Question

A far-sighted person has a near point at 100 cm. what must be the power of the correcting lens?

Answer 1

Power of the lens = $\frac{1}{focal length}$

To calculate focal length using the near point=100cm

f= $\frac{uv}{u+v}$

Where, u= object size

v= image size

∵ image is virtual, it is with a - sign.

The near point of a normal eye=u=25 cm

Using this,

f= $\frac{(25)*(-100)5}{25+(-100)}$

= +33.3 cm=+ $\frac{33.3}{100}$

Also, Power=P=$\frac{1}{f}$ where f is in meters

⇒P=+ $\frac{100}{33.3}$= + 3 D