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= 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 `triangle=lambda`. Thus shown figure the path difference at P can be given as
`triangle= S_1P - S_2 P cong S_1 Q`
`=d cos theta = 2 lambda cos theta`
And we have at point P, path difference `triangle = lambda`, thus

`triangle = 2 lambda cos theta = lambda`
or, `cos theta = (1)/(2)`
or, ` theta= (pi)/(3)`
Thus the value of x can be written as `x= D tan theta = D tan (pi/3) = sqrt(3) D`.