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

Step 1: Given data

Magnifying power, $m=100$
Focal length of the objective,$f_0=0.5cm$
Tube length,$l=6.5cm$

Step 2: Formula used

$m=\frac{v_0}{u_0}\times \frac{D}{f_e}$

Where, $m$ is magnifying power, $u_0$ is the distance of the object from the lens, $v_0$ is the distance of the image from the lens, $D$ is the distance of distinct vision and $f_e$ is focal length of the eyepiece

Step 3: Find the focal length of the eyepiece.

Since the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eyepiece.
$\therefore v_0+f_e=6.5cm...\left(i\right)$

The magnifying power for normal adjustment is given by

$m=\frac{v_0}{u_0}\times \frac{D}{f_e}$

$$\begin{gathered} \Rightarrow m=-\left[1-\frac{v_0}{f_0}\right]\times \frac{D}{f_e}\left[\because \frac{v_0}{u_0}=-\left[1-\frac{v_0}{f_0}\right]\right] \\ \Rightarrow 100=-\left[1-\frac{v_0}{0.5}\right]\times \frac{25}{f_e}\left[\mathrm{Taking}D=25\right] \\ \Rightarrow 2v_0-4f_e=1...\left(ii\right) \end{gathered}$$
On solving above equations $\left(i\right)$ and $\left(ii\right)$,we get
$v_0=4.5cm$ and $f_e=2cm$

​Hence, the focal length of the eyepiece is $2cm$.