Q. To remove myopia (short sightedness) a lens of power 0.66 $D$ is required. The distance point of the eye is approximately
A) $100\ cm$
B) $150\ cm$
C) $50\ cm$
D) $25\ cm$
Explanation
(b) Far point of the eye = focal length of the lens
$= \frac{100}{P} = \frac{100}{0.66} = 151\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.