JEE PYQ: Electromagnetic Waves Question 19
Question 19 - 2020 (06 Sep 2020 Shift 1)
Suppose that intensity of a laser is $\left(\frac{315}{\pi}\right)$ W/m$^2$. The rms electric field, in units of V/m associated with this source is close to the nearest integer is ______. ($\epsilon_0 = 8.86 \times 10^{-12}$ C$^2$Nm$^{-2}$; $c = 3 \times 10^{8}$ ms$^{-1}$)
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Answer: 270
Solution
$I = \frac{1}{2} C \epsilon_0 E_{\text{rms}}^2$. Solving gives $E_{\text{rms}} \approx 270$ V/m.
BETA
