Question

he intensity of a cylindrical laser beam is $\mathbf{0}\mathbf{.}\mathbf{800}\mathbf{}\mathbf{W}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. The cross-sectional area of the beam is$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathbf{W}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$ and the intensity is uniform across the cross section of the beam. (a) What is the average power output of the laser? (b) What is the$\mathbf{rms}$ value of the electric field in the beam?

Short answer

a. The average power output of the lase is $2.4\times {10}^{-4}\mathrm{W}$.

b. The$\mathbf{rms}$ value of the electric field of the beam is $\mathbf{17}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{V}\mathbf{/}\mathbf{m}$.

Step-by-step solution

Define the intensity ( I ) and define the formulas.

The power transported per unit area is known as the intensity ( I ) .

The formula used to calculate the intensity ( I ) is:

$I=\frac{P}{A}$

Where, $A$is area measured in the direction perpendicular to the energy and $P$is the power in watts.

And also, $I=\frac{1}{2}{\epsilon }_{0}c{E}_{max}^2$. Where, $I$is the intensity in $W/m^2$and $c$is the speed of light that is equal to $3.0\times {10}^{8}m/s$and also

$$\begin{gathered} E_{rms}=\frac{E_{max}}{\sqrt{2}} \\ {\epsilon }_0=8.85\times {10}^{-12}{C}^2/N·m^2 \end{gathered}$$

So, the$\mathbf{rms}$ value of the electric field in terms of intensity ( I ) is written as

$E_{rms}=\sqrt{\frac{I}{{\epsilon }_{0}c}}$

Determine the average power.

Given that,

$$\begin{gathered} I=0.800\mathrm{W}/{\mathrm{m}}^2 \\ A=3\times {10}^{-4}\mathrm{m} \end{gathered}$$

The formula used to calculate the intensity ( I ) is:

$$\begin{gathered} I=\frac{P_{avg}}{A} \\ \Rightarrow P_{avg}=IA \end{gathered}$$

Substitute the values

$$\begin{gathered} P=\left(0.800\right)\left(3\times {10}^{-4}\right) \\ =2.4\times {10}^{-4} \end{gathered}$$

Hence, average power output of the lase is $2.4\times {10}^{-4}$.

Determine the value of electric field.

The $\mathbf{rms}$value of the electric field is

$E_{rms}=\sqrt{\frac{I}{{\epsilon }_{0}c}}$

Substitute the values

$$\begin{gathered} E_{rms}=\sqrt{\frac{0.800}{\left(8.85\times {10}^{-12}\right)\left(3\times {10}^8\right)}} \\ =17.4\mathrm{V}/\mathrm{m} \end{gathered}$$

Hence, the role="math" localid="1664349239967" $\mathbf{rms}$value of the electric field of the beam is role="math" localid="1664349211670" $17.4\mathrm{V}/\mathrm{m}$.