Two slits S_1 and S_2 , arranged symmetrically around the central...

Let the intensity of light from each of the slits $S_1 and$S_2$be$I_0$.<br><br>The intensity at a point on screen$\Sigma_1$at a distance$y$from the central line is given by:<br>$I(y) = 4I_0 \cos^2 \left(\dfrac{\pi d y}{\lambda D}\right)

For slit $S_3$, positioned at the central line ($y = 0$), the intensity is:
$I_3 = 4I_0 \cos^2(0) = 4I_0$

For slit $S_4$, positioned at $y = z =$\dfrac{D\lambda}{4d}, the phase difference is:
$\Delta \phi =$\dfrac{2\pi}{\lambda} \dfrac{d}{D} \left(\dfrac{D\lambda}{4d}\right) = \dfrac{\pi}{2}

The intensity at $S_4$ is:
$I_4 = 4I_0 \cos^2\left(\dfrac{\pi}{4}\right) =$2I_0$<br><br>Slits$S_3$and$S_4$act as coherent secondary sources for the screen$\Sigma_2$. The maximum and minimum intensities on$\Sigma_2$are:<br>$I_\max = $(\sqrt{I_3} + \sqrt{I_4})^2 =$(\sqrt{4I_0} + \sqrt{2I_0})^2 = I_0(2 + $\sqrt{2})^2$
I_\min =(\sqrt{I_3} - \sqrt{I_4})^2 = $(\sqrt{4I_0} - \sqrt{2I_0})^2 = I_0(2 -$\sqrt{2})^2$<br><br>The ratio of maximum to minimum intensity on$\Sigma_2is:<br>\dfrac{I_\max}{I_\min} = \left(\dfrac{2 + \sqrt{2}}{2 - \sqrt{2}}\right)^2 = \left(\dfrac{\sqrt{2} + 1}{\sqrt{2} - 1}\right)^2

Rationalizing the fraction:
\dfrac{\sqrt{2} + 1}{\sqrt{2} - 1} = (\sqrt{2} +$1)^2 = 3 +$2\sqrt{2}

Squaring this result to find the intensity ratio:
\dfrac{I_\max}{I_\min} = ($3 +$2\sqrt{2})^2 = 9 + 8 + $12\sqrt{2} =$17 + $12\sqrt{2}<br><br>Using the approximation \sqrt{2} \approx$1.414:<br>\dfrac{I_\max}{I_\min} \approx17 + 12(1.414) = 17 + 16.968 \approx 34$<br><br>Answer:$34