In a microscope the objective is having focal length $f_o=2 \mathrm{~cm}$ and eye-piece is having fo

Question

In a microscope the objective is having focal length $f_o=2 \mathrm{~cm}$ and eye-piece is having focal length $f_e=4 \mathrm{~cm}$. The tube length is 32 cm . The magnification produced by this microscope for normal adjustment is $\_\_\_\_$。

Professor PC Thomas Classes · Answer

For a compound microscope in normal adjustment (final image at infinity), the magnifying power is given by $M = \left(\frac{L}{f_o}ight) \left(\frac{D}{f_e}ight)$, where $L$ is the tube length and $D = 25 	ext{ cm}$ is the least distance of distinct vision. Substituting the given values: $M = \left(\frac{32}{2}ight) 	imes \left(\frac{25}{4}ight) = 16 	imes 6.25 = 100$.