You are given two converging lenses of focal lengths, $1.5\text{cm}$ and $4\text{cm}$ to design a compound microscope. If it is desired to have a magnifying power of $50$, in case of normal adjustment, then find out the separation between the objective lens and the eyepiece.

The correct option is B $12\text{cm}$
Given:
$f_o=1.5\text{cm}$
$f_e=4\text{cm}$
$m=50$

We know that, in normal adjustment, the final image is formed at infinity.

So, $|u_e|=|f_e|$

Now, magnifying power,

$m=\frac{v_o}{|u_o|}\left(\frac{D}{f_e}\right)$

$\Rightarrow m=\frac{L}{f_o}\left(\frac{D}{f_e}\right)$

$\Rightarrow 50=\frac{L}{1.5}\left(\frac{25}{4}\right)$

$\Rightarrow L=12\text{cm}$

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