Question

The near point of an elderly person in 50 cm from the eye. Find the focal length and power of the corrective lens that will correct his vision.

Answer 1

When object is placed at 25 cm, then a virtual image should form at 50 cm

So $u=-25cm$

$v=-50$

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

$\frac{1}{f}=\frac{1}{-50}-\frac{1}{-25}$

$f=50cm=0.5m$

Power $P=\frac{1}{f}=\frac{1}{0.5}=2D$

A convex is needed.