Q. The far point of a myopia eye is at $40\ cm$. For removing this defect, the power of lens required will be
A) 40 $D$
B) $- 4\ D$
C) $- 2.5\ D$
D) 0.25 $D$
Explanation
(c) For myopic eye $f = -$ (defected far point)
$\Rightarrow f = - 40\ cm\ \Rightarrow P = \frac{100}{- 40} = - 2.5\ D$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.