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

Case-I
If final image is at least distance of clear vision
$\begin{array}{l}M.P.=\frac{L}{f_0}\left(1+\frac{D}{f_e}\right)\text{}\\ \Rightarrow 375=\frac{150}{5}\left[1+\frac{25}{f_e}\right] \text{}\\ \Rightarrow \frac{375}{30}=1+\frac{25}{f_e} \text{}\\ \Rightarrow \frac{345}{30}=\frac{25}{f_e}\end{array}$     
$\begin{array}{l}\Rightarrow f_e=\frac{750}{345}=2.17cm\text{}\\ \Rightarrow f_e\approx 22mm\end{array}$ 
Case-II
If final image is at infinity
$\begin{array}{l}M.P=\frac{L}{f_0}\left(\frac{D}{f_e}\right)=375\\ \text{}\Rightarrow \frac{375}{30}=\frac{25}{f_e} \text{}\\ \Rightarrow f_e = \frac{750}{375} \text{} \\ \Rightarrow f_e=2 cm\text{}\\ \Rightarrow f_e\approx 22 mm\text{}\end{array}$