InterviewSolution
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InterviewSolution
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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 semicircularcut 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» Solution :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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