He–Ne lasers are often used in physics demonstrations. They produce light of wavelength 633 nm and a power of 0.500 mW spread over a cylindrical beam 1.00 mm in diameter (although these quantities can vary). What is the average energy density in the laser beam?

Definition of Intensity of light

Intensity can be defined as power per unit area .

Formula of intensity

$\mathrm{I}=\frac{\mathrm{p}}{\mathrm{A}}$

Where, P is the power and A is the area of surface .

Intensity of electromagnetic wave

Intensity of electromagnetic wave in terms of electric field is given by

$I=\frac{1}{2}{\epsilon }_{0}c{E^2}_{max}$

Where, ${\epsilon }_0$is the permittivity of vacuum and ${\mathrm{E}}_{\max }$is the maximum amplitude of electric field.

Average energy density of electromagnetic wave associated with electric field

It can be defined as energy per unit volume of electromagnetic wave associated with electric field.

It can be given by

$u_{avg}=\frac{1}{2}{\epsilon }_{0}c{E^2}_{max}$

Where, $u_{avg}$is the average energy density of EM wave.

Using

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

Put the values of constants in above equation

$$\begin{gathered} I=\frac{0.5\times {10}^{-3}W}{7.85\times {10}^{-7}{m}^2} \\ I=637W/m^2 \end{gathered}$$

Thus, the value of intensity of light is$637W/m^2$.

Using

$$\begin{gathered} I=\frac{1}{2}{\epsilon }_{0}c{E^2}_{max} \\ {E}_{max}=\sqrt{\frac{2I}{{\epsilon }_{0}c}} \end{gathered}$$

Put the values of constants in above equation

$$\begin{gathered} {\mathrm{E}}_{\max }=\sqrt{\frac{2\left(637\mathrm{W}/{\mathrm{m}}^2\right)}{8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2(3\times {10}^8\mathrm{m}/\mathrm{s})}} \\ {\mathrm{E}}_{\max }=693\mathrm{V}/\mathrm{m} \end{gathered}$$

Thus, the maximum amplitude of electric field in EM wave is$693\mathrm{V}/\mathrm{m}$$$" localid="1664427146988">

Using

$u_{avg}=\frac{1}{2}{\epsilon }_0{E^2}_{max}$

Now, put the values of constants in above equation

${\mathrm{u}}_{\mathrm{avg}}=\frac{1}{2}(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2){(693\mathrm{V}/\mathrm{m})}^2{\mathrm{u}}_{\mathrm{avg}}=2.12\times {10}^{-6}\mathrm{J}/\mathrm{m}{3}^{}$

Thus, the average energy density in the laser beam is $2.12\times {10}^{-6}\mathrm{J}/\mathrm{m}{3}^{}$.