Q. The length of the compound microscope is $14\ cm$. The magnifying power for relaxed eye is 25. If the focal length of eye lens is $5\ cm$, then the object distance for objective lens will be
A) $1.8\ cm$
B) $1.5\ cm$
C) $2.1\ cm$
D) $2.4\ cm$
Explanation
(a) $L_{\infty} = v_{o} + f_{e} \Rightarrow 14 = v_{o} + 5 \Rightarrow v_{o} = 9\ cm$
Magnifying power of microscope for relaxed eye
$m = \frac{v_{o}}{u_{o}}.\frac{D}{f_{e}}\ or\ 25 = \frac{9}{u_{o}}.\frac{25}{5}\ or\ u_{o} = \frac{9}{5} = 1.8\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.