Q 12-10-119JEE MainJEE Main 2020 (9 Jan, Shift 1)Medium
Three harmonic waves having equal frequency $\nu$ and same intensity $I_0$, have phase angles $0$, $\dfrac{\pi}{4}$ and $-\dfrac{\pi}{4}$ respectively. When they are superimposed the intensity of the resultant wave is close to:
Answer: (A) $5.8I_0$
Equal intensities mean equal amplitudes $a$. Add the phasors: the components perpendicular to the first wave cancel ($+\pi/4$ and $-\pi/4$), and along it
$$A = a + 2a\cos\frac{\pi}{4} = a(1 + \sqrt{2}) \approx 2.414a$$
$$I = (2.414)^2 I_0 \approx 5.8I_0$$
Solution by Sreeraj P, M.Sc Physics