Question #43569

Two plane electromagnetic waves, Wave 1 and Wave 2, propagate in vacuum along the $+z$-axis. Their electric field vectors are given by: $\vec{E}_1(z,t) = E_0 \cos(k_1 z - \omega_1 t) \hat{i}$ $\vec{E}_2(z,t) = E_0 \cos(k_2 z - \omega_2 t) \left( \cos\theta \hat{i} + \sin\theta \hat{j} \right)$ where $E_0$ is the electric field amplitude, $\theta = 60^\circ$, and $\omega_1, \omega_2$ are optical frequencies such that $\omega_1 - \omega_2 = \delta > 0$. The waves fall normally on a perfectly absorbing detector of area $A$ placed at $z = 0$. The detector has a response time much larger than $2\pi/\omega_1$ but much smaller than $2\pi/\delta$, so it measures the instantaneous intensity $I(t)$ averaged over the fast optical cycles. If $I_0 = \frac{1}{2}\epsilon_0 c E_0^2$ is the intensity of a single wave, what is the total electromagnetic energy absorbed by the detector during the time interval from $t = 0$ to $t = \frac{\pi}{2\delta}$?