Q. The magnifying power of a microscope with an objective of $5\ mm$ focal length is 400. The length of its tube is $20\ cm$. Then the focal length of the eye-piece is
A) $200\ cm$
B) $160\ cm$
C) $2.5\ cm$
D) $0.1\ cm$
Explanation
(c) If nothing is said then it is considered that final image is formed at infinite and $m_{\infty} = \frac{\left( L_{\infty} - f_{o} - f_{e} \right).D}{f_{o}f_{e}} \simeq \frac{LD}{f_{0}f_{e}}$
$\Rightarrow 400 = \frac{20 \times 25}{0.5 \times f_{e}} \Rightarrow f_{e} = 2.5\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.