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A plane light wavewith wavelength lambda = 0.65mum falls normally on a large glass plate whose opposite side has a large and an opaque strip of width a = 0.30mm (Fig.) A screen is placed at a distance b = 110 cm from the plate. The height h of the ledge is such that the intensity of light at point 2 of the screen is the heighest possible. making use of Fig. find the ratio of intensities at point 1 and 2. |
Answer» Solution : Using the method of problem `5.103` we can immediately write down the amplitudes as `1` and `2`. We get: At `1` amplitude `A_(1)~ const [underset(-oo)OVERSET(0)int E^(- ipiu^(2)//2)DU + e^(-I delta)underset(v)overset(oo)int e^(- ipiu^(2)//2)du ]` At `2`amplitude `A_(2)~ const [underset(-oo)overset(0)int e^(- ipiu^(2)//2)du + e^(-I delta)underset(0)overset(oo)int e^(- ipiu^(2)//2)du]` where `v = a sqrt((2)/(b lambda))` is the parameter of CORNU's spiral and consatnt factor is common to `1` and `2`. With the sual notation `C = C(v) = underset(0)overset(v)int cos. (pi u^(2))/(2) du` `S = S(v) = underset(0)overset(v)int (pi u^(2))/(2) du` and the result `underset(0)overset(oo)int cos .(pi u^(2))/(2) du = underset(0)overset(oo)int sin .(pi u^(2))/(2) du = (1)/(2)` We find the ratio of intensituies as `(I_(2))/(I_(1)) = |(((1)/(2)-C)-i((1)/(2)-S) + e^(-i delta)((1 - i))/(2))/((1-i)/(2) + e^(-i delta) {((1)/(2) - C) - i((1)/(2) - S)})|^(2)` (The constant in `A_(1)` and `A_(2)` must be the same by symmetry) In our csse, `a = 0.30mm, lambda = 0.65 mu m, b = 1.1 m` `v = 0.30 xx sqrt((2)/(1.1xx 0.65)) = 0.50` `C(0.50) = 0.48 S(0.50) = 0.06` `(I_(2))/(I_(1)) = |(0.02 - 0.44i + e^(-i delta)((1-i))/(2))/((1-i)/(2)e^(i delta) + 0.02 - 0.44i)|^(2) = |(1+(0.02 - 0.44i)sqrt(2)e^(i delta)+(ipi)/(4))/(1+(0.02-0.44i)sqrt(2)e^(-i delta) +(ipi)/(4))|^(2)` But `0.02 - 0.44i = 0.44e^(i alpha), alpha = 1.525 rad (~~ 87.4^(@))` So `(I_(2))/(I_(1)) = |(1+ 0.44xx sqrt(2) xx e^(i(delta - 0.740)))/(1+0.44 xx sqrt(2) xx e^(-i(delta + 0.740)))|^(2) =(1+2 (0.44)^(2) + 2sqrt(2) xx 0.44 cos (delta - 0.740))/(1+2(0.44)^(2) + 2sqrt(2) xx 0.44 cos (delta + 0.740))` `i_(2)` is maximum when `delta - 0.740 = 0` (modulo `2pi`) Thus in that case `(I_(2))/(I_(1)) = (1.387 + 1.245)/(1.387 + 1.245 cos (1.48)) = (2.632)/(1.5) ~~ 1.75` |
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