Question

A compound microscope consists of an objective of focal length $2$cm and an eye piece of focal length $5$cm. When an object is kept $2.4$cm from the objective, final image formed is virtual and $25$cm from the eye piece. Determine magnifying power of this compound microscope in this set up i.e., in normal use.

Answer 1

Given: Focal length of objective, $f_o=2$cm, Focal length of eyepiece, $f_e=5$cm

Object distance, $u_o=2.4$cm, Image distance, $v=25$cm, Magnifying power, m$=?$

We know, $\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}$

$\frac{1}{2}=\frac{1}{v_o}+\frac{1}{2.4}$

$\frac{1}{v_o}=\frac{1}{2}-\frac{1}{2.4}$

$\frac{1}{v_o}=\frac{2.4-2}{2\times 2.4}$

$\therefore v_o=\frac{2\times 2.4}{0.4}$

$v_o=12$cm.

$\therefore$ Magnifying power, $m=\frac{v_o}{u_o}(1+\frac{D}{f_e})$

$m=\frac{-12}{2.4}(1+\frac{25}{5})$

$=\frac{-12}{2.4}\times \frac{30}{5}$

$m=-30$.