"AKC SCIENCE CLASSES"
CLASS 10 TH (CBSE AND MP BOARD)
CHAPTER 11
THE HUMAN EYE AND THE COLORFUL WORLD
EXERCISE QUESTIONS AND ANSWERS
Q.01:- 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
Ans:- (b).
Q.02:- The human eye forms the image of an object at its
(a) cornea
(b) iris
(c) pupil
(d) retina
Ans:- (d).
Q.03:- 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).
Q.04:- 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).
Q.05:- 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 distance viewing part of the lens, P₁ = -5.5 D
Focal length of this part,
f₁ = 1/P₁ = 1/(-5.5) m = -18.73 cm. Ans.
(ⅱ) As power of the near-vision part is measured relative to the main part of the lens of power -5.5 D, so we use
P₁ + P₂ = P Or -5.5 + P₂ = +1.5 Or P₂ = +6.5 D
Focal length of near-vision part,
f₂ = 1/P₂ = 1/+6.5 m = +15.4 cm. Ans.
Q.06:- The fair point of an myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Ans:- The remedial lens should make the objects at infinity appear at the far point.
ஃ For object at infinity, u = -∞
Far point distance of the defected eye, v = -80 cm
By lens formula,
1 1 1 1 1
As, ------ = -------- -- ---------- = ---------- -- ---------
f v u -80 -∞
1 1
= -- -------- + 0 = -- --------
80 80
Or f = -80 cm
Power, P = 100/f(in cm) = 100/(-80) = -1.25 D. Ans.
Negative sign shows that the remedial lens is a concave lens.
Q.07:- Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropia is corrected. The near point of a hypermetropia 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.
Ans:-
The object is placed at 25 cm from the correcting lens must produce a virtual image at 1 m or 100 cm.
ஃ u = -25 cm, v = -100 cm
By lens formula,
1 1 1 1 1
As, ------ = -------- -- ---------- = ---------- -- ---------
f v u -100 -25
1 1 3
= -- --------- + ---------- = + --------
100 25 100
Or 100 1
f = + ---------- cm = + ------- m
3 3
Power, P = 1/f = + 3/1 = +3 D. Ans.
Q.08:- Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Or
We cannot see an object clearly if it is placed very close to the eye. Give reason.
Ans:- 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.
Q.09:- What happens to the image distance in the eye when we increase the distance of an object from the eye?
Ans:- The image distance remains same, equal to the distance of the retina from the eye lens. The change in object distance is compensated by the change in the focal length of the eye lens due to the action of ciliary muscles so that a clear image is formed on the same retina.
Q.10:- Why do stars twinkle?
Ans:- 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.
Q.11:- Explain why the planets do not twinkle ?
Ans:- 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 nullified and they do not twinkle.
Q.12:- Why does the Sun appears reddish early in the morning?
Ans:- Early in the morning, the sun is near the horizon. Sunlight reaches us after covering a large thickness of the atmosphere. So shorter waves of blue region are almost completely scattered away by the air molecules. Red waves of longer wavelength are least scattered and reach our eyes. The sun appears red.
Q.13:- Why does the sky appear dark instead of blue to an astronaut?
Ans:- The atmosphere is quite thin at very high altitudes. There is almost no scattering of sunlight. So the sky appears dark to an astronauts.
