Abstract

A method is described for modeling the effects of spatial apertures on optical sensor systems. The method consists of defining a set of basis functions that is obtained by partitioning the aperture image plane into a series of rectangular regions and replacing the field in each rectangular subregion with an orthogonal function series approximation. Each orthogonal function has a finite extent that is matched to the aperture image. The individual functions are propagated by application of the Fresnel approximation of the Rayleigh–Sommerfeld diffraction formula to other ranges, and the resultant functions are shown to be valid basis functions for defining a field at any other range. The technique is applied to a scattering problem using complex Fourier series.

© 2006 Optical Society of America

References

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1. J. W. Goodman, Introduction to Fourier Optics (McGraw-Hill, 1996), pp. 66-67, 104-106.
[PubMed]

Goodman, J. W.

J. W. Goodman, Introduction to Fourier Optics (McGraw-Hill, 1996), pp. 66-67, 104-106.
[PubMed]

Other (1)

J. W. Goodman, Introduction to Fourier Optics (McGraw-Hill, 1996), pp. 66-67, 104-106.
[PubMed]

Cited By

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Figures (2)

Fig. 1

Partitioning arrangement of the object plane surface.

Fig. 2

Optical layout.

Equations (37)

$E ( r , z ) = exp ( j k z ) j λ z ∫ d r 0 2 E 0 ( r 0 ) exp ( j k ∣ r − r 0 ∣ 2 2 z ) ,$
$∫ d r 2 q n ( r ) q m * ( r ) = δ n , m ,$
$E 0 ( r 0 ) = ∑ n c n q n ( r 0 ) ,$
$c n = ∫ d r 0 2 E 0 ( r 0 ) q n * ( r 0 ) .$
$E ( r , z ) = ∑ n c n Q n ( r , z )$
$Q n ( r , z ) = exp ( j k z ) j λ z ∫ d r 0 2 q n ( r 0 ) exp ( j k ∣ r − r 0 ∣ 2 2 z ) .$
$b n , m = ∫ d r 2 Q n ( r , z ) Q m * ( r , z ) ,$
$b n , m = ∫ d r 0 2 ∫ d r 1 2 q n ( r 0 ) q m * ( r 1 ) exp [ k ( r 0 2 − r 1 2 ) 2 z ] ∫ d r 2 λ 2 z 2 exp [ j k r ∙ ( r 1 − r 0 ) z ] .$
$b n , m = δ n , m .$
$c n = ∫ d r 2 E ( r , z ) Q n * ( r , z )$
$p n x , n y ( r 0 ) = rect ( x 0 − n x a a ) rect ( y 0 − n y a a )$
$rect ( x ∕ a ) = { 1 , for − a ∕ 2 ⩽ x < a ∕ 2 0 , elsewhere } .$
$f n ( r 0 ) = p n ( r 0 ) f ( r 0 ) = ∑ m C n , m exp ( j κ m ∙ r 0 ) = p n ( r 0 ) ∑ m C n , m exp ( j κ m ∙ r 0 ) ,$
$C n , m = 1 a 2 ∫ d r 0 2 p n ( r 0 ) f ( r 0 ) exp ( − j κ m ∙ r 0 ) .$
$g n ( r , z ) = exp ( j k z ) j λ z ∫ d r 0 2 f n ( r 0 ) exp ( j k ∣ r − r 0 ∣ 2 2 z ) = ∑ m C n , m exp ( j k z ) j λ z ∫ d r 0 2 p n ( r 0 ) exp ( j κ m ∙ r 0 ) exp ( j k ∣ r − r 0 ∣ 2 2 z ) .$
$h n , m ( r , z ) = exp ( j k z ) j λ z ∫ d r 0 2 p n ( r 0 ) exp ( j κ m ∙ r 0 ) exp ( j k ∣ r − r 0 ∣ 2 2 z ) ;$
$g n ( r , z ) = ∑ m C n , m h n , m ( r , z ) .$
$f ( r 0 ) = ∑ n f n ( r 0 ) ,$
$g ( r , z ) = exp ( j k z ) j λ z ∫ d r 0 2 f ( r 0 ) exp ( j k ∣ r − r 0 ∣ 2 2 z ) = ∑ m g n ( r , z ) = ∑ n ∑ m C n , m h n , m ( r , z ) .$
$b n , n ′ , m , m ′ ≡ ∫ d r 2 h n , m ( r ) h n ′ , m ′ * ( r ) = 1 ( λ z ) 2 ∫ d r 0 2 ∫ d r 0 ′ 2 p n ( r 0 ) p n ′ ( r 0 ′ ) exp [ j ( κ m ∙ r 0 − κ m ′ ∙ r 0 ′ ) ] × ∫ d r 2 exp ( j k ∣ r − r 0 ∣ 2 2 z − j k ∣ r − r 0 ′ ∣ 2 2 z ) = a 2 δ n , n ′ δ m , m ′ ,$
$∫ d r 2 exp ( j k ∣ r − r 0 ∣ 2 2 z − j k ∣ r − r 0 ′ ∣ 2 2 z ) = ( λ z ) 2 δ ( r 0 − r 0 ′ )$
$p n ( r 0 ) p n ′ ( r 0 ′ ) = δ n , n ′ p n ( r 0 ) .$
$ε B ( r ) = ε z ( r ) s ( r ) exp [ j ϕ ( r ) ] = ∑ n ∑ m E n , m h n , m ( r ) .$
$E n , m = 1 a 2 ∫ d r 2 ε B ( r ) h n , m * ( r ) .$
$ρ n , m , n ′ , m ′ = ⟨ ∣ E n , m ∣ 2 ∣ E n ′ , m ′ ∣ 2 ⟩ − ⟨ ∣ E n , m ∣ 2 ⟩ ⟨ ∣ E n ′ , m ′ ∣ 2 ⟩ { [ ⟨ ∣ E n , m ∣ 4 ⟩ − ( ⟨ ∣ E n , m ∣ 2 ) ⟩ 2 ] [ ⟨ ∣ E n ′ , m ′ ∣ 4 ⟩ − ( ⟨ ∣ E n ′ , m ′ ∣ 2 ⟩ ) 2 ] } 1 ∕ 2 .$
$I n , m = ⟨ ∣ E n , m ∣ 2 ⟩ = ∫ d r 0 2 ∫ d r 1 2 h n , m * ( r 0 ) h n , m ( r 1 ) ε z ( r 0 ) ε z * ( r 1 ) ⟨ s ( r 0 ) s ( r 1 ) exp { j [ ϕ ( r 0 ) − ϕ ( r 1 ) ] } ⟩$
$I z ( r 0 ) = ∣ ε z ( r 0 ) ∣ 2 .$
$I n , m = ∫ d r 0 2 I z ( r 0 ) S ( r 0 ) ∣ h n , m ( r 0 ) ∣ 2 ,$
$⟨ s ( r 0 ) s ( r 1 ) exp { j [ ϕ ( r 0 ) − ϕ ( r 1 ) ] } ⟩ = S ( r 0 ) δ ( r 0 − r 1 ) .$
$B n , n ′ , m , m ′ ≡ ⟨ ∣ E n , m ∣ 2 ∣ E n ′ , m ′ ∣ 2 ⟩ = ∫ d r 3 2 ∫ d r 2 2 ∫ d r 1 2 ∫ d r 0 2 h n , m * ( r 0 ) h n , m ( r 1 ) h n ′ , m ′ * ( r 2 ) h n ′ , m ′ ( r 3 ) ε z ( r 0 ) ε z * ( r 1 ) ε z ( r 2 ) ε z * ( r 3 ) ⟨ s ( r 0 ) s ( r 1 ) s ( r 2 ) s ( r 3 ) exp { j [ ϕ ( r 0 ) − ϕ ( r 1 ) + ϕ ( r 2 ) − ϕ ( r 3 ) ] } ⟩ = I n , m I n ′ , m ′ + ∫ d r 0 2 h n , m * ( r 0 ) h n ′ , m ′ ( r 0 ) S ( r 0 ) I z ( r 0 ) × ∫ d r 1 2 h n ′ , m ′ * ( r 1 ) h n , m ( r 1 ) S ( r 1 ) I z ( r 1 ) ,$
$⟨ s ( r 0 ) s ( r 1 ) s ( r 2 ) s ( r 3 ) exp { j [ ϕ ( r 0 ) − ϕ ( r 1 ) + ϕ ( r 2 ) − ϕ ( r 3 ) ] } ⟩$
$= S ( r 0 ) δ ( r 0 − r 1 ) S ( r 2 ) δ ( r 2 − r 3 ) + S ( r 0 ) δ ( r 0 − r 3 ) S ( r 2 ) δ ( r 1 − r 2 ) .$
$H n , n ′ , m , m ′ ≡ ∫ d r 0 2 h n , m * ( r 0 ) h n ′ , m ′ ( r 0 ) S ( r 0 ) I z ( r 0 )$
$ρ n , n ′ , m , m ′ = ∣ H n , n ′ , m , m ′ ∣ 2 [ ∣ H n , n , m , m ∣ 2 ∣ H n ′ , n ′ , m ′ , m ′ ∣ 2 ] 1 ∕ 2 .$
$S ( r ) = S 0 = c o n s t a n t .$
$H n , n ′ , m , m ′ = I 0 S 0 a 2 δ n , n ′ δ m , m ′$
$ρ n , n ′ , m , m ′ = δ n , n ′ δ m , m ′ .$