A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. Find the focal length of the eye- piece.

When final image is at infinity the magnification produced by eye piece is $\frac{\mathrm{D}}{f_e}\mathrm{and}{\text{ \hspace{0.17em}u}}_{\mathrm{e}}={\mathrm{f}}_{\mathrm{e}}$

$$\begin{gathered} \mathrm{L}={\mathrm{v}}_{\mathrm{o}}+{\mathrm{u}}_{\mathrm{e}} \\ 6.5={\mathrm{V}}_{\mathrm{o}}+{\mathrm{f}}_{\mathrm{e}} \end{gathered}$$

$$\begin{gathered} \frac{1}{{\mathrm{V}}_{\mathrm{o}}}+\frac{1}{{\mathrm{u}}_{\mathrm{o}}}=\frac{1}{{\mathrm{f}}_{\mathrm{o}}} \\ \frac{{\mathrm{v}}_{\mathrm{o}}}{{\mathrm{u}}_0}+1=\frac{{\mathrm{v}}_{\mathrm{o}}}{{\mathrm{f}}_{\mathrm{o}}} \end{gathered}$$

                     $\left(\frac{v_0}{u_{\mathrm{o}}}+1\right)=\frac{\left(6.5-{\mathrm{f}}_{\mathrm{e}}\right)}{0.5}$

Overall magnification  $\frac{{\mathrm{v}}_0}{{\mathrm{u}}_{\mathrm{o}}}\left(\frac{\mathrm{D}}{{\mathrm{f}}_{\mathrm{e}}}\right)=100$

                                            $\frac{{\mathrm{v}}_{\mathrm{o}}}{{\mathrm{u}}_{\mathrm{o}}}=4{\mathrm{f}}_{\mathrm{e}}$

                                    $\left(4{\mathrm{f}}_{\mathrm{e}}+1\right)=2\left(6.5-{\mathrm{f}}_{\mathrm{e}}\right)$

                                      $4{\mathrm{f}}_{\mathrm{e}}+1=13-2{\mathrm{f}}_{\mathrm{e}}$

                                            $6{\mathrm{f}}_{\mathrm{e}}=12$

                                             ${\mathrm{f}}_{\mathrm{e}}=2\mathrm{cm}$