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

C, D

Explanation

Since `S_1, S_2` are in same phase at O there wil be maximum intensity.
Given that there will be a maximum intensity at P.
`rarr path difference =/_\x=nl`

From the ure (in question)
`(S_1P)^2-(S_2P)^2`
=(sqrtD^2+x_2)^2-[(sqrtD-2lamda^2+x^2]^2`
`=4lamdaD=4lamda^2=4lamdaD`
`(lamda^2` is so small and can neglected )
`S_1P-S_2P=(4lamdaD)/(2sqrt(x^2+D^2))=nlamda`
`=(2D)/(sqrt(x^2+D^2))=n`
`rarr n^2(x^2+D^2)=4D^2`
`rarr x=D/n sqrt4-n^2`
When `n=1, x=sqrt3D(1st order)`
n=2,
`x=0 (2nd order)
`:.Wen x=sqrt3D`
at P there will be maximum intensity