# Chapter 11 Human Eye and Colourful World

Question 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

(b) Accommodation

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

(d) Retina

Question 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

(c) 25 cm

Question 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

(c) Ciliary muscles

Question 5
A person needs a lens of power -5.5 diopters for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 diopter. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?

Solution:
(i)
Power of distant viewing part of the lens, P1 = -5.5 D
Focal length of this part, f1 = = m = -0.182 m = -18.2 cm

(ii) For near vision,

Question 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:
The remedial lens should make the objects at infinity appear at the far point.
Therefore, for object at infinity, u = ∞
Far point distance of the defected eye, ν = – 80 cm

Negative sign shows that the remedial lens is a concave lens.

Question 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 the defect ? Assume that the near point of the normal eye is 25 cm.

Solution:
(i) The near point N of hypermetropic eye is farther away from the normal near point N.

(ii) In a hypermetropic eye, the image of nearby object lying at normal near point N (at 25 cm) is formed behind the retina.

(iii) Correction of hypermetropia : The convex lens forms a virtual image of the object (lying at normal near point N) at the near point N’ of this eye.

The object placed at 25 cm from the correcting lens must produce a virtual image at 1 m or 100 cm.
Therefore, u = – 25 cm, ν = 100 cm

The positive sign shows that it is a convex lens.

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

At distance less than 25 cm, the ciliary muscles cannot bulge the eye lens any more, the object cannot be focused on the retina and it appears blurred to the eye, as shown in the given figure.

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

The eye lens of a normal eye forms the images of objects at various distances on the same retina. Therefore, the image distance in the eye remains the same.

Question 10
Why do stars twinkle ?

Stars appear to twinkle due to atmospheric refraction. The light of star after the entry of light in earth’s atmosphere undergoes refraction continuously till it reaches the surface of the earth. Stars are far away. So, they are the point source of light. As the path of light coming from stars keep changing, thus the apparent position of stars keep changing and amount of light from stars entering the eye keeps twinkling. Due to which a star sometimes appear bright and sometimes dim, which is the effect of twinkling.

Question 11
Explain why the planets do not twinkle ?