Two coherent narrow slits emitting wavelength lambda in the same phase are placed parallel to each other at a small separation of2lambda, the sound is detected by moving a detector on the screen S at a distance D (gtgt lambda) from the slit S_(1) as shown in Fig. 7.76. Find the distance x such that the intensity at P is equal to the intensity at O.

Two coherent narrow slits emitting wavelength lambda in the same phase

1.	Two coherent narrow slits emitting wavelength lambda in the same phase are placed parallel to each other at a small separation of2lambda, the sound is detected by moving a detector on the screen S at a distance D (gtgt lambda) from the slit S_(1) as shown in Fig. 7.76. Find the distance x such that the intensity at P is equal to the intensity at O.
Answer» Solution :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`