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 eyepiece.

For the compound microscope, we have:
Magnifying power, m = 100
Focal length of the objective, f_o = 0.5 cm
Tube length, l = 6.5 cm
Since the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eyepiece.
∴ v_o + f_e = 6.5 cm …(1)
The magnifying power for normal adjustment is given by
$$\begin{gathered} m=\frac{v_o}{u_o}\times \frac{D}{f_e} \\ =-\left[1-\frac{v_o}{f_o}\right]\frac{D}{f_e} \\ \Rightarrow 100=-\left[1-\frac{v_o}{0.5}\right]\times \frac{25}{f_e} \\ \Rightarrow 2v_o-4f_e=1\dots \left(2\right) \end{gathered}$$
On solving equations (1) and (2), we get:
V_o = 4.5 cm and f_e = 2 cm
Thus, the focal length of the eyepiece is 2 cm.