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

Step-by-step 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((\phi)/(2))=0\\\\\cos((\phi)/(2))=\cos((\pi)/(2))\\\\\phi=\pi`

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