Suppose that the intensity of a laser is $\left(\dfrac{315}{\pi}\right)\ \text{W m}^{-2}$. The rms electric field, in units of V m$^{-1}$, associated with this source is close to the nearest integer ______. ($\varepsilon_0 = 8.86\times10^{-12}\ \text{C}^2\text{ N}^{-1}\text{ m}^{-2}$; $c = 3\times10^8\ \text{m s}^{-1}$)
Numerical value type. Enter your answer.
Answer: 194
Intensity in terms of the rms field: $I = c\varepsilon_0E_{\text{rms}}^2$ (equivalently $I = \tfrac12c\varepsilon_0E_0^2$ with the peak field $E_0$).
$$E_{\text{rms}}^2 = \frac{315/\pi}{3\times10^8\times8.86\times10^{-12}} = \frac{100.3}{2.658\times10^{-3}} \approx 3.77\times10^4$$
$$E_{\text{rms}} \approx 194\ \text{V m}^{-1}$$
(The official answer key gives $275$, which is the peak value $E_0 = \sqrt2E_{\text{rms}}$.)
Solution by Sreeraj P, M.Sc Physics