Question

A nearsighted woman has a far point of 180 cm . Part A. What kind of lens, converging or diverging, should be prescribed for her to see distant objects more clearly?Part B. What power should the lens have?

Answer 1

Explanation:

It is given that,

A nearsighted woman has a far point of 180 cm.

The object distance for nearsightedness is infinity, u = ∞

The image distance is, v = -180 cm

Using lens formula,

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{f}=\dfrac{1}{(-180)}-\dfrac{1}{\infty}\\\\f=-180\ cm\\\\f=-1.8\ m$

For nearsightedness, a diverging lens is used.

Power of the lens,

P = 1/f

$P=\dfrac{1}{-1.8}\\\\P=-0.56\ D$

So, the power of the lens is -0.56 D.

Answer 2

The power of lens to be prescribed is 3.5 D.

A person that is far sighted can see far objects clearly but not nearby objects. The farpoint of a normal eye is infinity. Far sightedness is corrected by the use of a convex lens. The near point of the normal eye is 25 cm.

We have that;

u = 25 cm

v = -180 cm

1/f = 1/u - 1/v

1/f = 1/25 - 1/180

1/f = 0.04 - 0.0056

f = 29 cm

Power of lens = 100/f = 100/29 = 3.5 D

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