Chapter – 11 The Human Eye and the Colourful World
Q 1. What is meant by power of accommodation of the eye?
Ans. The ability of the eye lens to adjust its focal length, so as to clearly focus rays coming from
distant as well a near objects on the retina, is called the power of accommodation of the eye.
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?
Ans. A person with a myopic eye should use a concave lens of focal length 1.2 m so as to restore
proper vision.
Q 3. What is the far point and near point of the human eye with normal vision?
Ans. For a human eye with normal vision the far point is at infinity and the near point is at 25 cm from
the eye.
Q 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?
Ans. The student is suffering from myopia or short-sightedness. The defect can be corrected by the
use of concave (diverging ) lens of an appropriate power.
Q 5. The human eye can focus objects at different distance by adjusting the focal length of the eye
lens. This is due to (a) presbyopia (b) accommodation (c) near-sightedness
(d) far-sightedness
Ans. (b) accommodation. [ Hint: While seeing distant object eye lens is thinner and its focal length is
more (about 2.5 cm). While looking near objects eye lens assumes a more round shape, its focal
length decreases and nearly objects are clearly focussed at the retina.]
Q 6. The human eye forms the image of an object at its (a) cornea (b) iris (c) pupil (d) retina
Ans. (d) retina.
Q 7. 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 Ans. (c) 25 cm.
Q 8. 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
Ans. (c) ciliary muscles.
Q 9. 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?
Ans. (i) Power of lens needed for correction distant vision of the person P1 = 5.5 D
Focal length of lens required for correcting distant vision
f1 = 1/p1 = 1/-5.5 = -18cm
(ii) For correcting near vision power required P2 = + 1.5 D
Focal length of lens required for correcting near vision f2 = 1/p2 = 1/-1.5 = 0.67 m=66.7 cm
Q10. The far point of a myopic the person is 80 cm in front of the eye. What is the nature and power
of the lens required to correct the problem?
Ans. To correct the myopia the person concerned should use concave lens (diverging lens) of focal
length = 80 cm, so that for an object at infinity (u = ), the virtual image is formed at the far point
of myopic person (v = 80 cm).
From lens formula
,
1 1 1
v u f
we have
80
1 1 1
( )
1
( 80 )
1
f f
or
f 80 cm = -0.8m
Power of the lens P =1/ 0. 80= 1.25 D.
Q11. Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye
is 1m. What is the power of the lens required to correct this defect? Assume that the near point of the
normal eye is 25 cm.
Ans. The relevant diagram has been shown in adjoining fig. Here the convex lens used forms virtual
image of object placed at N (near point of normal
eye i.e., 25 cm from eye) at N, the near point of
defective eye (at a distance x).
Now, the defective eye can focus the rays from N at
the retina.
In the problem it is given that the near point of the
normal eye is 25 cm, hence u = 25 cm. The lens used forms its virtual image at near point of
hypermetropic eye i.e., v = 1m = 100 cm.
Using lens formula, we have
Power of correcting lens P = 1/ in m = 1/(1/3) = + 3 D.
Q12. Why is a normal eye not able to see clearly the objects placed closer than 25 cm ?
Ans. Due to a power of accommodation, the focal length of the eye lens cannot be decreased below
a certain minimum limit.Consequently, a normal eye cannot see clearly the objects placed closer than
a minimum distance, called near point of eye.
For a young adult value of near point is 25 cm. So, we cannot see clearly objects placed closer than
25 cm.
Q13. What happens to the image distance in the eye when we increase the distance of an object from
the eye?
Ans. The image is formed on the retina even on increasing the distance of an object from the eye. For
this eye lens becomes thinner and its focal length increases as the object is moved away from the
eye.
Q14. Why do stars twinkle?
Ans. Stars twinkle due to atmospheric refraction of starlight. As the stars are very far away, they
behave as almost point sources of light. A son account of atmospheric refraction, the path of rays of
light coming from the star goes on varying slightly, the apparent position of the star fluctuates and the
amount of starlight entering the eye flickers. So, sometimes, the star appears brighter and at some
other time, fainter. Thus, the stars twinkle.
Q15. Explain why the planets do not twinkle.
Ans. Planets are much closer to the earth and are seen as extended sources. So, a planet may be
considered as a collection of a large number of point-sized light sources. Although light coming from
individual point-sized sources flickers but the total amount of light entering our eye from all the
individual point-sized sources average out to be constant. Thereby, planets appear equally bright and
there is no twinkling of planets
Q17. Why does the sky appear dark instead of blue to an astronaut?
Ans. Colour of the sky is on account of scattering of light of shorter wavelengths by particles in the
atmosphere of earth. If the earth had no atmosphere, there would not have been any scattering and
the sky would have looked dark. When an astronaut in his spacecraft goes above the atmosphere of
earth, sky appears dark to him because there is no scattering of light.
Q 1. What is meant by power of accommodation of the eye?
Ans. The ability of the eye lens to adjust its focal length, so as to clearly focus rays coming from
distant as well a near objects on the retina, is called the power of accommodation of the eye.
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?
Ans. A person with a myopic eye should use a concave lens of focal length 1.2 m so as to restore
proper vision.
Q 3. What is the far point and near point of the human eye with normal vision?
Ans. For a human eye with normal vision the far point is at infinity and the near point is at 25 cm from
the eye.
Q 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?
Ans. The student is suffering from myopia or short-sightedness. The defect can be corrected by the
use of concave (diverging ) lens of an appropriate power.
Q 5. The human eye can focus objects at different distance by adjusting the focal length of the eye
lens. This is due to (a) presbyopia (b) accommodation (c) near-sightedness
(d) far-sightedness
Ans. (b) accommodation. [ Hint: While seeing distant object eye lens is thinner and its focal length is
more (about 2.5 cm). While looking near objects eye lens assumes a more round shape, its focal
length decreases and nearly objects are clearly focussed at the retina.]
Q 6. The human eye forms the image of an object at its (a) cornea (b) iris (c) pupil (d) retina
Ans. (d) retina.
Q 7. 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 Ans. (c) 25 cm.
Q 8. 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
Ans. (c) ciliary muscles.
Q 9. 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?
Ans. (i) Power of lens needed for correction distant vision of the person P1 = 5.5 D
Focal length of lens required for correcting distant vision
f1 = 1/p1 = 1/-5.5 = -18cm
(ii) For correcting near vision power required P2 = + 1.5 D
Focal length of lens required for correcting near vision f2 = 1/p2 = 1/-1.5 = 0.67 m=66.7 cm
Q10. The far point of a myopic the person is 80 cm in front of the eye. What is the nature and power
of the lens required to correct the problem?
Ans. To correct the myopia the person concerned should use concave lens (diverging lens) of focal
length = 80 cm, so that for an object at infinity (u = ), the virtual image is formed at the far point
of myopic person (v = 80 cm).
From lens formula
,
1 1 1
v u f
we have
80
1 1 1
( )
1
( 80 )
1
f f
or
f 80 cm = -0.8m
Power of the lens P =1/ 0. 80= 1.25 D.
Q11. Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye
is 1m. What is the power of the lens required to correct this defect? Assume that the near point of the
normal eye is 25 cm.
Ans. The relevant diagram has been shown in adjoining fig. Here the convex lens used forms virtual
image of object placed at N (near point of normal
eye i.e., 25 cm from eye) at N, the near point of
defective eye (at a distance x).
Now, the defective eye can focus the rays from N at
the retina.
In the problem it is given that the near point of the
normal eye is 25 cm, hence u = 25 cm. The lens used forms its virtual image at near point of
hypermetropic eye i.e., v = 1m = 100 cm.
Using lens formula, we have
Power of correcting lens P = 1/ in m = 1/(1/3) = + 3 D.
Q12. Why is a normal eye not able to see clearly the objects placed closer than 25 cm ?
Ans. Due to a power of accommodation, the focal length of the eye lens cannot be decreased below
a certain minimum limit.Consequently, a normal eye cannot see clearly the objects placed closer than
a minimum distance, called near point of eye.
For a young adult value of near point is 25 cm. So, we cannot see clearly objects placed closer than
25 cm.
Q13. What happens to the image distance in the eye when we increase the distance of an object from
the eye?
Ans. The image is formed on the retina even on increasing the distance of an object from the eye. For
this eye lens becomes thinner and its focal length increases as the object is moved away from the
eye.
Q14. Why do stars twinkle?
Ans. Stars twinkle due to atmospheric refraction of starlight. As the stars are very far away, they
behave as almost point sources of light. A son account of atmospheric refraction, the path of rays of
light coming from the star goes on varying slightly, the apparent position of the star fluctuates and the
amount of starlight entering the eye flickers. So, sometimes, the star appears brighter and at some
other time, fainter. Thus, the stars twinkle.
Q15. Explain why the planets do not twinkle.
Ans. Planets are much closer to the earth and are seen as extended sources. So, a planet may be
considered as a collection of a large number of point-sized light sources. Although light coming from
individual point-sized sources flickers but the total amount of light entering our eye from all the
individual point-sized sources average out to be constant. Thereby, planets appear equally bright and
there is no twinkling of planets
Q17. Why does the sky appear dark instead of blue to an astronaut?
Ans. Colour of the sky is on account of scattering of light of shorter wavelengths by particles in the
atmosphere of earth. If the earth had no atmosphere, there would not have been any scattering and
the sky would have looked dark. When an astronaut in his spacecraft goes above the atmosphere of
earth, sky appears dark to him because there is no scattering of light.
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