Explanation

(b) As $S_{3},\ \mathrm{\Delta}x = S_{1}S_{3} - S_{2}S_{3} = 0$
$\therefore\ \ \phi = \frac{2\pi}{\lambda}\mathrm{\Delta}x = 0$
$\therefore I_{3} = I_{0} + I_{0} + 2\sqrt{I_{0} \times I_{0}\left( \cos{0{^\circ}} \right)}$
$\therefore I_{3} = 4I_{0}$
The path difference at $S_{4}$ is
$\mathrm{\Delta}x^{'} = S_{1}S_{4} - S_{2}S_{4} = \frac{xd}{D}$ $\left( here,\ x = \frac{\lambda D}{2d} \right)$
$= \frac{d}{D} \times \frac{\lambda D}{2d} = \frac{\lambda}{2}$
$\therefore\ \ \phi^{'} = \frac{2\pi}{\lambda}\frac{\lambda}{2} = \pi$
$\therefore I_{4} = I_{0} + I_{0} + 2I_{0}\cos\pi = 0$
$\therefore\frac{I_{3}}{I_{4}} = \frac{4I_{0}}{0} = \infty$