InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
`S_1 and S_2` are two coherent sources of radiations separated by distance `100.25 lambda`, where `lambda` is the wave length of radiation. `S_1` leads `S_2` in phase by `pi//2`.A and B are two points on the line joining `S_1 and S_2` as shown in figure.The ratio of amplitudes of component waves from source `S_1 and S_2` at A and B are in ratio 1:2. The ratio of intensity at A to that of B `(I_A/I_B)` is A. `oo`B. `(1)/(9)`C. `0`D. `9` |
|
Answer» Correct Answer - B For interference at `A : S_(2)` is behind of `S_(1)` by a distance of `100 lambda = lambda//2`. Hence the waves from `S_(1)` and `S_(2)` interfere at `A` with phase difference of `200.5 pi + 0.5 pi = 201 pi = pi`. Hence the net amplitude at `A is 2a - a = a` . For interference at `B : S_(2)` is ahead of `S_(1)` by a distance of `100 lambda + lambda//4` (equal to phase difference `pi//2`) .Further `S_(2)` lags `S_(1)` by `pi//2`. Hence the waves from `S_(1)` and `S_(2)` interfere at `B` with a phase difference of `200.5 pi - 0.5 pi = 200 pi = 0 pi` . Hence the net amplitude at `A is 2 a + a = 3 a`. Hence , `((I_(A))/(I_(B))) = ((a)/( 3 a))^(2) = (1)/(9)` |
|