Question

In the arrangement shown in the figure, slits $S_{3}$ and $S_{4}$ are having a variable separation $Z$ . Point $O$ on the screen is at the common perpendicular bisector of $S_{1}S_{2}$ and $S_{3}S_{4}$ . (Assume $D>>d$ )

The minimum value of $Z$ for which the intensity at $O$ is zero is

• A. $\frac{3 \lambda D}{2 d}$
• B. $\frac{\lambda D}{2 d}$
• C. $\frac{\lambda D}{3 d}$
• D. $\frac{\lambda D}{d}$

Answer 1

Answer: (D)

Intensity at $O$ is due to $S_{3}$ and $S_{4}$ .
$\therefore $ Intensity at $S_{3}$ = zero and Intensity at $S_{3}$ is given by
$I=K \cos ^2\left(\frac{\phi}{2}\right)$
Intensity at $O$ is zero if $\frac{\phi}{2}=\frac{\pi }{2}$
$\phi=\frac{2 \pi }{\beta }\times \frac{Z}{2}=\pi \Rightarrow Z=\beta $ ;
$Z=\beta =\frac{\lambda D}{d}$
$\phi=$ phase difference between waves at $S_{3}$ and $S_{4}$