InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Give the theory of interference of light by considering waves of equal amplitude and hence arrive at the conditions for constructive and destructive interference in terms of path difference. |
|
Answer» Solution :Theory of interference : Consider TWO harmonic light waves of same amplitude. Let `phi` be the constant phase difference between them. The displacement of these waves can be represented as. `y_1 =a cos omega t""rarr(1)` `y_2 = a cos (omega t+ phi)""rarr(2)` According superposition principle `y=y_1 +y_2` `y= a cos omega t + a cos (omega t+phi)` `y = [cos omega t + cos (omega t +phi )]` We know that `cos C+cos D=2cos((C+D)/(2))cos((C-D)/(2))` So, `y=a[2cos((omega t+omega t+phi)/(2))cos((omega t-omega t-phi)/(2))]` `y=2a cos(omega t+phi/2)cos(- phi/2)` `y=[2acos(phi/2)]cos(omega t+phi/2)` `y=A cos(omega t+phi/2)` Where `A = 2a cos(phi/2)` called amplitude of resultant wave. So intensity of resultant wave, `I = A^2 = 4a^2 cos^2(phi/2)` Condition for constructive interference For MAXIMUM intensity `cos^2(phi/2) =+-1` `cos(phi/2)=+-1=cos(2n pi)` where `n=0,1,2,3,....,phi = 0, +- 2pi, +- 4pi, +-6pi.....` `:.` Path difference `delta=n lambda` `delta=0,lambda,2lambda,3lambda,......` Condition for destructive interference For MINIMUM intensity `cos(phi/2)=0` so, `phi = +-pi,+-3PI,+-5PI......` `:.` Path difference `delta= (2n+1) (lambda)/(2)` `delta = (lambda)/(2), (3lambda)/(2), (5lambda)/(2) ....` |
|