PHYS 408 HOMEWORK 6 SOLUTIONS

# A delta function model of a slit is dierent in that

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Unformatted text preview: 2 −02 −02 d k2 −(2π /Λ)2 d + 1i e 2i ￿ (2π /Λ)·x+0·y sin(2π /Λ) (56) ￿√ ￿ ￿ 1 i −(2π/Λ)·x+0·y i√k2 −(−2π/Λ)2 −02 d 2 2 2 ei k −(2π/Λ) −0 d − e e (57) 2i (58) single plane wave f (x, y ) = ei ￿ (π /2Λ)·x+(π /2Λ)·y ￿ (59) ˜ f (kx , ky ) = δ (kx − (π /2Λ))δ (ky − (π /2Λ)) g (x, y ) = ei((π/2Λ)·x+(π/2Λ)·y) ei =e i((π /2Λ)·x+(π /2Λ)·y ) i 11.3 e √ √ (60) k2 −(π /2Λ)2 −(π /2Λ)2 d (61) k2 −2(π /2Λ)2 d (62) delta-function slit, y -axis f (x, y ) = δ (αx) δ (ky ) ˜ f (kx , ky ) = |α| g (x, y ) = = ￿ dkx ￿ 1 |α| (63) (64) ￿ dky ei dkx ei √ √ 2 2 k2 −kx −ky d i(kx x+ky y ) e δ (ky )/|α| (65) 2 k2 −kx d ikx x (66) e 2 In the case of a vanishingly small slit, the transmitted beam intensity is proportional to sinc(x) , and approaches a constant, vanishingly small intensity about x = 0 (the pattern widens, with nodes going to inﬁnity). A delta-function model of a slit is diﬀerent in that it always admits some constant amount of light, due to the delta function having a ﬁxed area. Consequently, what we expect for g (x, y ) in the far-ﬁeld is purely harmonic – corresponding to constant intensity – and ﬁnite amplitude, here relating to α, which is the inverse ‘width’ of the delta-function. In the Fresnel approximation, the output function is g (x, y ) ≈ = ≈ ￿ 2 2 1 dkx eik(1−kx /2k )d eikx x |α| ￿ 2 1 ikd e dkx e−ikx d/2k eikx x |α| ￿ ￿...
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## This document was uploaded on 03/11/2014 for the course PHYS 408 at University of British Columbia.

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