Consider two harmonic oscillations being superimposed. Let $I$ be the time-averaged energy (intensit

Question

Consider two harmonic oscillations being superimposed. Let $I$ be the time-averaged energy (intensity) of the resultant oscillation, and $I_1$, $I_2$ be the time-averaged energies of the constituent oscillations. Under which of the following conditions is the resultant time-averaged energy exactly equal to the sum of the energies of the constituent oscillations ($I = I_1 + I_2$)?

Professor PC Thomas Classes · Answer

For two waves $\vec{E}_1$ and $\vec{E}_2$, the resultant intensity $I \propto \langle (\vec{E}_1 + \vec{E}_2)^2 \rangle = \langle \vec{E}_1^2 \rangle + \langle \vec{E}_2^2 \rangle + 2\langle \vec{E}_1 \cdot \vec{E}_2 \rangle$. For $I = I_1 + I_2$, the interference term $2\langle \vec{E}_1 \cdot \vec{E}_2 \rangle$ must be zero. (A) If incoherent and phase difference is uniformly distributed, $\langle \cos \delta \rangle = 0$. (B) If mutually perpendicular, the dot product $\vec{E}_1 \cdot \vec{E}_2 = 0$ instantaneously at all times. (C) If coherent with phase difference $\pi/2$, $\cos(\pi/2) = 0$. (D) If phase difference is $0$, the term is $2\sqrt{I_1 I_2}$, which is non-zero.