Question

The near point of a hypermetropic person is $75\, cm$ from the eye. What is the power of the lens required to enable the person to read clearly a book held at $25\, cm$ from the eye?

• A. +2 D
• B. +2.5 D
• C. +2.67 D
• D. +3.33 D

Answer 1

Answer: (C)

$u=-25\, cm , v=-75\, cm$
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{(-75)}-\frac{1}{(-25)}$
$=\frac{1}{-75}+\frac{3}{75}=\frac{2}{75 cm }$
$P=\frac{1}{f( m )}=\frac{2}{0.75 m }=+2.67 D$