Question

The separation between the objective and the eyepiece of a compound microscope can be adjusted between 9'8 cm to 11.8 cm. If the focal lengths of the objective and the eyepiece are 1.0 cm and 6 cm respectively, find the range of the magnifying power if the image is always needed at 24 cm from the eye.

Answer 1

Here, $f_0$ = 1.0 cm, $f_e$ = 6 cm,

$V_e$ = 24 cm,

Separation = 9.8 cm to 11.8 cm.

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

$\frac{1}{-24}-\frac{1}{u_e}=\frac{1}{6}$

$\Rightarrow -\frac{1}{u_e}=\frac{1}{6}+\frac{1}{24}=\frac{4+1}{24}=\frac{5}{24}$

$\therefore u_e=-\frac{24}{5}cm=-4.8cm$

$\therefore v_0$ = 9.8 - 4.8 = 5cm

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

$\frac{1}{5}-\frac{1}{u_0}=\frac{1}{1}$

$\Rightarrow -\frac{1}{u_0}=1-\frac{1}{5}=\frac{5-1}{5}$

$\Rightarrow u_0=-\frac{5}{4}cm=-1.25cm$

So,

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

$=\frac{5}{1.25}(1+\frac{24}{6})=20$

Similarly, when, l = 11.8 cm, m = 30

So, required range is 20 to 30.