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A plane monochromatic light wave falls normally on an opaque screen shaped as a long strip with a round hole in the middle. For the observation point P the hole corresponds to half the Fresnel zone, with the hole diameter being eta = 1.07 times less then the width of the strip. Using Fig. find the intensity of light at the point P provided that the intensity of the incident light is equal to I_(0). |
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Answer» Solution :The radius of the first HALF Fresnel zone is `sqrt(b LAMBDA//2)` and the amplitude at `P` is obtained using problem `A = (a_(0))/(b) [[-etasqrt(b lambda//2) oo],[int + int],[-oo etasqrt(b lambda//2)]] E^(-ikb-(kx^(2))/(2b)) dx` `underset(-oo)overset(oo)int e^(-iky^(2)/(2b)) dy + (a_(0))/(b)e^(-ikb) underset(-oo)overset(sqrt(b lambda//2))int e^(-ikrho^(2)//2b) 2PI rho d rho` We use `underset(-oo)overset(-etasqrt(b lambda//2))int e^(-ikx^(2)//2b) dx` `= underset(eta sqrt(b lambda//2))overset(oo)int e^(-ikx^(2)//2b) dx =underset(eta sqrt(b lambda//2))overset(oo)int e^(-i(pix^(2))/(b lambda)) dx` `= underset(eta)overset(oo)int e^(-ik piu^(2)//2) sqrt((b lambda)/(2)) DU = sqrt((blambda)/(2))(underset(0)overset(oo)int -underset(0)overset(eta)int) e^(-i piu^(2)//2) du` `= sqrt((b lambda)/(2)) (((1)/(2)-C(eta)) -i(1)/(2)-S(eta)))` Thus `A = a_(0) (lambda)/(cancel(2)) xx cancel(2) xx (1 - i) e^(-ikb) [((1)/(2)-C(eta))` `-i((1)/(2)-S(eta))] + a_(0)lambda(1 - i)e^(-ikb)` where we have used `underset(0)overset(sqrt(b lambda//2))int e^(-ikrho^(2)/(2b)) 2pi rho d rho = (2pi ib)/(k) (-1-i) = (2pib)/(k)(1-i) = lambda b(1-i)` Thus the intensity is `I = |A|^(2) = a_(0)^(2)lambda^(2) xx 2[(3//2-C(eta))^(2) + ((1)/(2)-S(eta))^(2)]` From Cornu's Sprial, `C(eta) = C(1.07) = 0.76` `S(eta) = S(1.07) = 0.50` `I = a_(0)^(2)lambda^(2) xx 2 xx (0.74)^(2) = 1.09 a_(0)^(2) lambda^(2)` As before `i_(0) = a_(0)^(2)lambda^(2)` do `I ~~ I_(0)`. 
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