Solve this following

Question:

Suppose that intensity of a laser is

$\left(\frac{315}{\pi}\right) \mathrm{W} / \mathrm{m}^{2}$. The rms electric field, in units

of $\mathrm{V} / \mathrm{m}$ associated with this source is close to the nearest integer is

$\left(\epsilon_{0}=8.86 \times 10^{-12} \mathrm{C}^{2} \mathrm{Nm}^{-2} ; \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right)$

 

Solution:

$\mathrm{I}=\epsilon_{0} \mathrm{E}_{\mathrm{rms}}^{2} \mathrm{C}$

$\mathrm{E}_{\mathrm{rms}}^{2}=\frac{\mathrm{I}}{\epsilon_{0} \mathrm{C}}$

$=\frac{315}{\pi \in_{0}} \times \frac{1}{\mathrm{C}}$

$=\frac{4 \times 315}{4 \pi \in_{0}} \times \frac{1}{3 \times 10^{8}}$

$=\frac{4 \times 315 \times 9 \times 10^{9}}{3 \times 10^{8}}$

$\mathrm{E}_{\mathrm{rms}}^{2}=4 \times 315 \times 30$

$E_{\mathrm{rms}}=2 \sqrt{315 \times 30}$

$=194.42$

Ans. $194.00$

 

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