If we need a magnification of $375$ from a compound microscope of tube length $150mm$ and an objective of focal length $5mm$, the focal length of the eye-piece, should be close to :

Given $M=375,L=150mm,f_0=5mm,fe=?,d=25cm$.

$M$ for a compound microscope is given by

$M=\frac{v_0}{+u_0}\left(1+\frac{d}{fe}\right)$

focal length of objective lens is small $u_0\simeq f_0$. Alos as focal length of eye is small $v_0\simeq L$

$\therefore M=\frac{L}{+f_0}\left(1+\frac{d}{f_e}\right)$

$\Rightarrow 375=\frac{150}{5}\left(1+\frac{250}{f_e}\right)$

$\Rightarrow \frac{375}{30}=\left(1+\frac{250}{f_e}\right)$

$\Rightarrow \frac{375}{30}-1=\frac{250}{f_e}$

$\Rightarrow \frac{345}{30}=\frac{250}{f_e}$

$\Rightarrow f_e=2.2cm$

$\boxed{F_e=22mm}$