A small laser used as a pointer produces a beam of red light $\mathbf{5}\mathbf{}\mathit{m}\mathit{m}$ in diameter and has a power output of $\mathbf{5}\mathbf{}\mathit{m}\mathit{W}$. What is the magnitude of the electric field in the laser beam?

The beam of the red light,$d=5\text{mm}$

The output power,$P=5\text{mW}$

The energy density for flux of electromagnetic radiation is defines as the ratio of the product of electric field and magnetic field and permeability of the vacuum.

The expression to calculate the energy density for flux of electromagnetic radiation is given as follows.

$\mathit{S}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}\mathit{E}\mathbf{\times }\mathit{B}$ …(i)

Here,${\mathit{\mu }}_{\mathbf{0}}$ is the permeability of vacuum, E is the electric field and B is the magnetic field.

The expression to calculate the energy density in terms of power and area is given as follows.

$\mathit{S}\mathbf{=}\frac{\mathbf{P}}{\mathbf{A}}$ …(ii)

Here, A is the area.

The magnetic field for the slab of radiation is given by,

$\mathit{B}\mathbf{=}\frac{\mathbf{E}}{\mathbf{c}}$

Calculate the energy density.

Substitute$\pi r^2$ for A into equation (ii).

$S=\frac{P}{\pi r^2}$

Substitute $5\text{mWforP}$and $\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.forr$into above equation.

$$\begin{gathered} S=\frac{5\times {10}^{-3}\text{W}}{\pi \times {\left(\frac{5}{2}\times {10}^{-3}\text{m}\right)}^2} \\ =\frac{5\times {10}^{-3}\text{W}}{1.9\times {10}^{-5}{\text{m}}^2} \\ =263.15W/m^2 \end{gathered}$$

Calculate the energy density for flux of electromagnetic radiation.

Substitute $\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$c$}\right.forB$into equation (i).

$$\begin{gathered} S=\frac{1}{{\mu }_0}E\times \frac{E}{c} \\ =\frac{1}{{\mu }_0}\frac{E^2}{c} \\ ={\mu }_{0}cS \\ =\sqrt{{\mu }_{0}cS} \end{gathered}$$

Substitute $263.15\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$m^2$}\right.forS,4\pi \times {10}^{-7}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$A$}\right.for{\mu }_0,$and $3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.forc$into above equation.

$$\begin{gathered} E=\sqrt{\left(4\pi \times {10}^{-7}T\raisebox{1ex}{$.m$}\!\left/ \!\raisebox{-1ex}{$A$}\right.\right)\times \left(3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)\times \left(263.15\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$m^2$}\right.\right)} \\ E=\sqrt{99205.2T.\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$A.s$}\right.} \\ E=314.9\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right. \end{gathered}$$

Hence the energy density for flux of electromagnetic radiation is $314.9\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$.