Saved Bookmarks
| 1. |
In Young.s double-slit experiment using monochromatic light of wavelength lambda, the intensity of light at a point on the screen where path difference is lambda, is K units. What is the intensity of light at a point where path difference is lambda"/"3? |
|
Answer» Solution :Phase difference corresponding to `lambda` is `2PI`. Phase difference corresponding to `(lambda)/(3)" is "(2pi)/(3)` `I= I_(1)+I_(2)+2 sqrt(I_(1)I_(2)) cos phi""` ASSUME that `I_(1)= I_(2)= I_(0)` In the first case, `K= I_(0)+I_(0)+2I_(0) cos 2pi = 4I_(0)` In the second case, `K= I_(@)+I_(0)+2I_(0) cos ""(2pi)/(3)` `I_(@)+I_(0)-2I_(0)(1/2)= I_(0)` Now `(K.)/(K)= (I_0)/(4I_0)= (1)/(4)" or "K.= (K)/(4)`. |
|