1 A helium–neon laser,radiating at, $\mathbf{632}\mathbf{.8}\mathbf{\text{nm}}$has a power output of $\mathbf{3}\mathbf{.0}\mathbf{}\mathbf{\text{mW}}$. The beam diverges (spreads) at angle$\mathit{\theta }\mathbf{=}\mathbf{0}\mathbf{.17}\mathbf{\text{mrad}}$(Fig. 33-72). (a) What is the intensity of the beamfrom the laser? (b) What is the power of a point source providing that intensity at that distance?

i) The wavelength of the radiating helium-neon laser,$\lambda =632.8\text{nm}$

ii) Output power$P=3.0\text{mWor}3.0\times {10}^{-3}\text{W}$

iii) Beam angle of divergence,$\theta =0.17\text{mrad}$

iv) Distance of laser and the source,$L=40\text{m}$

First, we have to convert the beam divergence angle to rad. Then, find the diameter of the circle. From that, we can find the radius of the circle. Using the radius, we can find the intensity at that distance. We can find the power by using the formula of intensity for the point source.

Formula:

The intensity of a light for itspower, $\mathit{I}\mathbf{=}\frac{\mathbf{P}}{\mathbf{A}}$(i)

The intensity of a point source, $\mathit{I}\mathbf{=}\frac{\mathbf{P}}{\mathbf{4}\mathbf{\pi }{\mathbf{r}}^{\mathbf{2}}}$ (ii)

The small angle approximation for tangent case for a laser source,

(iii)

$\mathit{\theta }\mathbf{=}\frac{\mathbf{D}\mathbf{i}\mathbf{s}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{c}\mathbf{e}}{\mathbf{D}\mathbf{i}\mathbf{a}\mathbf{m}\mathbf{e}\mathbf{t}\mathbf{e}\mathbf{r}\mathbf{o}\mathbf{f}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{s}\mathbf{o}\mathbf{u}\mathbf{r}\mathbf{c}\mathbf{e}}$

(a)

First, we have to convert the beam divergence into rad. That is given by,

$\begin{array}{c}\theta =0.17\text{mrad}\times \frac{\text{1m}}{\text{1000mm}}\\ =0.00017\text{rad}\end{array}$

Now,the diameter of the laser source using the given data using equation (iii) is given by,

data-custom-editor="chemistry" $\begin{array}{c}\text{0.00017rad}=\frac{D}{\text{40m}}\\ D=6.8\times {10}^{-3}\text{m}\end{array}$

So, the radius of circle is given as follows:

$\begin{array}{c}r=\frac{D}{2}\\ =\frac{6.8\times {10}^{-3}\text{m}}{2}\\ =3.4\times {10}^{-3}\text{m}\end{array}$

Now, the intensity is given by the formula of equation (i) as follows:

$\begin{array}{c}I=\frac{P}{\pi r^2}\\ =\frac{3\times {10}^{-3}\text{W}}{\pi \times {(3.4\times {10}^{-3}\text{m})}^2}\\ =82.6\\ \approx 83{\text{W/m}}^{\text{2}}\end{array}$

Hence, the intensity of the beam is.$83{\text{W/m}}^{\text{2}}$

(b)

For the point source providing the intensity of$83{\text{W/m}}^{\text{2}}$, the power can be calculated as using equation (ii) and the given data as follows:

$\begin{array}{c}P=I\times 4\pi \times r^2\\ =83\times 4\pi \times {40}^2\\ =1.67\times {10}^6\text{W}\\ =1.67\text{MW}\end{array}$

$\approx 1.7\text{MW}$

Hence, the value of the power is$1.7\text{MW}$