Question

The near point of an elderly person is $50$ cm from the eye. Find the power of the corrective lens that will correct his vision.

• A. $+0.5dioptre$
• B. $0.2dioptre$
• C. $-2dioptre$
• D. $+2dioptre$

Answer 1

Answer: (D)

$+2dioptre$

The near point of an elderly person $=50cm$ from the eye.For minimum to see object placed at normal point, its image should be formed 50 cm from the eye.

Hence, $v=-50cm,u=-25cm$

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

$=\frac{-1+2}{50}=\frac{1}{50cm}$

$\therefore f=50cm$ or $0.5m$

Power of the lens, $P=\frac{1}{f}=\frac{1}{0.50cm}=+2dioptre$