Question #43938

In a modified Young's Double Slit Experiment, slit $S_1$ is illuminated by a steady coherent light source of constant intensity $I_0$. Slit $S_2$ is illuminated by a second coherent light source whose emitted electric field amplitude is directly proportional to the instantaneous current $i(t)$ in an adjacent $L-R$ circuit. The $L-R$ circuit (with inductance $L$ and resistance $R$) initially carries a steady current $i_0$. At $t=0$, the DC voltage source is short-circuited, and the current begins to decay. At $t=0$, the intensity of light from $S_2$ is exactly $I_0$. A photographic plate at the screen records the total integrated light energy per unit area (exposure) from $t=0$ to $t=\tau$, where $\tau = L/R$ is the time constant of the circuit. Find the visibility $V$ of the resulting interference fringes on the photographic plate. (Visibility is defined as $V = \frac{E_{max} - E_{min}}{E_{max} + E_{min}}$, where $E$ is the total integrated exposure).