The objective lens of a compound microscope produces magnification of $10$. In order to get an overall magnification of $100$ when image is formed at $25cm$ from the eye, the focal length of the eye piece should be

The correct option is C $\frac{25}{9}cm$
For final image to be formed at $25cm$, the image formed by the objective has to lie within the focal length of the eyepiece.


In this case,

Magnification $m=m_o(1+\frac{D}{f_e})$ where $m_0$ is the magnification of the objective, $D$ is the near point of the eye ($25cm$) and $f_e$ is th focal length of the eyepiece.

Given $m=100$,
$m_0=10$
Substituting in the equation,

$100=10(1+\frac{25}{f_e})$
From this $f_e=\frac{25}{9}cm$