Jump to ContentJump to Main Navigation
Coherent X-Ray Optics$

David Paganin

Print publication date: 2006

Print ISBN-13: 9780198567288

Published to Oxford Scholarship Online: September 2007

DOI: 10.1093/acprof:oso/9780198567288.001.0001

Show Summary Details
Page of

PRINTED FROM OXFORD SCHOLARSHIP ONLINE (www.oxfordscholarship.com). (c) Copyright Oxford University Press, 2017. All Rights Reserved. Under the terms of the licence agreement, an individual user may print out a PDF of a single chapter of a monograph in OSO for personal use (for details see http://www.oxfordscholarship.com/page/privacy-policy). Subscriber: null; date: 24 February 2017

(p.401) Appendix C Reciprocity theorem for monochromatic scalar fields

(p.401) Appendix C Reciprocity theorem for monochromatic scalar fields

Source:
Coherent X-Ray Optics
Publisher:
Oxford University Press

Here, we outline a certain reciprocity theorem governing forward-propagating monochromatic scalar electromagnetic fields, a result due to Lukosz (1968) and Shewell and Wolf (1968). Use is made of this theorem in the discussion on in-line holography given in Section 4.3.1.

With reference to Fig. C.1 let us consider a set of sources A lying in the half-space z < -▵, which radiate monochromatic scalar electromagnetic waves. With specified optic axis z, the volume z ≥ -▵ is assumed to be vacuum. Denote by ψ ω ( 1 ) ( x , y , z ) the spatial part of the monochromatic field radiated by the sources. Here, ω denotes the angular frequency of the radiation corresponding to wave-number k, and (x, y) are Cartesian coordinates in planes perpendicular to the optic axis z. The radiated field obeys the Helmholtz equation, given in eqn (1.16):

(C.1)
( 2 + k 2 ) ψ ω ( 1 ) ( x , y , z ) = 0.

Assuming the field to be forward propagating, we saw in Section 1.3 that the problem of diffracting the field from z = 0 to z = ▵ ≥ 0 is given by the following angular-spectrum representation which exactly solves the Helmholtz equation (see eqn (1.23)):

Appendix C Reciprocity theorem for monochromatic scalar fields

Fig. C.1. Sources A radiate a monochromatic scalar electromagnetic field into a vacuum-filled space. The field is assumed to be forward propagating over the plane z = 0, which is perpendicular to a specified optic axis z. Further, the field is assumed to be free from evanescent waves. According to the reciprocity theorem, if one propagates the field from the plane z = 0 to the plane z = ▵, the intensity of the resulting field will be equal to the intensity obtained if one propagates the complex conjugate of the field, from the plane z = 0 to the plane z = -▵.

(p.402)

(C.2)
ψ ω ( 1 ) ( x , y , z = Δ ) = 1 2 π ψ ω ( 1 ) ( k x , k y , z = 0 ) exp [ i Δ k 2 k x 2 k y 2 ] × exp [ i ( k x x + k y y ) ] d k x d k y .
Here ψ ω ( 1 ) ( k x , k y , z = 0 ) represents the Fourier transform of the unpropagated field ψ ω ( 1 ) ( x , y , z = 0 ) with respect to x and y, using the convention for two-dimensional Fourier transforms outlined in Appendix A. As specified there, the Fourier variables conjugate to x and y are denoted by kx and ky, respectively.

Next we use eqn (A.5) to write ψ ω ( 1 ) ( k x , k y , z = 0 ) as:

(C.3)
ψ ω ( 1 ) ( k x , k y , z = 0 ) = 1 2 π ψ ω ( 1 ) ( x , y , z = 0 ) exp [ i ( k x x + k y y ) ] d x d y ,
which may be substituted into eqn (C.2) to give:
(C.4)
ψ ω ( 1 ) ( x , y , z = Δ ) = 1 ( 2 π 2 ) ψ ω ( 1 ) ( x , y , z = 0 ) exp [ i Δ k 2 k x 2 k y 2 ] × exp { i [ k x ( x x ) + ( y y ) ] } d k x d k y d x d y .

Next, consider a second forward-propagating monochromatic scalar electromagnetic field ψ ω ( 2 ) ( x , y , z ) , which also obeys the Helmholtz equation. By definition we assume that the boundary value of this field, over the plane z = 0, is equal to the complex conjugate of the boundary value taken by ψ ω ( 1 ) ( x , y , z ) over the same plane. Thus:

(C.5)
ψ ω ( 1 ) ( x , y , z = 0 ) = [ ψ ω ( 2 ) ( x , y , z = 0 ) ] .
Substitute this into eqn (C.4) and then take the complex conjugate of the resulting expression, leaving:
(C.6)
[ ψ ω ( 1 ) ( x , y , z = Δ ) ] = 1 ( 2 π ) 2 ψ ω ( 2 ) ( x , y , z = 0 ) × exp [ i Δ ( k 2 k x 2 k y 2 ) ] × exp { i [ k x ( x x ) + k y ( y y ) ] } × d k x d k y d x d y .
In the above expression let us consider the integrals over x′ and y′ to be performed before those over kx and ky. These first two integrals will serve to map ψ ω ( 2 ) ( x , y , z = 0 ) into its two-dimensional Fourier transform ψ ω ( 2 ) ( k x , k y , z = 0 ) . Now suppose that evanescent waves (see Section 1.3) are absent, which amounts (p.403) to the requirement that ψ ω ( 2 ) ( k x , k y , z = 0 ) vanishes at every point (kx, ky) for which k x 2 + k y 2 > k 2 . There will therefore be a zero contribution, to the integral over kx and ky, from points (kx, ky) for which k x 2 + k y 2 > k 2 . The only non-zero contribution to the integral over kx and ky will be due to points (kx, ky) for which k x 2 + k y 2 k 2 . At all such points, the square root in the second line of eqn (C.6) is real. We may therefore drop the star from this square root, leaving:
(C.7)
[ ψ ω ( 1 ) ( x , y , z = Δ ) ] = 1 ( 2 π ) 2 ψ ω ( 2 ) ( x , y , z = 0 ) × exp ( i Δ k 2 k x 2 k y 2 ) × exp { i [ k x ( x x ) + k y ( y y ) ] } × d k x d k y d x d y .

To proceed further make the change of variables:

(C.8)
k x k x , k y k y ,
in eqn (C.7), to see that it reduces to:
(C.9)
[ ψ ω ( 1 ) ( x , y , z = Δ ) ] = 1 ( 2 π ) 2 ψ ω ( 2 ) ( x , y , z = Δ ) × exp ( i Δ k 2 k x 2 k y 2 ) × exp { i [ k x ( x x ) + k y ( y y ) ] } × d k x d k y d x d y .

Compare the right side of the above equation with the right side of the equation which results when ψ ω ( 1 ) is replaced by ψ ω ( 2 ) in eqn (C.4). This shows us that the right side of eqn (C.9) is equal to the field that results when ψ ω ( 2 ) ( x , y , z = 0 ) is inverse propagated from the plane z = 0, to the plane z = -▵ (see Fig. C.1). Thus we arrive at the desired reciprocity theorem:

(C.10)
[ ψ ω ( 1 ) ( x , y , z = Δ ) ] = ψ ω ( 2 ) ( x , y , z = Δ ) .

This theorem states that if one propagates through a certain positive distance ▵ a given forward-propagating monochromatic scalar electromagnetic field which is free of evanescent waves, the propagation being from plane to parallel plane, then the propagated field is equal to the complex conjugate of the field one would have obtained if the complex conjugate of the unpropagated field had been (inverse) diffracted through a distance equal to the negative of ▵.

Taking the squared modulus of eqn (C.10), and denoting the intensities of ψ ω ( 1 ) ( x , y , z ) and ψ ω ( 2 ) ( x , y , z ) by I ω ( 1 ) ( x , y , z ) = | ψ ω ( 1 ) ( x , y , z ) | 2 and I ω ( 2 ) ( x , y , z ) = | ψ ω ( 2 ) ( x , y , z ) | 2 , we see that: (p.404)

(C.11)
I ω ( 1 ) ( x , y , z = Δ ) = I ω ( 2 ) ( x , y , z = Δ ) .
Thus if one propagates the field from the plane z = 0 to the plane z = ▵, the resulting intensity will be equal to the intensity obtained if one were to propagate the complex conjugate of the field, from the plane z = 0 to the plane z = -▵ (see Fig. C.1).