The central fringe in the interference pattern obtained in Young's double slit experiment will be a dark…
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
zero
\(\frac{\pi}{2}\)
\(\pi\)
\(\frac{\pi}{3}\)
Solution
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,
$\begin{aligned}
\Delta \phi & =
\frac{2 \pi}{\lambda}\times \Delta x \\
& =\frac{2 \pi}{\lambda}\times\frac{\lambda}{2}
=\pi
\end{aligned}$