Question

In a compound microscope focal length of objective and eyepiece lens is $4$ cm and $8$ cm respectively. If length of tube is $30$ cm then magnifying power for normal adjustment will be:-

Answer 1

$f_e=8cm$

$f_0=4cm$

$u_0=30cm$

$v_e=\infty$

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

$\frac{1}{v_0}=\frac{1}{f_0}-\frac{1}{u_0}=\frac{1}{4}-\frac{1}{30}=\frac{26}{4}=6.5$

$v_0=\frac{1}{6.5}$

$m=\frac{v_0}{u_0}(1+\frac{D}{f_e})=\frac{0.153}{30}(1+\frac{25}{3})=0.04$