PHYS 408 HOMEWORK 6 SOLUTIONS

A delta function model of a slit is dierent in that

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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| ￿ ￿...
View Full Document

This document was uploaded on 03/11/2014 for the course PHYS 408 at University of British Columbia.

Ask a homework question - tutors are online