Question

The nearer point of hypermetropic eye is $40cm$. The lens to be used for its correction should have the power

• A. $+1.5D$
• B. $-1.5D$
• C. $+2.5D$
• D. $+0.5D$

Answer 1

Answer: (A)

Near point of a hypermetropic eye $=40cm$
Near point of a normal eye $=25cm$
Object distance, $u=-25cm$
Image distance, $v=-40cm$
Using lens formula,
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{-40}+\frac{1}{25}$
$\frac{1}{f}=\frac{-5+8}{200}=\frac{3}{200cm}=\frac{300}{200m}$
$P=\frac{1}{f(inm)}=\frac{300}{200}=\frac{3}{2}$
$P=1.5D$