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A plane monochromatic light wave falls normally on an opaque screen with a long slit whose shape is shown in Fig. Making use of Fig. find the ratio of intensities of light at points 1, 2 and 3 located behind the screen at equal distances from it. For point 3 the rounded-off edge of the slit coincides with the boundary line of the first Fresnel zone. |
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Answer» Solution :From the statement of the problem we know that the width of the slit `=` diameter of the first FRESNEL zone `= 2sqrt(b lambda)` where `b` is the distance of the observation point from the slit. We calaculate the amplitudes by evaluating the intergral of problem `5.103` We get `A_(1) = (a_(0))/(b) underset(-sqrt(b lambda))overset(sqrt(b lambda))int e^(-ikb) e^(-IK(x^(2))/(2b)) dx underset(0)overset(oo)int e^(-ik(y^(2))/(2b)) dy` `= (a_(0))/(b) e^(-ikb) (b lambda)/(2) underset(-sqrt(2))overset(sqrt(2))int e^(-ipiu^(2)//2) du xx underset(0)overset(oo)int e^(-iku^(2)//2) du` `= (a_(0))/(2)(1 - i) e^(-ikb) (C(sqrt(2)) - iS(sqrt(2)))` `A_(2) = (a_(0))/(b) underset(-sqrt(b lambda))overset(sqrt(b lambda))int e^(-ikb) e^(-ik(x^(2))/(2b)) dx underset(0)overset(oo)int e^(-ik(y^(2))/(2b)) dy` `= 2A_(1)` `A_(3) =-ia_(0)lambda e^(-ikb) + (a_(0)lambda(1-i))/(2) (C(sqrt(2)) - iS (sqrt(2)))e^(-ikb)` where the COMTRIBUTION of the `1^(st)` half Fresnel zone (in `A_(3)`, first term) has been obtained from the least peoblem `I_(1) = a_(0)^(2)lambda^(2) |((1 - i)(0.53 - 0.72i))/(2)|^(2)` (on using `(C(sqrt(2)) = 0.53, S(sqrt(2)) = 0.72)` `= a_(0)^(2)lambda^(2) |-0.095 - 0.625i|^(2) = 0.3996a_(0)^(2)lambda^(2)` `I_(2) = 4I_(1)` `I_(3) = a_(0)^(2)lambda^(2) |-0.095 - 0.625i - i|^(2)` `= a_(0)^(2)lambda^(2) |-0.095 - 1.625i|^(2)` `= 2.6496 a_(0)^(2)lambda^(2)` So `I_(3) = 6.6I_(1)` Thus `I_(1):I_(2):I_(3) ~~1:4:7` 
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