Q. A man can see the object between $15\ cm$ and $30\ cm$. He uses the lens to see the far objects. Then due to the lens used, the near point will be at
A) $\frac{10}{3}cm$
B) $30\ cm$
C) $15\ cm$
D) $\frac{100}{3}cm$
Explanation
(b) For improving far point, concave lens is required and for this concave lens $u = \infty,\ v = - 30\ cm$
So $\frac{1}{f} = \frac{1}{- 30} - \frac{1}{\infty} \Rightarrow f = - 30\ cm$
For near point $\frac{1}{- 30} = \frac{1}{- 15} - \frac{1}{u}$
$\Rightarrow u = - 30\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.