Question

When a central dark fringe is observed in reflection in a circular interference pattern, waves reflected from the upper and lower surfaces of the medium must have a phase difference, in radians, of

Answer 1

Explanation:

Let the first wave is :

$y_1=A\sin\omega t$

And another wave is :

$y_2=A\sin (\omega t+\phi)$

$\phi$ is phase difference between waves

Let y is the resultant of these two waves. So,

$y =y_1+y_2$

The waves reflected from the upper and lower surfaces of the medium, it means that the resultant to be zero. So,

$\cos(\dfrac{\phi}{2})=0\\\\\cos(\dfrac{\phi}{2})=\cos(\dfrac{\pi}{2})\\\\\phi=\pi$

So, the phase difference between the two waves is $\pi$.