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Monochromatic light waves of constant phase difference phi and amplitues A_(1) and A_(2) produce an interference pattern. State an expression for the resultant amplitude at a point in the pattern.Hence deduce the conditions for (i) constructive interference with maximum intensity (ii) destructive interference with minimum intensity. Also show that the ratio of the maximum and minimum intensities(I_(max))/(I_(min))=((A_(1)+A_(2))/(A_(1)-A_(2)))^(2). |
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Answer» Solution :Consider a two-source interference pattern produced by monochromatic light waves of angular frequency `omega`, wavelength `lambda`, amplitudes `A_(1) and A_(2)`,and a constant phase difference `phi`. LET the individualdisplacements due to the waves at a point P in the interference pattern be `y_(1)=A_(1) SIN omega t and y_(2) = A_(2) sin(omega t+phi)` Let `A_(1)+A_(2) cos phi = R cos theta and A_(2) sin phi = R sin theta ` It can be shown that, by the superposition principle, the resultant displacement at P is the algebraic sum, `y=y_(1)+y_(2)=A_(1) sin omega t +A_(2) sin (omega t +phi)` `=R sin( omega t +theta)` where the resultant amplitude is `|R|= sqrt(A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2) cos phi ) "" ` ...(1) Since the intensity of a wave is proportional to the square of its amplitude, the resultant intensity at P, `I prop R^(2)` `therefore I prop A_(1)^(2) +A_(2)^(2) +2A_(1)A_(2) cos phi "" ` ...(2) that is, the intensity depends on `cos phi` Hence, P will be a point of constructive interference with MAXIMUM intensity when `cos phi` is maximum, equal to 1, i.e., when`phi =2n pi ""(n=0,1,2,3, ...) "" ` ...(3) P will be a point of destructive interference with minimum intensity when `cos phi` is minimum, equal to -1, i.e., when `phi =(2m-1)pi""(m=1,2,3,...) "" ` ...(4) Expressions (3) and (4) GIVE the conditions of constructive and destructive interference, respectively. At POINTS where R and I are maximum, `R_(max)=A_(1)+A_(2) "" ` ...(5) ` and I_(max) prop A_(1)^(2) +A_(2)^(2) +2A_(1)A_(2)` `therefore I_(max) prop (A_(1) +A_(2))^(2) "" `...(6) At points where R and I are minimum, `R_(min) =|A_(1)-A_(2)| "" `...(7) ` and I_(min) prop A_(1)^(2)+A_(1)^(2) -2A_(1)A_(2) ` `therefore I_(min) prop (A_(1) -A_(2))^(2) "" ` ...(8) `therefore` From Eqs. (6) and (8), `(I_(max))/(I_(min))=((A_(1)+A_(2))/(A_(1)+A_(2)))^(2) "" ` ...(9) |
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