The separation $L$ between the objective $(f=0.5cm)$ and the eyepiece $(f=5cm)$ of a compound microscope is $7cm$.

If a small object be placed so that the eye is least strained to see the image, find the angular magnification produced by the microscope.

The correct option is C $15$
When the eye is least strained, the eyelens is completely relaxed and it focuses parallel rays from infinity.

Therefore in this situation, the image formed by the eye piece has to be at infinity. For that, the image formed by the objective has to lie at the focal length of the eye piec (i.e $5cm$ to the left of the eye-piece).


Using lens formula,

$\frac{1}{f_0}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{0.5}=\frac{1}{2}-\frac{1}{u}$
$\Rightarrow u=\frac{-2}{3}cm$

When the final image is produced at infinity, magnification is given by
$m=m_0\frac{D}{f_e}$
$m_0=\frac{v}{u}=\frac{2}{-2/3}=-3$
$\frac{D}{f_e}=\frac{25cm}{5cm}=5$

Therefore net magnification $m=3\ast 5=15$