Question

A doctor prescribes a corrective lens of power - 0.5 D to a person. Find the
focal length of the lens. Is this lens diverging or converging ? Give reason.
How does the property of this lens can be used to correct eye defects ?​

Answer 1

Answer:

Explanation:

Power = 1/f

-0.5 = 1/f

f = -1/0.5

f = -2m

f = -200cm

since focal length is negative, lens is concave (diverging)

The cure for the nearsighted/myopic eye is to equip it with a diverging lens. Since the nature of the problem of nearsightedness/myopia is that the light is focused in front of the retina, a diverging lens will serve to diverge light before it reaches the eye

hope this helps ☺️

Answer 2

Answer:

The focal length of the lens is -2m, the lens is a diverging lens and hence used to correct myopia.

Explanation:

Given the power of a corrective lens, $D=0.5m$  

Power of a lens is defined as reciprocal of focal length in meters.

$P=\frac{1}{f}$

Therefore,  $f=\frac{1}{P}$

$=\frac{1}{0.5}=-2m$

Therefore, the focal length of the lens is -2m.

Since the focal length is negative, lens is a diverging lens i.e., a concave lens.

Concave lens is used to correct myopia (or short-sightedness).

• Myopia is the defect of vision where a person cannot see the distant objects clearly but can see the nearby objects clearly.
• The far point of an eye will be less than infinity, hence the image of a distant object is formed in front of the retina .    
• When a concave lens of suitable power is placed in front of a myopic eye, the parallel rays coming from the distant object gets diverged.
• A virtual image of the distant object is formed at the far point of eye.
• The rays now appear to be coming from the eye’s far point, and hence gets focused by the eye-lens on the retina.
• The concave lens used for correcting myopia should produce a virtual image of the distant object at the far point of the myopic eye.