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 is at 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

$f_o=2cm=$Focal length of the objective

$f_e=5cm=$Focal length of the eye piece

$M_e=$Magnification of the eye piece

$M_o=$Magnification of the objective

$M_e=1+\frac{D}{f_o}$

$M_e=1+\frac{25}{5}$

$M_e=6$

For the objective

$u=-2.4cm,v=?f_o=2cm$

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

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

$\frac{1}{v}=\frac{12}{24}-\frac{10}{24}$

$\frac{1}{v}=\frac{1}{12}$

$v=12cm$

$M_o=\frac{v}{u}=\frac{12}{2.4}=5$

Magnifying power of the compound microscope

$M=M_o\times M_e$

$M=5\times 6$

$M=30$