Question

In the given electromagnetic wave $E_y = 600 \sin(\omega t - kx) \, \text{V/m},$ intensity of the associated light beam is (in W/m$^2$); (Given $\epsilon_0 = 9 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2}$).

• A. 486
• B. 243
• C. 729
• D. 972

Answer 1

Answer: (A)

486

Answer 2

The intensity $I$ of an electromagnetic wave is given by:
$I = \frac{1}{2} \varepsilon_0 E_0^2 c$
where $E_0 = 600 \, \text{Vm}^{-1}$ and $c = 3 \times 10^8 \, \text{m/s}$.
Substitute the values:
$I = \frac{1}{2} \times 9 \times 10^{-12} \times (600)^2 \times 3 \times 10^8$
$= \frac{9}{2} \times 36 \times 3 = 486 \, \text{W/m}^2$