Question #43245

In a modified Young's Double Slit Experiment, monochromatic light of wavelength $\lambda$ illuminates two identical slits $S_1$ and $S_2$ separated by a distance $d$. The screen is placed at a distance $D \gg d$. Slit $S_1$ is fixed, while slit $S_2$ is made to oscillate along the axis perpendicular to the plane of the slits with a displacement $z(t) = A \sin(\omega t)$, where $A \ll \lambda$. The frequency of oscillation $f = \omega / 2\pi$ is much higher than the response rate of the detector on the screen, so the detector records the time-averaged intensity. If the intensity produced by a single slit on the screen is $I_0$, what is the fringe visibility $V = \frac{I_{\text{max}} - I_{\text{min}}}{I_{\text{max}} + I_{\text{min}}}$ of the observed interference pattern?