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A plane monochromatic light wave of intensity `I_(0)` falls normally on an opaque screen with a long slit having a semicircular cut on one side (Fig.) The edge of the cut coincides with the boundary line of the first Fresnel zone for the observation point `P`. Thw width of the slit measure `0.90` of the radius of the cut. using Fig. find the intensity of light at the point `P`. |
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Answer» we apply the formula of problem `5.103` and calculate `underset("aperture")int (a_(0))/(r ) e^(-ikr) dS = underset("semicircle")int + underset("slit")int` The contribution of the full `1^(st)` Fresnel zone has been evaluated in `5.103`. The contribution the semi-circule is one half of it and is `-(2pi)/(k)i a_(0)e^(-ikb) =- ia_(0) lambda e^(-ikb)` The contribution of the slit is `(a_(0))/(b) underset(0)overset(0.90sqrt(b lambda))int e^(-ikb) e^(-ik(x^(2))/(2b)) dx underset(-oo)overset(oo)int e^(-iky^(2)//2b) dy` Now `underset(-oo)overset(oo)int e^(-iky^(2)//2b) dy = underset(-oo)overset(oo)int e^(-i(piy^(2))/(b lambda)) dy` `sqrt((b lambda)/(2)) underset(-oo)overset(oo)int e^(-ipiy^(2)//2) du = sqrt(b lambda) e^(-ipi//4)` Thus the contribution of the slit is `(a_(0))/(b) sqrt(b lambda) e^(-ik b-i pi//4) underset(-oo)overset(0.9 xx sqrt(2))int e^(-i piu^(2)//2) du sqrt((b lambda)/(2))` `= a_(0)lambda e^(-ikb-ipi//4) (1)/(sqrt(2)) underset(0)overset(1.27)int e^(-ipiu^(2)//2) du` Thus the intensity at the observation point `P` on the screen is `a_(0)^(2)lambda^(2) |-i+(1 - i)/(2)(C(1.27) - iS(1.27))|^(2) = a_(0)^(2)lambda^(2) |-i + ((1-i)(0.67 - 0.65i))/(2)|^(2)` (on using `(C(1.27) = 0.67` and `S(1.27) = 0.65)` `= a_(0)^(2) lambda^(2)|-i + 0.01 - 0.66i|^(2)` `a_(0)^(2)lambda^(2)|0.01 - 1.66i|^(2)` `= 2.76 a_(0)^(2)lambda^(2)` Now `a_(0)^(2)lambda^(2)` is the intensity due to half of `1^(st)` Fresnel zone and is therefore equal to `I_(0)`. (It can also be obtained by doing the `x`-intergal over `-oo` to `+oo`). Thus `I = 2.76 I_(0)`. |
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