphere before reaching our eyes. In this journey, the shorter wavelengths of
lights are scattered out and only longer wavelengths are able to reach our eyes. Since blue
colour has a shorter wavelength and red colour has a longer wavelength, the red colour is able
to reach our eyes after the atmospheric scattering of light. Therefore, the Sun appears reddish
early in the morning.

Q13 Why does the sky appear dark instead of blue to an astronaut?

Answer. The sky appears dark instead of blue to an astronaut because there is no atmosphere
in the outer space that can scatter the sunlight. As the sunlight is not scattered, no scattered
light reach the eyes of the astronauts and the sky appears black to them.