Question

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$f_1=-26.0\mathrm{cm}$?

Short answer

1. The width of laser beam expander$6.0\mathrm{mm}$is.
2. The intensity of laser beam expander islocalid="1663000775812" role="math" $1.6kW/m^2$.
3. The value of needed for expander if lens 1 is replaced with the diverging lens of $f_1=-26.0\mathrm{cm}$is$4\mathrm{cm}$.

Step-by-step solution

The given data

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,$f_2=30\mathrm{cm}$

Understanding the concept of properties of the lens

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)

Calculation of the width of the laser beam expander

(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$6.0mm$.

Calculation of the intensity of the beam expander

(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" $1.6kW/m^2$

Calculation of the value of  

(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$4cm$.