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Show that two waves interfere constructively when the path difference them is an integral multiple of wave length. |
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Answer» Solution :Let , `Y_1=A_1 SIN (omega t- phi_1) and Y_2 =A_2 sin (omega t-phi_2)` when these WAVES interfere, the resultant DISPLACEMENT will be `Y=Y_1+Y_2` i.e. `Y=A_1 sin (omega t-phi_1)+A_2 sin (omega t-phi_2)`. i.e., `Y=A sin (omega t-phi)` where, `A=sqrt(A_1^2+A_2^2+2A_1 A_2 cos (phi_1-phi_2))` and `phi=tan^(-1){(A_1 sin phi_1+A_2 sin phi_2)/(A_ cos phi_1+A_2 cos phi_2)}` For a constructive interference , `A=A_("MAX")`. This implies that `cos (phi_1-phi_2)=+1,` So that `A=A_1+A_2` i.e. `phi_1-phi_2=0,2 pi, 4pi..........2npi` or `(phi_1-phi_2)=2npi` where `n=0,1,2...........` For a constructive interference, PATH difference `delta=nlambda` |
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