Q. The focal lengths of the objective and of the eye-piece of a compound microscope are $f_{0}$ and $f_{e}$ respectively. If $L$ is the tube length and D, the least distance of distinct vision, then its angular magnification, when the image is formed at infinity, is
A) $\left( 1 - \frac{L}{f_{0}} \right)\left( \frac{D}{f_{e}} \right)$
B) $\left( 1 + \frac{L}{f_{0}} \right)\left( \frac{D}{f_{e}} \right)$
C) $\frac{L}{f_{0}}\left( 1 - \frac{D}{f_{e}} \right)$
D) $\frac{L}{f_{0}}\left( \frac{D}{f_{e}} \right)$
Explanation
The correct answer is Option D: $\frac{L}{f_{0}}\left( \frac{D}{f_{e}} \right)$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.