Suppose that intensity of a laser
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:

(270)

Using, intensity, $I=\frac{1}{2} C \epsilon_{0} E_{\mathrm{rms}}^{2}$

$\Rightarrow \frac{1}{2} C \in_{0} E_{\mathrm{rms}}^{2}=\frac{315}{\pi}$

$E_{r m s}=\sqrt{\frac{315 \times 2}{\pi \times 3 \times 10^{8} \times 8.86 \times 10^{-12}}}$

$=\sqrt{\frac{630}{83.4612 \times 10^{-4}}}=\sqrt{7.5484 \times 10^{4}}$

$=2.75 \times 10^{2}=275 \mathrm{~V} / \mathrm{m}$