Question

A person uses a lens of power $+3D$ to normalize vision. Find the near point of the hypermetropic eye (in m).

Answer 1

Suppose near point of hypermetropic eye is $v.$

Then $u$ = minimum distance of distinct

Vision $=-D=-25cm$

Focal length of the lens is $f=\frac{100}{3}cm$.

From lens formula

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

$\Rightarrow \frac{3}{100}=\frac{1}{v}-\frac{1}{-25}$

$\Rightarrow v=-100cm$

Near point of the hypermetropic eye is 1 m.