Question

Suppose that intensity of a laser is $\frac{315}{\mathrm{\pi }}W{m}^{-2}$. The RMS electric field, in units of V/m associated with this source is close to the nearest integer is __________. $\left({\epsilon }_{\circ }=8·86\times {10}^{-12}{C}^{2}N{m}^{-2},c=3·0\times {10}^{8}m{s}^{-1}\right)$

Answer 1

Step 1: Given data

The intensity of a laser $\left(I\right)=\frac{315}{\mathrm{\pi }}W{m}^{-2}$.

The value of permittivity ${\epsilon }_{\circ }=8.86\times {10}^{-12}{C}^{2}N{m}^{-2}$

Speed of light $c=3·0\times {10}^{8}m{s}^{-1}$

Step 2: Formula used

The relationship between the intensity of the laser and electric field is given by the formula,

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

Where, $c$ is the speed of light

${\epsilon }_{\circ }$is permittivity

$E_o$ is the absolute electric field

The RMS electric field value can be given by-

$E_{rms}=\frac{E_{\circ }}{\sqrt{2}}$

Step3: Calculating the RMS electric field

Using the formula, the RMS electric field can be calculated as-

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

It is understood that $E_{rms}=\frac{E_{\circ }}{\sqrt{2}}$. So,

$\begin{array}{rcl}{E}_{rms}\times \sqrt{2}& =& \sqrt{\frac{2I}{{\epsilon }_{\circ }c}}\\ E_{rms}& =& \sqrt{\frac{I}{{\epsilon }_{\circ }c}}\\ & =& \sqrt{\frac{{\displaystyle \frac{315}{\mathrm{\pi }}}}{\left(8·86\times {10}^{-12}\right)\left(3·0\times {10}^8\right)}}\\ & =& 194V/m\end{array}$

Hence, the RMS electric field associated with this source is $194\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$.