Q. A student can distinctly see the object upto a distance $15\ cm$. He wants to see the black board at a distance of $3\ m$. Focal length and power of lens used respectively will be
A) $- 4.8\ cm,\ - 3.3\ D$
B) $- 5.8\ cm,\ - 4.3\ D$
C) $- 7.5\ cm,\ - 6.3\ D$
D) $- 15.8\ cm,\ - 6.3\ D$
Explanation
(d) $v = - 15cm,\ u = - 300cm$
From lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\Rightarrow \frac{1}{f} = \frac{1}{- 15} - \frac{1}{- 300} = \frac{- 19}{300}$
$\Rightarrow f = \frac{- 300}{19} = - 15.8\ cm$
and power $P = \frac{100}{f_{in\ cm}} = \frac{- 100 \times 19}{300}$
$= - 6.33\ D$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.