Figure 34-56 shows a beam expander made with two coaxial converging lenses of focal lengths $f_1$and $f_1$and separation$d=f_1+f_2$. The device can expand a laser beam while keeping the light rays in the beam parallel to the central axis through the lenses. Suppose a uniform laser beam of width $W_i=2.5\mathrm{mm}$and intensity $I_i=9.0kW/m^2$enters a beam expander for which$f_1=12.5\mathrm{cm}$and $f_2=30.0\mathrm{cm}$.What are (a) $W_f$and (b) $l_f$of the beam leaving the expander? (c) What value of d is needed for the beam expander if lens 1 is replaced with a diverging lens of focal length$f_1=-26.0\mathrm{cm}$?

1. Laser beam width of lens 1, $W_i=2.5\mathrm{mm}$
2. Laser intensity of lens 1, $I_i=9.0kW/m^2$
3. Focal length of laser 1, $f_1=12.5\mathrm{cm}$

4. Focal length of laser 2,$f_2=30\mathrm{cm}$

The width of the beam is defined as the distance between the points of the measured curve. The beam-width of lens 2 is determined and that is used to get the intensity of lens 2 with the intensity of lens 1.

Formula:

Since the triangles that meet at the coincident focal point are similar, the width focal length relation is given by, $\left(W_f\right)/W_i=\left(f_2\right)/f_1$ $\stackrel{}{\to }$(i)

The intensity of a beam,$1=\frac{P}{A}$ $\stackrel{}{\to }$ (ii)

(a)

Rays are converged at focal point of lens1. The rays coming from focal point $f_2$of lens2 are diverged. Since the tringle made by the focal point has the same angle, the beam width of lens 2 can be given using the data in equation (i) as follows:

localid="1663000869987" $$\begin{gathered} W_{f=}\frac{f_2}{f_1}{W}_i \\ =\frac{30.0}{12.5}\times 2.5mm \\ =6.0mm \end{gathered}$$

Hence, the value of the beam width is$6.0mm$.

(b)

Area is proportional to square of the laser width$A\alpha W^2$. Thus, the intensity of the lens 2 can be given using the data in equation (ii) as follows:

role="math" localid="1663000228706" $$\begin{gathered} \frac{I_f}{l_i}=\frac{P/W^2}{P/W^2} \\ =\frac{W{i}^2}{W_{f^2}} \\ =\frac{{f_i}^2}{{f^2}_f}\left(\because Fromequation\left(i\right)\right) \end{gathered}$$

$\begin{array}{l}{l}_{f_{}}=\frac{f_i^2}{f_f^2}{l}_i\\ =\frac{12.5^2}{30.0^2}\times 9.0\times {10}^3\\ =1.56\times {10}^3\\ =1.6kW/m^2\end{array}$

Hence, the value of the intensity is.role="math" localid="1663000528435" $1.6kW/m^2$

(c)

The focal point of the first lens coincides withthefocal point ofthesecond lens. The distance between the two lenses (d) in this case can be given as follows:

$\begin{array}{l}d=f_2-\left|f_1\right|\\ =30\mathrm{cm}-26\mathrm{cm}\\ =4\mathrm{cm}\end{array}$

Hence, the value of $d$is$4cm$.