Two slits S_1 and S_2 , located symmetrically about the central...

The path difference for the light waves from slits $S_1$ and $S_2$ reaching a point on screen $\Sigma_1$ at a distance $y$ from the central axis is given by:
$\Delta x = \dfrac{y d}{D}$

The intensity at any point on $\Sigma_1$ is:
$I = I_0 \cos^2$\left(\dfrac{\pi \Delta x}{\lambda}\right)
where $I_0$ is the maximum intensity on $\Sigma_1$.

For slit $S_3$, which is on the central axis ($y = 0$):
$\Delta x_3 = 0$
$I_3 = I_0 \cos^2(0) = I_0$

For slit $S_4$, which is at a distance $y = z =$\dfrac{D\lambda}{d} from the central axis:
$\Delta x_4 =$\dfrac{z d}{D} = \dfrac{\left(\dfrac{D\lambda}{d}\right) d}{D} = \lambda
$I_4 = I_0 \cos^2\left(\dfrac{\pi \lambda}{\lambda}\right) =$I_0 \cos^2(\pi) = I_0$<br><br>Since$I_3 = I_4$, the slits$S_3$and$S_4$act as coherent secondary sources of equal intensity for the interference pattern on screen$\Sigma_2$.<br><br>For two coherent sources of intensities$I_3$and$I_4$, the maximum and minimum intensities on the screen$\Sigma_2$are:<br>$I_\max = $(\sqrt{I_3} + \sqrt{I_4})^2 =$(\sqrt{I_0} + \sqrt{I_0})^2 = 4I_0$<br>$I_\min = $(\sqrt{I_3} - \sqrt{I_4})^2 =$(\sqrt{I_0} - \sqrt{I_0})^2 = 0$<br><br>Therefore, the ratio of the maximum to minimum intensity observed on$\Sigma_2is:<br>\dfrac{I_\max}{I_\min} = \dfrac{4I_0}{0} = \infty