Find the maximum magnifying power of a compound microscope having a $25\text{D}$ objective lens, a $5\text{D}$ eyepiece and a separation of $30\text{cm}$ between the two lenses.

The correct option is C $8.5$
$f_o=\frac{1}{P_o}=\frac{1}{25}\text{m}=4\text{cm}$

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

Magnifying power is maximum when the final image is formed at least distance of clear vision.

For the eye-piece :

$v_e=-25\text{cm}$ and $f_e=20\text{cm}$

Using lens formula,

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

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

$\Rightarrow u_e=-11.1\text{cm}$

Also, length of the compound microscope,

$L=|v_0|+|u_e|$

$\Rightarrow |v_0|=L-|u_e|=30-11.1=18.9\text{cm}$

For the objective lens:

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

Using lens formula,

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

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

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

The maximum magnifying power is given by,

$m=\frac{v_o}{|u_o|}(1+\frac{D}{f_e})$

$\Rightarrow m=\frac{18.9}{5}(1+\frac{25}{20})=8.5$

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