Question

Suppose that intensity of a laser is $\left(\frac{315}{\pi }\right){\text{W/m}}^2$. The rms electric field, in units of $\text{V/m}$ associated with this source is close to the nearest integer is $({\in }_0=8.86\times {10}^{-12}{\text{C}}^2{\text{Nm}}^{-2};c=3\times {10}^8{\text{ms}}^{-1})$

Answer 1

Using formula of intensity, we have
$I=\frac{1}{2}c{\in }_0{E}_{rms}^2$
$\Rightarrow \frac{1}{2}c{\in }_0{E}_{rms}^2=\frac{315}{\pi }$
$\Rightarrow E_{rms}=\sqrt{\frac{315\times 2}{\pi \times 3\times {10}^8\times 8.86\times {10}^{-12}}}$
$=\sqrt{\frac{630}{83.4612\times {10}^{-4}}}=\sqrt{7.5484\times {10}^4}$
$=2.75\times {10}^2=275\text{V/m}$
Hence, the rms electric field value is, $E_{rms}=275$