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

2. The human eye forms an image of an object at its

(a) cornea

(b) iris

(c) pupil

(d) retina

(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

(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

(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?

(i) Given

Power of the lens used for correcting distant vision = – 5.5 D

We know that P =\frac{1}{f}

  \therefore f =\frac{1}{P}

f =\frac{1}{-5.5}m

f =\frac{1}{-5.5} \times 100 cm = -18.18 cm

So focal length   f = -18.18 cm

So the focal length of lens required for correcting distant vision is,    -18.18 cm cm. Minus sign of focal length tells us that it is a concave lens.

(ii) for near vision : power of lens, P=+ 1.5D

P =\frac{1}{f}

  \therefore f =\frac{1}{P}

f =\frac{1}{+5.5}m

f =\frac{1}{+5.5} \times 100 cm = +66.66 cm

 So the focal length of lens required for correcting near vision is +66.7 cm. Plus sign of focal length tells us that it is a convex lens.

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?

The person is suffering from myopia in which the image is formed in front of the retina and requires a concave lens to correct the defect. 

Here f =80 cm = -0.8 m

Power of lens is given as

P =\frac{1}{f}

P =\frac{1}{-0.8}

P = -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.

Hypermetropia can be corrected by using a convex lens.

Given,

Object distance u =-25 cm

Image distance v =-100 cm 

We know the lens formula 

\frac{1}{v}-\frac{1}{u} = \frac{1}{f}

\frac{1}{-100}-\frac{1}{-25} = \frac{1}{f}

f=\frac{1}{3}m

Power, P =\frac{1}{f}=\frac{1}{1/3} =3D

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

The objects placed closer than 25 cm appears blurred because the ciliary muscles of the eyes are unable to contract beyond a certain limit.

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

Image distance is the distance between eye lens and retina. The image distance in the eye does not change when the object’s distance from the eye is increased.

10. Why do stars twinkle?

It is due to the refraction of light by atmosphere. When starlight enters our atmosphere it is affected by various factors like wind, temperature in the atmosphere.

11. Explain why the planets do not twinkle?

Planets appears to us larger than the stars because they are close to us. So, a planet may be considered as a collection of a large number of point-sized light source and appears as a tiny disks. The light from these little disks is also refracted by Earth’s atmosphere. But the light from one part of disk is cancelled out by the light from other parts of the disk . That’s why planet do not twinkle. 

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

This happens due to scattering of light when sunlight travels through the atmosphere. Red colour which is having the longest wavelength reaches our eye while all other colour scatters during its travel to our eye. 

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

Scattering of light is not possible in the outer space due to the absence of atmosphere. So the sky appears dark instead of blue to an astronaut.