Consider the setup shown in the figure. Two slits S_1 and S_2 are...

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

Since slit $S_3$ is on the central line ($y = 0$), the path difference is $\Delta x_3 = 0$. Thus, the intensity at $S_3$ is maximum, let it be $I_0$.

Slit $S_4$ is at a distance $z = \dfrac{D\lambda}{2d}$ from the central line. The path difference at $S_4$ is:

$\Delta x_4 = \dfrac{z d}{D} = \left(\dfrac{D\lambda}{2d}\right) \dfrac{d}{D} = \dfrac{\lambda}{2}$

A path difference of $\dfrac{\lambda}{2}$ corresponds to destructive interference. Therefore, the intensity at $S_4$ is $0$.

Since slit $S_4$ receives no light, it acts as a source of zero intensity for the second screen $\Sigma_2$. Only the light from slit $S_3$ reaches screen $\Sigma_2$.

With only one coherent source ($S_3$) illuminating screen $\Sigma_2$, no interference pattern is formed. The intensity distribution on $\Sigma_2$ will be uniform.

Therefore, I_\max = I_\min, which gives the ratio \dfrac{I_\max}{I_\min} = 1.