Question #42680

In a Young's Double Slit Experiment (YDSE), the maximum intensity on the screen is $I_m$. A parallel plate capacitor of length $L$ and plate separation $d$ is placed immediately behind one of the slits. The space between the plates is filled with an electro-optic crystal. Light of wavelength $\lambda$ passes through the crystal parallel to the capacitor plates. The other slit has no medium in front of it. The refractive index of the crystal responds to the electric field $E$ between the plates as $n(E) = n_0 + \alpha E$, where $n_0$ and $\alpha$ are constants with appropriate SI units.

An alternating voltage $V(t) = V_0 \sin(\omega t)$ is applied across the capacitor. Assume the low-frequency capacitance remains constant, and the frequency $\omega$ is much higher than the response rate of the intensity detector. The static phase difference introduced by the crystal (when $V=0$) at the central point $O$ on the screen (where the geometric path difference is zero) is $\phi_0 = \frac{\pi}{3}$.

Given the integral identities $\frac{1}{2\pi} \int_0^{2\pi} \cos(z \sin \theta) d\theta = J_0(z)$ and $\frac{1}{2\pi} \int_0^{2\pi} \sin(z \sin \theta) d\theta = 0$, what is the time-averaged intensity $\langle I \rangle$ detected at point $O$?

(Define $\beta = \frac{2\pi \alpha V_0 L}{\lambda d}$)