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Diffraction from an Array of Apertures

Module by: Paul Padley

Summary: We derive the diffraction pattern from an array of rectangular apertures.

An array of rectangular apertures

Say we have an array of rectangular apertures sitting in the x − y x − y plane and light hits this aperture traveling in the positive z direction. There are N N apertures arranged vertically (in the y y direction). Each aperture has a width in the x x direction of a a and a height in the y y direction of b b . For convenience, the apertures are aligned with their centers at x = 0 x = 0 . The apertures are equally spaced by a distance d d .

Figure 1
Figure 1 (Array-of-Apertures.png)
The electric field at some point P P away from the array is E = ∑ n y = 1 N E n y ( a , b ) E = ∑ n y = 1 N E n y ( a , b ) where E n y ( a , b ) E n y ( a , b ) is the field from the n y n y slit at that point. The y y position of the center of the aperture is ( n y − 1 ) d ( n y − 1 ) d so we write y = ( n y − 1 ) d + y ′ y = ( n y − 1 ) d + y ′ and ( x , y ′ ) ( x , y ′ ) is the position of a point in the aperture with respect to the center of the aperture. We can write E n y ( a , b ) = ∫ − b / 2 b / 2 ⅆ y ′ ∫ − a / 2 a / 2 ⅆ x ɛ A R e − i ( k r n y − ω t ) E n y ( a , b ) = ∫ − b / 2 b / 2 ⅆ y ′ ∫ − a / 2 a / 2 ⅆ x ɛ A R e − i ( k r n y − ω t ) If the point of observation is ( x P , y P , z P ) ( x P , y P , z P ) then r n y = [ ( x P − x ) 2 + ( y P − y ) 2 + ( z P − z ) 2 ] 1 / 2 r n y = [ ( x P − x ) 2 + ( y P − y ) 2 + ( z P − z ) 2 ] 1 / 2 but we take z to be zero at the aperture so r n y = [ ( x P − x ) 2 + ( y P − y ) 2 + z P 2 ] 1 / 2 = [ x P 2 − 2 x x P + x 2 + y P 2 − 2 y y P + y 2 + z P 2 ] 1 / 2 = R [ 1 − 2 x x P / R 2 − 2 y y P / R 2 + ( x 2 + y 2 ) / R 2 ] 1 / 2 r n y = [ ( x P − x ) 2 + ( y P − y ) 2 + z P 2 ] 1 / 2 = [ x P 2 − 2 x x P + x 2 + y P 2 − 2 y y P + y 2 + z P 2 ] 1 / 2 = R [ 1 − 2 x x P / R 2 − 2 y y P / R 2 + ( x 2 + y 2 ) / R 2 ] 1 / 2 where R 2 = x P 2 + y P 2 + z P 2 R 2 = x P 2 + y P 2 + z P 2 , the distance from the origin. In the far field approximation ( x 2 + y 2 ) / R 2 = 0 ( x 2 + y 2 ) / R 2 = 0 and we can write: r n y ≈ R [ 1 − 2 x x P / R 2 − 2 y y P / R 2 ] 1 / 2 . r n y ≈ R [ 1 − 2 x x P / R 2 − 2 y y P / R 2 ] 1 / 2 . We use the first two terms in the binomial expansion and get r n y ≈ R [ 1 − x x P / R 2 − y y P / R 2 ] = R − x x P / R − y y P / R = R − x x P / R − [ ( n y − 1 ) d + y ′ ] y P / R = R − x x P / R − ( n y − 1 ) ⅆ y P / R − y ′ y P / R r n y ≈ R [ 1 − x x P / R 2 − y y P / R 2 ] = R − x x P / R − y y P / R = R − x x P / R − [ ( n y − 1 ) d + y ′ ] y P / R = R − x x P / R − ( n y − 1 ) ⅆ y P / R − y ′ y P / R so now we have E n y ( a , b ) = ∫ − b / 2 b / 2 ⅆ y ′ ∫ − a / 2 a / 2 ⅆ x ɛ A R e − i ( k ( R − x x P / R − ( n y − 1 ) ⅆ y P / R − y ′ y P / R ) − ω t ) . E n y ( a , b ) = ∫ − b / 2 b / 2 ⅆ y ′ ∫ − a / 2 a / 2 ⅆ x ɛ A R e − i ( k ( R − x x P / R − ( n y − 1 ) ⅆ y P / R − y ′ y P / R ) − ω t ) . We rearrange to get E n y ( a , b ) = ɛ A R e − i ( k R − ω t ) e i k ( n y − 1 ) ⅆ y P / R ∫ − b / 2 b / 2 ⅆ y ′ e i k y ′ y P / R ∫ − a / 2 a / 2 ⅆ x e i k x x P / R . E n y ( a , b ) = ɛ A R e − i ( k R − ω t ) e i k ( n y − 1 ) ⅆ y P / R ∫ − b / 2 b / 2 ⅆ y ′ e i k y ′ y P / R ∫ − a / 2 a / 2 ⅆ x e i k x x P / R . We define k x P / R = k x k x P / R = k x and k y P / R = k y k y P / R = k y so that E n y ( a , b ) = ɛ A R e − i ( k R − ω t ) e i k y ( n y − 1 ) d ∫ − b / 2 b / 2 ⅆ y ′ e i k y y ′ ∫ − a / 2 a / 2 ⅆ x ′ e i k x x ′ E n y ( a , b ) = ɛ A R e − i ( k R − ω t ) e i k y ( n y − 1 ) d ∫ − b / 2 b / 2 ⅆ y ′ e i k y y ′ ∫ − a / 2 a / 2 ⅆ x ′ e i k x x ′ E = ɛ A R e − i ( k R − ω t ) ∑ n y = 1 N e i k y ( n y − 1 ) d ∫ − b / 2 b / 2 ⅆ y ′ e i k y y ′ ∫ − a / 2 a / 2 ⅆ x e i k x x . E = ɛ A R e − i ( k R − ω t ) ∑ n y = 1 N e i k y ( n y − 1 ) d ∫ − b / 2 b / 2 ⅆ y ′ e i k y y ′ ∫ − a / 2 a / 2 ⅆ x e i k x x . We see that each piece of this is something we did before E = ɛ A R e − i ( k R − ω t ) e i k y ( N − 1 ) d / 2 sin ( N k y d / 2 ) sin ( k y d / 2 ) b s i n c ( k y b / 2 ) a s i n c ( k x a / 2 ) E = ɛ A R e − i ( k R − ω t ) e i k y ( N − 1 ) d / 2 sin ( N k y d / 2 )