Obtain the equation of magnification for a compound microscope.

(N/A) The linear magnification due to the objective lens is given by $m_{0} = \frac{h^{\prime}}{h}$,where $h^{\prime}$ is the size of the intermediate image and $h$ is the size of the object.
From the geometry of the ray diagram,$\tan \beta = \frac{h}{f_{0}} \Rightarrow h = f_{0} \tan \beta$ $(1)$ and $\tan \beta = \frac{h^{\prime}}{L}$,where $L$ is the tube length (distance between the focal point of the objective and the eyepiece).
Thus,$h^{\prime} = L \tan \beta$ $(2)$.
Therefore,the magnification of the objective is $m_{0} = \frac{h^{\prime}}{h} = \frac{L \tan \beta}{f_{0} \tan \beta} = \frac{L}{f_{0}}$ $(3)$.
The angular magnification of the eyepiece is $m_{e} = \frac{D}{f_{e}}$ when the final image is formed at infinity,and $m_{e} = 1 + \frac{D}{f_{e}}$ when the final image is formed at the near point $(D)$.
Thus,the total magnification $m$ of the compound microscope is given by $m = m_{0} \times m_{e} = \left( \frac{L}{f_{0}} \right) \left( \frac{D}{f_{e}} \right)$ for the final image at infinity.