Two coherent narrow slits emitting sound of wavelength `lambda` in the same phase are placed parallet to each other at a small separation of `2 lambda` . The sound is delected by maving a delector on the screen at a distance `D(gt gt lambda)` from the slit `S_(1)` as shows in figure. Find the distance `y` such that the intensity at `P` is equal to intensity at `O` .

Explanation

When detector is at O, we can see that the path difference in the two waves reaching O is `d = 2lambda` thus at O detector receives a maximum sound. When it reachesP and again there isa maximum sound detected at P the path difference between two waves must be `Delta = lambda` Thus from figure-6.32 the path difference at P can be given as
`Delta = S_(1)P - S_(2)P cong S_(1) Q`
= `d cos theta`
= `2lambda cos thete`

And we have at point P, path difference `Delta = lambda, Thus
`Delta = 2 lambda cos theta = lambda`