Question

An optical instrument used for angular magnification has a 25 D objective and a 20 D eyepiece. The tube length is 25 cm when the eye is least strained.

(a) Whether it is a microscope or a telescope?

(b) What is the angular magnification produced?

Answer 1

Microscope

For given microscope,

$f_0=\frac{1}{25D}=0.04m=4cm$

$f_e=\frac{1}{20D}=0.05m=5cm$

Tube length = 25 cm, $v_e$ = 25 cm

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

$-\frac{1}{25}-\frac{1}{u_e}=\frac{1}{5}$

$\Rightarrow -\frac{1}{u_e}=\frac{1}{5}+\frac{1}{25}=\frac{5+1}{25}=\frac{6}{25}$

$\Rightarrow u_e=-\frac{25}{6}cm$

$\therefore v_0=25-\frac{25}{6}$

$=25\times \frac{5}{6}=20.84cm$

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

$\frac{1}{20.83}-\frac{1}{u_0}=\frac{1}{4}$

$\Rightarrow -\frac{1}{40}=\frac{1}{4}-\frac{1}{20.84}$

$=\frac{5.21-1}{20.84}=\frac{4.21}{20.84}$

$\Rightarrow -\frac{v_0}{f_0}=4.21$

So, required magnification (m)

$=\frac{v_0}{u_0}\left(\frac{D}{f_e}\right)=4.21\times \frac{25}{5}=21$