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 $30\text{cm}$ between the two lenses. The least distance for clear vision is $25\text{cm}$.

The correct option is C $8.5$
Given,
Power of objective, $P_o=25\text{D}$
Power of eye piece, $P_e=5\text{D}$
Length of compound microscope, $L=30\text{cm},D=25\text{cm}$

$f_o=\frac{1}{P_o}=\frac{1}{25}\text{m}=4\text{cm}$

$f_e=\frac{1}{P_e}=\frac{1}{5}\text{m}=20\text{cm}$

Magnifying power is maximum when the image formed by the eye piece is at least the distance of clear vision, $D=25\text{cm}$.

For the eye-piece
$v=-25\text{cm},f_e=20\text{cm}$

Using lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u_e}$

$\Rightarrow \frac{1}{20}=\frac{1}{-25}-\frac{1}{u_e}$

$\Rightarrow \frac{1}{u_e}=\frac{-1}{25}-\frac{1}{20}$

$\therefore u=11.1\text{cm}$


So, for the objective lens, the image distance should be,

$v_o=30-\left(11.1\right)=18.9\text{cm}$

Now, for the objective lens:

$v_o=18.9\text{cm},f_o=4\text{cm}$

Using the formula,

$\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_0}$

$\Rightarrow \frac{1}{4}=\frac{1}{18.9}-\frac{1}{u_o}$

$\Rightarrow u_o\approx -5\text{cm}$

The maximum magnifying power is given as:

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

$=\frac{-18.9}{-5}(1+\frac{25}{20})\approx 8.5$

Hence, $\left(C\right)$ is the correct answer.

Key concept: Maximum magnified image is obtained when the image formed is at least a distance of clear vision from the eye-piece.