Question

The central fringe in the interference pattern obtained in Young's double slit experiment will be a dark fringe when the phase difference between the waves from the two slits is

• A. $0$
• B. $\frac{\pi}{2}$
• C. $\pi$
• D. $\frac{\pi}{3}$

Answer 1

Answer: (C)

When the central fringe in the interference pattern in YDSE be a dark fringe, then path difference, will be $\frac{\lambda}{2}$.
i.e., $\Delta x=\frac{\lambda}{2}$
$\therefore $ Phase difference,
$\Delta \varphi=\frac{2 \pi}{\lambda} \times \Delta x$
$=\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}=\pi$