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Show that two waves interfere destructively when the path difference between them is an odd multiple of half wavelength. |
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Answer» Solution :Let `Y_1=A sin (omega t-phi_1) and Y_2 = A sin (omega t-phi_2)` represent two interfering WAVES. Their resultant displacement `Y=A sin (omega t-phi)` where, `A=sqrt(A_1^2+A_2^2+2A_1 A_2 cos (phi_1-phi_2))` `phi=tan^(-1)((A_1 sin phi_1+A_2 sin phi_2)/(A_1 cos phi_1+A_2 cos phi_2))` For a destructive interference, `A=A_("min") and cos (phi_1-phi_2)=-1` so that `A=A_1-A_2` i.e., `phi_1-phi_2=1pi,3pi,5pi,..........(2n+1)PI` or `phi_1-phi_2=(2n+1)pi` where `n=0,1,2,3,..........` In terms of path difference, `delta=(lambda)/(2pi)(2n+1) pi ` so that `delta =(2n+1)(lambda)/(2)` For a constructive interference `delta =2n((lambda)/(2))` and for a destructive interference path difference between the waves `delta=(2n+1)(lambda)/(2)` |
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