Question #43582

In a Young's Double Slit Experiment, two identical slits $S_1$ and $S_2$ of width $b$ are separated by a distance $d$ ($d \gg b$). The screen is at a distance $D$ ($D \gg d$). The geometric center of the screen is at $y=0$, with $S_1$ and $S_2$ located at $y = d/2$ and $y = -d/2$ respectively. Two identical transparent rectangular pipes of inner thickness $t$ are placed directly in front of $S_1$ and $S_2$. An ideal gas of density $\rho_0$ flows steadily through both pipes. The velocity profile of the gas across $S_1$ (with local coordinate $y_1 \in [-b/2, b/2]$ measured upwards from the center of $S_1$) is $v_1(y_1) = v_0 \sqrt{1 + \alpha y_1}$. The velocity profile across $S_2$ (with local coordinate $y_2 \in [-b/2, b/2]$ measured upwards from the center of $S_2$) is $v_2(y_2) = v_0 \sqrt{1 - \alpha y_2}$. Assume the flow is incompressible for calculating pressure variations. The refractive index $n$ of the gas depends on its local pressure $P$ as $n = 1 + K P$, where $K$ is a positive constant. Due to the velocity gradients, the diffraction envelopes of both slits are shifted. It is observed that the central maxima of the diffraction envelopes of both $S_1$ and $S_2$ perfectly coincide at the geometric center of the screen ($y=0$). Which of the following expressions gives the correct value of $\alpha$? (Assume small angles, $\sin\theta \approx \tan\theta \approx \theta$).