Q. A man’s near point is 0.5 m and far point is 3 m. Power spectacle lenses repaired for (i) reading purposes (ii) seeing distant objects, respectively
A) $- 2\ $D and + 3 D
B) +2 D and $-$3 D
C) +2 D and – 0.33 D
D) $-$2 D and + 0.33D
Explanation
(c) For reading purpose
$u = - 25\ cm,v = - 50\ cm,\ f = ?$
$\therefore\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{50} + \frac{1}{25} = \frac{1}{50}$
$P = \frac{100}{f} = + 2D$
For distinct vision,
$f^{'} =$distance of far point$\ = - 3$m
$P\ $=$\ \frac{1}{f'} = - \frac{1}{3}$D =$-$0.33 D
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.