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Unformatted text preview: 2 + αy/2)) 10.3 (52) (54) explanation All worked in part (a) because we were able to factor out all but the accumulated phase exp(ikz d). In part b, we have two
terms in the numerator, each of which accumulated a diﬀerent phase exp(ikz,A d) or exp(ikz,B d), and thus we cannot simply
factor out a single factor of the accumulated phase (because no such single factor exists). 11 freespace propagation ˜
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The recipe: calculate f (kx , ky ), calculate g (kx , ky ) = H (kx , ky )f (kx , ky ) for the freespace transfer function H (kx , ky ) =
˜
exp(ikz d), then calculate the output function in realspace (g (x, y )) by taking the inverse fourier transform of the output
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function in k space, i.e. g (x, y ) = F −1 {H f }.
10 11.1 three plane waves The input function in real space is the sum of three plane waves, where each plane wave has a distinct :
k
f (x, y ) = ei 0·x+0·y 1i
e
2i + (2π /Λ)·x+0·y − 1i
e
2i −(2π /Λ)·x+0·y (55) 1
1
˜
f (kx , ky ) = δ (kx )δ (ky ) +
δ (kx − (2π /Λ))δ (ky ) −
δ (kx − (−2π /Λ))δ (ky )
2i
2i
g (x, y ) = ei 0·x+0·y = eikd + ei 11.2 √ ei √ k...
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This document was uploaded on 03/11/2014 for the course PHYS 408 at University of British Columbia.
 Fall '11
 KirkW.Madison
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