Q. A person’s near point is $50\ cm$ and his far point is $3\ m$. Power of the lenses he requires for (i) reading and (ii) for seeing distant stars are
A) $- 2\ D$ and $0.33\ D$
B) $2\ D$ and $- 0.33\ D$
C) $- 2\ D$ and $3\ D$
D) $2\ D$ and $- 3\ D$
Explanation
(b) For correcting the near point, required focal length
$f = \frac{50 \times 25}{(50 - 25)} = 50cm$
So power $P = \frac{100}{50} = + 2D$
For correcting the far point, required focal length
$f = -$ (defected far point) $= - 3\ m$
$\therefore P = - \frac{1}{3}D = - 0.33\ D$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.