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

At point `O` on the screen the path difference between the sound waves reaching from `S_(1)` and `S_(1)` is `2 lambda` , i.e. constructive interference is obtained at `O` . At a very large distance from point `O` on the screen the path difference is zero.

Thus, we can conlude that as we move away from point `O` on the screen path difference decrease from `2 lambda` at zero. At `O` constructive interference is obtained (where `Deltax = 2 lambda`) . So, next constructive interference will be obtained where `Delta x = lambda` . Hence,
`S_(1)P - S_(2)P = lambda`
or `sqrt(D^(2) +y^(2))- sqrt(y^(2) +(D - 2 lambda)^(2))= lambda`
`:. sqrt(D^(2) +y^(2))- lambda= sqrt(y^(2) +(D - 2 lambda)^(2))`
On squaring both sides, we get
`D^(2) + y^(2)+ lambda^(2) - 2lambda sqrt(D^(2) +y^(2))=y_(2) + D^(2)+4lambda^(2)- 4lambda D`
`:. 2sqrt(D^(2)+ y^(2))=4D- 3lambda`
as `Dgt gtlambda` , `4D- 3lambda~~4D`
`:. 2sqrt (D^(2)+ y^(2))=4D`
or `sqrt(D^(2)+ y^(2))=2D`
Again squaring both sides, we get
`D^(2) + y^(2) = 4D^(2)`
or `y= sqrt((3))D`
Let `Deltax = lambda` at angle theta as shows . Path difference between the waves is `S_(1)M = 2lambda cos theta` because `S_(2)P ~~ MP`
:. `2lambda cos theta = lambda`
or `theta = 60^(@) (Deltax = lambda)`
Now, `PO= S_(1)O tan theta = S_(1) O tan 60^(@)`
or `y= sqrt((3))D`