Example: The separation $L$ between the objective ($f_o=0.5cm$) and the eyepiece ($f_o=5cm$) of a compound microscope is $7cm$. Where should a small object be placed so that the eye is least strained to see the image and what is the angular magnification produced?

Solution:
The eye is least strained if the final image is formed at infinitely. In such a case the image formed by objective should fall at the focus of eyepiece.
As $f_o=5cm,L=7cm$, this first image is formed at a distance = $7-5=2cm$ from objective.
Thus $v=+2cm$. Also  $f_o=0.5cm$
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f_{\mathrm{o}}}$
or, $\frac{1}{u}=\frac{1}{v}-\frac{1}{f_{\mathrm{o}}}=\frac{1}{2}-\frac{1}{0.5}$
or, $u=-\frac{2}{3}cm$
The object should be placed at $\frac{2}{3}cm$ from objective.
Angular magnification = $\frac{v}{u}\times \frac{D}{f_e}$$=\frac{2}{-\frac{2}{3}}\times \frac{25}{5}$$=-15$