Consider a 'beam expander' which consists of two converging lenses of focal lengths $40\mathrm{cm}$ and $100\mathrm{cm}$ having a common optical axis. A laser beam of diameter $2\mathrm{mm}$ is incident on the $40\mathrm{cm}$ focal length lens. Then what is the diameter (in mm) of the final beam (see figure)?

Step 1: Given information

Diameter of the beam, $2{\mathrm{r}}_1=2\mathrm{mm}$

The focal length of lenses,

$$\begin{gathered} {\mathrm{f}}_1=40\mathrm{cm} \\ {\mathrm{f}}_2=100\mathrm{cm} \end{gathered}$$

Step 2: To find

The diameter of the final beam.

Step 3: Finding the diameter of the final beam

An objective lens plus an imaging lens together with a beam expander. As we've seen, the ratio of the first lens's radius and focal length is identical to the proportion of the second lens's radius and focal length.

Therefore,

$\Rightarrow \frac{{\mathrm{r}}_1}{{\mathrm{f}}_1}=\frac{{\mathrm{r}}_2}{{\mathrm{f}}_2}$

By substituting the given values, we get

$$\begin{gathered} \Rightarrow \frac{1}{40}=\frac{{\mathrm{r}}_2}{100} \\ \Rightarrow {\mathrm{r}}_2=\frac{100}{40} \\ \Rightarrow 2{\mathrm{r}}_2=5\mathrm{mm} \\ \Rightarrow Diameter=5mm \end{gathered}$$

Therefore, the diameter of the final beam will be $5\mathrm{mm}$.