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
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
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.
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