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Diffraction from a Rectangular Aperture

Module by: Paul Padley

Summary: Derive the expression for diffraction from a rectangular aperture.

We consider diffraction from apertures other than a slit. For example consider a rectangular aperture as shown below. If ɛ A ⃗ ɛ A ⃗ is the source strength per unit area (assumed to be constant over the entire area in this example) and d S = ⅆ y ⅆ z d S = ⅆ y ⅆ z is an infinitesmal area at a point in the aperture then we have:

Figure 1
Figure 1 (RectangularAperture.png)
d E ⃗ = ɛ A ⃗ r e i ( k r − ω t ) ⅆ y ⅆ z d E ⃗ = ɛ A ⃗ r e i ( k r − ω t ) ⅆ y ⅆ z

We see from the figure that r = [ x 2 + ( Y − y ) 2 + ( Z − z ) 2 ] r = [ x 2 + ( Y − y ) 2 + ( Z − z ) 2 ] and that R 2 = x 2 + Y 2 + Z 2 . R 2 = x 2 + Y 2 + Z 2 . Thus we use x 2 = R 2 − Y 2 − Z 2 x 2 = R 2 − Y 2 − Z 2 to write r = [ R 2 − Y 2 − Z 2 + Y 2 + y 2 − 2 y Y + Z 2 + z 2 − 2 z Z ] 1 / 2 r = [ R 2 − Y 2 − Z 2 + Y 2 + y 2 − 2 y Y + Z 2 + z 2 − 2 z Z ] 1 / 2 or r = [ R 2 − 2 Y y − 2 Z z + y 2 + z 2 ] 1 / 2 r = [ R 2 − 2 Y y − 2 Z z + y 2 + z 2 ] 1 / 2 r = R [ 1 − 2 Y y / R 2 − 2 Z z / R 2 + ( y 2 + z 2 ) / R 2 ] 1 / 2 . r = R [ 1 − 2 Y y / R 2 − 2 Z z / R 2 + ( y 2 + z 2 ) / R 2 ] 1 / 2 . We are only considering Fraunhofer diffraction so R , Z , Y R , Z , Y are much larger than y y and z z and we can rewrite r ≈ R [ 1 − 2 ( Y y + Z z ) / R 2 ] 1 / 2 r ≈ R [ 1 − 2 ( Y y + Z z ) / R 2 ] 1 / 2 and then finally expanding using the binomial theorem and taking only the most significant terms r ≈ R [ 1 − ( Y y + Z z ) / R 2 ] . r ≈ R [ 1 − ( Y y + Z z ) / R 2 ] .

So we have d E ⃗ = ɛ A ⃗ r e i ( k r − ω t ) ⅆ y ⅆ z d E ⃗ = ɛ A ⃗ r e i ( k r − ω t ) ⅆ y ⅆ z or using the fact that R R is large d E ⃗ = ɛ A ⃗ R e i ( k r − ω t ) ⅆ y ⅆ z d E ⃗ = ɛ A ⃗ R e i ( k r − ω t ) ⅆ y ⅆ z which we integrate to get the field

E ⃗ = ∫ − b / 2 + b / 2 ∫ − a / 2 + a / 2 ɛ A ⃗ R e i ( k r − ω t ) ⅆ z ⅆ y = ɛ A ⃗ e i ( k R − ω t ) R ∫ − a / 2 + a / 2 e − i k y Y / R ⅆ y ∫ − b / 2 + b / 2 e − i k z Z / R ⅆ z = ɛ A ⃗ e i ( k R − ω t ) R [ e − i k y Y / R − i k Y / R | − a / 2 + a / 2 [ e − i k z Z / R − i k Z / R | − b / 2 + b / 2 = ɛ A ⃗ e i ( k R − ω t ) R [ e − i k a Y / 2 R − e i k a Y / 2 R − i k Y / R ] [ e − i k b Z / 2 R − e i k b Z / 2 R − i k Z / R ] E ⃗ = ∫ − b / 2 + b / 2 ∫ − a / 2 + a / 2 ɛ A ⃗ R e i ( k r − ω t ) ⅆ z ⅆ y = ɛ A ⃗ e i ( k R − ω t ) R ∫ − a / 2 + a / 2 e − i k y Y / R ⅆ y ∫ − b / 2 + b / 2 e − i k z Z / R ⅆ z = ɛ A ⃗ e i ( k R − ω t ) R [ e − i k y Y / R − i k Y / R | − a / 2 + a / 2 [ e − i k z Z / R − i k Z / R | − b / 2 + b / 2 = ɛ A ⃗ e i ( k R − ω t ) R [ e − i k a Y / 2 R − e i k a Y / 2 R − i k Y / R ] [ e − i k b Z / 2 R − e i k b Z / 2 R − i k Z / R ]

Now lets define α = k a Y / 2 R α = k a Y / 2 R β = k b Z / 2 R β = k b Z / 2 R and β = k b Z / 2 R β = k b Z / 2 R and we see that

E ⃗ = ɛ A ⃗ e i ( k R − ω t ) R [ e i α − e − i k α i k Y / R ] [ e i β − e − i β i k Z / R ] E ⃗ = ɛ A ⃗ e i ( k R − ω t ) R [ e i α − e − i k α i k Y / R ] [ e i β − e − i β i k Z / R ]

or rearranging

E ⃗ = ɛ A ⃗ e i ( k R − ω t ) R [ e i α − e − i α 2 i α / a ] [ e i β − e − i β 2 i β / b ] E ⃗ = ɛ A ⃗ e i ( k R − ω t ) R [ e i α − e − i α 2 i α / a ] [ e i β − e − i β 2 i β / b ]

E ⃗ = ɛ A ⃗ e i ( k R − ω t ) a b R [ e i α − e − i α 2 i α ] [ e i β − e − i β 2 i β ] E ⃗ = ɛ A ⃗ e i ( k R − ω t ) a b R [ e i α − e − i α 2 i α ] [ e i β − e − i β 2 i β ]

E ⃗ = ɛ A ⃗ e i ( k R − ω t ) a b R [ sin α α ] [ sin β β ] E ⃗ = ɛ A ⃗ e i ( k R − ω t ) a b R [ sin α α ] [ sin β β ]

So finally we can write the intensity as

I ( Y , Z ) = I ( 0 , 0 ) [ sin α α ] 2 [ sin β β ] 2 I ( Y , Z ) = I ( 0 , 0 ) [ sin α α