Question

A laser beam has diameter 1.20 mm. What is the amplitude of the electric field of the electromagnetic radiation in this beam if the beam exerts a force of $\mathbf{3}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}\mathbf{}}\mathit{N}$on a totally reflecting surface?

Short answer

The amplitude of the electric field of the electromagnetic radiation is$1.95\times {10}^4\mathrm{V}/\mathrm{m}$ .

Step-by-step solution

Radiation pressure exerted by electromagnetic wave

Radiation pressure is mechanical pressure exerted upon any surface due to the exchange of momentum between the object and the electromagnetic field.

The Radiation pressure exerted by electromagnetic wave is given by

$p_{rad}=\frac{2I}{c}$

Where, $p_{rad}$is the radiation pressure, I is the intensity and C is the speed of light.

Intensity of electromagnetic wave

Intensity of electromagnetic wave in terms of electric field

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

Where, ${\epsilon }_0$is the permittivity of vacuum ,role="math" localid="1664425325176" $E_{max}$is the maximum amplitude of electric field and C is the speed of light.

Calculation of amplitude of electric field in EM wave in vacuum

Using

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

$E_{max}=\frac{2I}{{\epsilon }_{0}c}$

Put the values of constants in above equation

$$\begin{gathered} {\mathrm{E}}_{\max }=\sqrt{\frac{2(5.04\times {10}^5\mathrm{W}/{\mathrm{m}}^2)}{8.85\times {10}^{-12}{\mathrm{C}}^2\mathrm{N}.{\mathrm{m}}^2(3\times {10}^8\mathrm{m}/\mathrm{s})}} \\ {\mathrm{E}}_{\max }=1.95\times {10}^4\mathrm{V}/\mathrm{m} \end{gathered}$$

Thus, the amplitude of the electric field of the electromagnetic radiation is $1.95\times {10}^4\mathrm{V}/\mathrm{m}$.