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A plane light wave with wavelength lambda = 0.65 mum falls normally on a large glass plate whose opposite side has a longrectangular recess 0.60mm wide. Using Fig. find the depth H of the recess at which the diffraction pattern on the screen 77 cm away from the plate has the maximum illuminance at its centre. |
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Answer» Solution :Let `a = `width of the reccess and `v = (a)/(2) sqrt((2)/(b lambda)) = (a)/(sqrt(2b lambda)) = (0.6)/(sqrt(2 XX 0.77 xx 0.65)) ~= 0.60` be the paremeter along Cornu's spiral corresponding to the half-width of the resecc. The amplitude of the diffracted wave is given by `_(~)const [E^(i delta) underset(-v)overset(v)int e^(-i pi u^(2)//2) du + underset(v)overset(oo)int e^(-i pi u^(2)//2) du + underset(-oo)overset(v)int e^(-i pi u^(2)/(2) du)]` where `delta = (2pi)/(lambda) (n - 1) H` is the extra phase due to the recess. (Actually an extra phase `e^(-i delta)` appears outside the recess. When we take it out and obsord it in the constant we GET the expression written). Thus the amplitude is  `~ const [(C(v) - iS(v))e^(i delta) + ((1)/(2) - C(v)) - i ((1)/(2)-S(v))]` From the cornu's spiral, the corrdinates corresponding to the parameter `v = 0.60` are `C(v) = 0.57, S(v) = 0.13` so the intensity at `o` is proportional to `|[(0.57 - 0.13i)e^(i delta) - 0.07 - i0.37]|^(2)` `= (0.57^(2) + 0.13)^(2) + 0.07^(2) + 0.37^(2)` `+ (0.57 - 0.13i) (-0.07 + 0.37 i)e^(i delta)` `+ (0.57 + 0.13i)(-0.07 - i0.37i)e^(-i delta)` We write `0.57 bar(+) 0.13i = 0.585e^(bar(+)ialpha) alpha = 12.8^(@)` `-0.07 +- 0.37i = 0.377 e^(+-i beta) beta = 100.7^(@)` Thus the CROSS term is `2 xx 0.585 xx 0.377 cos (delta + 88^(@))` `= 2 xx 0.585 xx 0.377 cos (delta + (pi)/(2))` For maximum intensity `delta+(pi)/(2) = 2k'pi, k' = 1, 2, 3, 4, .............` `=2(k + 1) pi, k = 0,1,2,3,.........` or `delta = 2kpi + (3pi)/(2)` so `h = (lambda)/(n - 1) (k + (3)/(4))` |
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