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` .

Answer

`sqrt(3) D`

Explanation

When detector is at `O`, we can see that the path difference in the two waves reaching `O is d = 2 lambda` thus at `O` detector receives a maximum sound . When it reaches `p` and again there is a maximum sound detected at `P` the path difference between two waves must be `Delta = lambda`. Thus from the figure the path difference at `P` can `Delta = lambda`. Thus from the figure the path difference at `P` can be given as
`Delta = S_(1) P - S_(2) P = S_(1) Q`
` = 2 lambda cos theta`
And we have at point `P` , path difference `Delta = lambda` , thus
`Delta = 2 lambda cos theta = lambda`
`cos theta = (1)/(2)`
`theta = (pi)/(3)`
Thus the value of `x` can be written as ` x = D tan theta`
` = D tan ((pi)/(3)) = sqrt(3) D`