Circular Aperture

The circular aperture is particularly important because it is used a lot in optics. A telescope typically has a circular aperture for example.

We can use the same expression for the E field that we had for the rectangular aperture for any possible aperture, as long as the limits of integration are appropriate. So we can write

E ⃗ = ɛ ⃗ A R e i ( k R − ω t ) ∫ ∫ a p e r t u r e e − i K ( Y y + Z z ) / R ⅆ y ⅆ z E ⃗ = ɛ ⃗ A R e i ( k R − ω t ) ∫ ∫ a p e r t u r e e − i K ( Y y + Z z ) / R ⅆ y ⅆ z

For a circular aperture this integration is most easily done with cylindrical coordinates. Look at the figure

Figure 1

Then we have z = ρ cos φ z = ρ cos φ y = ρ sin φ y = ρ sin φ Z = q cos Φ Z = q cos Φ Y = q sin Φ Y = q sin Φ Then Y y + Z z = ρ q cos φ cos Φ + ρ q sin φ sin Φ Y y + Z z = ρ q cos φ cos Φ + ρ q sin φ sin Φ or Y y + Z z = ρ q cos ( φ − Φ ) Y y + Z z = ρ q cos ( φ − Φ ) and the integral becomes E ⃗ = ɛ ⃗ A R e i ( k R − ω t ) ∫ 0 a ∫ 0 2 π e − i K ρ q cos ( φ − Φ ) / R ρ ⅆ ρ ⅆ φ E ⃗ = ɛ ⃗ A R e i ( k R − ω t ) ∫ 0 a ∫ 0 2 π e − i K ρ q cos ( φ − Φ ) / R ρ ⅆ ρ ⅆ φ

In order to do this integral we need to learn a little about Bessel functions.

J 0 ( u ) = 1 2 π ∫ 0 2 π e i u cos v ⅆ v J 0 ( u ) = 1 2 π ∫ 0 2 π e i u cos v ⅆ v Is the definition of a Bessel function of the first kind order 0. J m ( u ) = 1 2 π ∫ 0 2 π e i ( m v + u cos v ) ⅆ v J m ( u ) = 1 2 π ∫ 0 2 π e i ( m v + u cos v ) ⅆ v Is the definition of a Bessel function of the first kind order m.

They have a number of interesting properties such as the recurrence relations ⅆ ⅆ u [ u m J m ( u ) ) ] = u m J m − 1 ( u ) ⅆ ⅆ u [ u m J m ( u ) ) ] = u m J m − 1 ( u ) so that for example when m = 1 m = 1 ∫ 0 u u ′ J 0 ( u ′ ) ⅆ u ′ = u J 1 ( u ) . ∫ 0 u u ′ J 0 ( u ′ ) ⅆ u ′ = u J 1 ( u ) . In order to numerically calculate the value of a Bessel function one uses the expansion J m ( x ) = ∑ s = 0 ∞ ( − 1 ) s s ! ( m + s ) ! ( x 2 ) m + 2 s . J m ( x ) = ∑ s = 0 ∞ ( − 1 ) s s ! ( m + s ) ! ( x 2 ) m + 2 s .

Now we want to evaluate the integral E ⃗ = ɛ ⃗ A R e i ( k R − ω t ) ∫ 0 a ∫ 0 2 π e − i K ρ q cos ( φ − Φ ) / R ρ ⅆ ρ ⅆ φ E ⃗ = ɛ ⃗ A R e i ( k R − ω t ) ∫ 0 a ∫ 0 2 π e − i K ρ q cos ( φ − Φ ) / R ρ ⅆ ρ ⅆ φ which we can do at any value of Φ Φ since the problem is symmetric about Φ Φ . So we can simplify things greatly if we do the integral at Φ = 0 Φ = 0 E ⃗ = ɛ ⃗ A R e i ( k R − ω t ) ∫ 0 a ∫ 0 2 π e − i K ρ q cos ( φ ) / R ρ ⅆ ρ ⅆ φ E ⃗ = ɛ ⃗ A R e i ( k R − ω t ) ∫ 0 a ∫ 0 2 π e − i K ρ q cos ( φ ) / R ρ ⅆ ρ ⅆ φ which becomes E ⃗ = ɛ ⃗ A R e i ( k R − ω t ) 2 π ∫ 0 a J 0 ( − K ρ q / R ) ρ ⅆ ρ E ⃗ = ɛ ⃗ A R e i ( k R − ω t ) 2 π ∫ 0 a J 0 ( − K ρ q / R ) ρ ⅆ ρ

Now J 0 J 0 is an even function so we can drop the minus sign and rewrite the expression as E ⃗ = ɛ ⃗ A R e i ( k R − ω t ) 2 π ∫ 0 a J 0 ( K ρ q / R ) ρ ⅆ ρ E ⃗ = ɛ ⃗ A R e i ( k R − ω t ) 2 π ∫ 0 a J 0 ( K ρ q / R ) ρ ⅆ ρ

To do this integral we change variables w = k ρ q / R w = k ρ q / R ρ = w R k q ρ = w R k q d ρ = R k q ⅆ w d ρ = R k q ⅆ w so that ∫ 0 a J 0 ( K ρ q / R ) ρ ⅆ ρ = ∫ 0 k a q / R ( R k q ) 2 J 0 ( w ) w ⅆ w = ( R k q ) 2 ( k a q R ) J 1 ( k a q / R ) = a 2 ( R k a q ) J 1 ( k a q / R ) = a 2 J 1 ( k a q / R ) k a q / R ∫ 0 a J 0 ( K ρ q / R ) ρ ⅆ ρ = ∫ 0 k a q / R ( R k q ) 2 J 0 ( w ) w ⅆ w = ( R k q ) 2 ( k a q R ) J 1 ( k a q / R ) = a 2 ( R k a q ) J 1 ( k a q / R ) = a 2 J 1 ( k a q / R ) k a q / R

So finally we have the result E ⃗ = ɛ A ⃗ e i ( k R − ω t ) R 2 π a 2 J 1 ( k a q / R ) k a q / R E ⃗ = ɛ A ⃗ e i ( k R − ω t ) R 2 π a 2 J 1 ( k a q / R ) k a q / R Or recognizing that π a 2 π a 2 is the area of the aperture A A and squaring to get the intensity we write I = I 0 [ 2 J 1 ( k a q / R ) k a q / R ] 2 I = I 0 [ 2 J 1 ( k a q / R ) k a q / R ] 2 If you want to write this in terms of the angle θ θ then one uses the fact that q / R = sin θ q / R = sin θ I ( θ ) = I ( 0 ) [ 2 J 1 ( k a sin θ ) k a sin θ ] 2 I ( θ ) = I ( 0 ) [ 2 J 1 ( k a sin θ ) k a sin θ ] 2

Figure 2

Above is a plot of the function J 1 ( x ) / x J 1 ( x ) / x . Notice how it peaks at 1 / 2 1 / 2 which is why there is the factor of two in the expression for the irradiance. Below is a 3D plot of the same thing (ie. J 1 ( r ) / r J 1 ( r ) / r ). Notice the rings.

Figure 3

Figure 4

Above is a plot of ( J 1 ( r ) / r ) 2 ( J 1 ( r ) / r ) 2 which corresponds to the irradiance one sees. The central peak out to the first ring of zero is called the Airy disk. This occurs at J 1 ( r ) / r = 0 J 1 ( r ) / r = 0 which can be numerically evaluated to give r = 3.83 r = 3.83 for the first ring.

For our circular aperture above this means the first zero occurs at k a q 1 / R = 3.83 k a q 1 / R = 3.83 or 2 π λ a