InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Two coherent point sources `S_(1)` and `S_(2)` vibrating in phase emit light of wavelength `lambda`. The separation between the sources is `2lambda`. Consider a line passing through `S_(1)` and perpendicular to line `S_(1) S_(2)`. Find the position of farthest and nearest minima.. |
|
Answer» `Delta x_(min) = (2n - 1) (lambda)/(2)` The farthest mimima has path difference `lambda//2` while the nearest minima has path difference `(3//2)lambda`. For the nearest minima, `S_(1) P - S_(2) P = (3)/(2) lambda` [as maximum path difference is `2lambda`] `implies sqrt((2lambda)^(2) + D^(2)) - D = (3)/(2) lambda` `implies (2lambda)^(2) + D^(2) = ((3)/(2) lambda + D)^(2)` `implies 4lambda^(2) + D^(2) = (9)/(4) lambda^(2) + D^(2) xx 2 xx (3)/(2) lambda xx D` `implies 3D = 4lambda - (9lambda)/(4) = (7lambda)/(4) implies D = (7)/(12) lambda` For the farthest minima, `S_(1)P - S_(2)P = (lambda)/(2)` `impliessqrt (4 lambda^(2) + D^(2)) - D = (lambda)/(2)` `4lambda^(2) + D^(2) = (lambda^2)/(4) + D^(2) + D lambda implies D = 4 lambda - lambda//4 = (15 lambda)/(4)`. |
|