Question

A compound microscope has an objective and eye piece of focal lengths 2.5 cm and 6.0 cm respectively. The lenses are 20cm apart. A small object is kept at 3 cm from the objective . Find the distance of final image from eyepiece and the magnifing power of microscope.

Answer 1

Given, $u_0=3cm,f_0=2.5cm,v_0=$image distance for objective lens

$\frac{1}{f_0}=\frac{1}{v_0}-\frac{1}{u_0}$

$\frac{1}{v_0}=\frac{1}{f_0}+\frac{1}{u_0}=\frac{1}{2.5}-\frac{1}{3}$

$v_0=\frac{3\times 2.5}{3-2.5}=15cm$

Object distance for eye piece $=20-15=5cm$

Given, $u_e=-5cm,f_e=6cm,v_e=$image formed by eye piece

$\frac{1}{v_e}=\frac{1}{f_e}+\frac{1}{u_e}=\frac{1}{6}-\frac{1}{5}$

$v_e=-30cm$

Magnification $M=\frac{-L}{f_0}\times \frac{25}{f_e}$ where L is length of microscope

Here $L=20cm$,

$M=\frac{-20}{2.5}\times \frac{25}{6}=-33.33$