Question

The near point of a hypermetropic eye is 50cm. What is the nature and Power of the lens required to enable him to read a book at 25cm from the eye?

Answer 1

Given,

Image distance, v = -50 cm

Object distance, u = -25 cm

From the lens formula

We know,

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

Putting all the values

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

After solving

$f=50cm=0.5m$

Now for the finding power of the lens

We know,

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

Putting all the values

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

Hence the power required is 2 D.