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S.T. for a constructive interference of two identical and coherent light waves, the maximum intensity is four times the intensity of individual waves and zero for a destructive interference. |
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Answer» Solution :For an interference of two identical waves `R=sqrt((A)^2+(A)^2+2(A)(A)cos (phi_1-phi_2))` i.e., `R=sqrt(2A^2(1+cos (phi_1-phi_2)))` or `R=sqrt2 A(sqrt(1+cos phi))` where `phi_1=phi_1-phi_2` or `R=sqrt2 (A sqrt(2 cos^2 ((phi)/(2))))` or `R=2A cos (phi)/(2)` so that intensity `I=4A^2 cos^2. (phi)/(2)` For `C.I. cos^2 .(phi)/(2)=1 rArr (phi)/(2)=0, pm pi , pm2pi,....................pmnpi` or `phi=0,pm2pi,pm4pi,...................,pm2n pi`, where `n=0,1,2,3`. i.e., `I_("max")=4A_2=4I_0` where `I_0` is the maximum intensity of individual waves. For destructive interference `I=0 rArr cos^2. (phi)/(2)=0` i.e.., `phi =pm1pi,pm3pi,pm5pi,..................pm(12n+1)pi,` where, `n=0,1,2.` Note : if the PHASE DIFFERENCE b/w the two waves is not constant then the average intensity `=4I_(0) lt cos^2. (phi)/(2) GT` i.e., `lt I gt =2I_0` at all points. |
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