Question

Find the maximum magnifying power of a compound microscope having a 25 diopter lens as the objective, a 5 diopter lens as the eyepiece and the separation 30 cm between the two lenses. The least distance for clear vision is 25 cm.

Answer 1

For the given compound microscope,

$f_0=\frac{1}{\text{25 diopter}}$

= 0.04 m = 4 cm,

$f_0=\frac{1}{\text{5 diopter}}=0.2m=20cm$

D = 25 cm, separation between objective and eyepiece = 30 cm.

The magnifying power is maximum when the image is formed by the eye piece at the least distance of clear vision i.e., D = 25 cm.

For the eye piece,

$V_e=-25cm,f_e=20cm$

From lense formula

$\frac{1}{v_e}=\frac{1}{u_e}+\frac{1}{f_e}$

$\Rightarrow \frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}$

$=\frac{1}{-25}-\frac{1}{20}=-\frac{(4+5)}{100}$

$\Rightarrow u_e=\frac{-100}{9}=11.11cm$

So, the maximum magnifying power is given by,

$m=\frac{v_0}{u_0}=(1+\frac{D}{f_e})$

$=-\left(\frac{18.89}{-5.07}\right)(1+\left(\frac{25}{20}\right))$

$=3.7225\times 2.25=8.376$