Q 12-08-057JEE MainJEE Main 2024 (6 Apr, Shift 2)Easy
For the electromagnetic wave $E_y = 600\sin(\omega t - kx)\ \text{V m}^{-1}$, the intensity of the associated light beam is (in $\text{W/m}^2$) (Given $\varepsilon_0 = 9\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$)
Answer: (D) $486$
$$I = \frac12c\,\varepsilon_0E_0^2 = \frac12\times3\times10^8\times9\times10^{-12}\times(600)^2 = 486\ \text{W/m}^2$$
Solution by Sreeraj P, M.Sc Physics