2 April 2004 Physics 218, Spring 2004 1
Circular-aperture diffraction and the Airy patternCircular obstacles, and Poisson’s spot.
Today in Physics 218: diffraction by a circular aperture or obstacle
V773 Tau: AO off(and brightness turned way up)
V773 Tau: AO onNeptune orbit diameter,seen from same distance
2 April 2004 Physics 218, Spring 2004 2
The circular aperture
Most experimental situations in optics (e.g. telescopes) have circular apertures, so the application of the Kirchhoff integral to diffraction from such apertures is of particular interest. We start with a plane wave incident normally on a circular hole with radius a in an otherwise opaque screen, and ask: what is the distribution of the intensity of light on a screen adistance away? The field in the aperture is constant, spatially:
and the geometry is as follows:
R a>>
0( , , ) ,i tN NE x y t E e ω−′ ′ =
2 April 2004 Physics 218, Spring 2004 3
x
y
z,Z
da’r’
r
θ
Φ
φ
a
R
Y
X
q
The circular aperture (continued)
cossin
X qY q= Φ= Φ
cossin
x ry rda r dr d
φφφ
′ ′ ′=′ ′ ′=′ ′ ′ ′=
Small angles:cos
sin
x
y
kqXk kr r
kqk
r
Φ≅ =
Φ≅
2 April 2004 Physics 218, Spring 2004 4
The circular aperture (continued)
Thus,
( ) ( )
( )
( )( )
02
00
20
0 0
, , ( , , )
exp cos cos sin sin
exp cos ,
x yikr i k x k y
F x y N
aikr
i tN
ai kr tN
eE k k t E x y t e dx dyr
e dr rr
ikr qd E e
r
ikr qE e dr r dr r
πω
πω
λ
λ
φ φ φ
φ φλ
∞ ∞′ ′− +
−∞ −∞
−
−
′ ′ ′ ′=
′ ′=
′⎛ ⎞′ ′ ′− Φ + Φ⎜ ⎟⎝ ⎠
′⎛ ⎞′ ′ ′ ′= − −Φ⎜ ⎟⎝ ⎠
∫ ∫
∫
∫
∫ ∫
¥
2 April 2004 Physics 218, Spring 2004 5
The circular aperture (continued)
The aperture is symmetrical about the z axis, so we expect that the answer will be independent of the “screen” azimuthal coordinate Φ; without loss of generality, then, we can take Φ = 0. The integral over
Don’t try to integrate that directly; it’s a Bessel function of the first kind, order zero:
becomesφ′2
0
exp cos .ikr q
dr
πφ φ
′⎛ ⎞′ ′= −⎜ ⎟⎝ ⎠∫I
( ) ( )2
cos0 0 0
0
1 2 .2
iu v kr qJ u J u e dv J
r
ππ
π′⎛ ⎞− = = ⇒ = ⎜ ⎟
⎝ ⎠∫ I
2 April 2004 Physics 218, Spring 2004 6
Flashback: Bessel functions
The Bessel function of the first kind, of order m, can be represented by the integral
Bessel functions of different order are related by the recurrence relation
Recurrence relations of special functions are very useful when one has to integrate those special functions, as you’re about to see.
( ) ( )2
cos
0
.2
mi mv u v
miJ u e dv
π
π
−+= ∫
( ) ( ) ( )1 10
( ) .u
m m m mm m m m
d u J u u J u u J u v J v dvdu − −⎡ ⎤ = ⇔ =⎣ ⎦ ∫
2 April 2004 Physics 218, Spring 2004 7
10 5 0 5 10
0.5
0
0.5
1
Flashback: Bessel functions (continued)
( )0J u
u
( )1J u( )2J u
J0 J1 J2
2.405 0 0
5.520 3.832 5.136
8.654 7.016 8.417
11.792 10.174 11.620
14.931 13.324 14.796
Zeroes of the Bessel functions
2 April 2004 Physics 218, Spring 2004 8
The circular aperture (continued)
So the far field is
Now use the recurrence relation, with m = 1:
( )( )
( )( )
00
02 /
00
0
2,
2 .
ai kr tN
F
kaq ri kr tN
kqrE eE q t dr r Jr r
E e r vJ v dvr kq
ω
ω
πλ
πλ
−
−
′⎛ ⎞′ ′= ⎜ ⎟⎝ ⎠
⎛ ⎞= ⎜ ⎟
⎝ ⎠
∫
∫
( ) ( )1 00
;u
uJ u vJ v dv= ∫
( )( ) 2
01
2, .i kr t
NF
kaq kaqE e rE q t Jr kq r r
ωπλ
− ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
2 April 2004 Physics 218, Spring 2004 9
The circular aperture (continued)
Rearrange the field in a somewhat more convenient form:
This leaves a minor problem: the expression is indeterminate at But the recurrence relation can help us again:
( )( )
( )( ) ( )
( ) ( ) ( ) ( )
20
1
10
22 210
2 2
, 2 ,
2or , ,
2whence , , .
8 8
i kr tN
F
i kr tN
F
NF F F
kaqa E e rE q t Jr kaq r
J kaE AeE tr ka
J kacE AcI ka E t E tkar
ω
ω
πλ
θθ
λ θ
θθ θ θ
π θπλ
−
−
∗
⎛ ⎞= ⎜ ⎟⎝ ⎠
=
⎡ ⎤= = ⎢ ⎥
⎣ ⎦
0.kaθ =
2 April 2004 Physics 218, Spring 2004 10
The circular aperture (continued)
Take the recurrence relation at m = 1 and use the chain rule:
Note that ( ) ( )0 10 1 and 0 0 :J J= =
( ) ( ) ( )
( ) ( ) ( )
10 1 1
110
( )
.
dJduJ u uJ u J u u udu du
J udJJ u udu u
= ⎡ ⎤ = +⎣ ⎦
= +
( ) ( ) ( )( )
( )
( ) ( )
111 1
0 0
1
1 10
1 0 lim 0 lim1
2 0 , or
1lim 0 .2
u u
u
dJ uJ udJ dJ dudu u dudJdu
J u dJu du
→ →
→
= + = +
=
= =
2 April 2004 Physics 218, Spring 2004 11
The circular aperture (continued)
This resolves the indeterminacy:
Because also has zeroes at finite values of , has a set of concentric rings for which the intensity is zero (dark rings) The first of these lies at or
( )
( ) ( ) ( )
22 2 2 20 02 2 2 2
21
10 2 ,28 8
20 .
N NF
F F
cE A cE AIr r
J kaI ka I
ka
πλ πλ
θθ
θ
⎡ ⎤= =⎢ ⎥⎣ ⎦
⎡ ⎤= ⎢ ⎥
⎣ ⎦
J1 kaθ ( )FI kaθ
1 3.832,kaθ =
Airy pattern
13.832 3.832 1.22 .
2ka a Dλ λθ
π= = = First
dark ring
2 April 2004 Physics 218, Spring 2004 12
The Airy pattern
( )( )0
I kaI
θ
kaθ
Linear scale Logarithmic scale
10 5 0 5 1010 4
10 3
0.01
0.1
1
10 5 0 5 100
0.2
0.4
0.6
0.8
1
2 April 2004 Physics 218, Spring 2004 13
The Airy pattern (continued)
Linear scale Logarithmic scale
2 April 2004 Physics 218, Spring 2004 14
The Airy pattern (continued)
A young triple-star system, T Tau, observed at with adaptive optics on the Palomar 200-inch telescope. The brightest star has saturated the detector in the Airy disk. Note the extensive nest of concentric dark rings around it. (Linear scale.)
2.2 mλ µ=
2 April 2004 Physics 218, Spring 2004 15
The opaque circular obstacle
We can handle the case of diffraction by a circular obstaclequite easily, using the result just obtained. For the field,
( )( )
( )
( )
( )( )
00
00
0
00
02
00 1
2,
2
2
2 ,
i kr tN
Fa
i kr tN
ai kr tN
i kr ti kz t N
N
kqrE eE q t dr r Jr r
kqrE e dr r Jr r
kqrE e dr r Jr r
kaqE a e rE e Jr kaq r
ω
ω
ω
ωω
πλ
πλ
πλ
πλ
∞−
∞−
−
−−
′⎛ ⎞′ ′= ⎜ ⎟⎝ ⎠
′⎛ ⎞′ ′= ⎜ ⎟⎝ ⎠
′⎛ ⎞′ ′− ⎜ ⎟⎝ ⎠
⎛ ⎞= − ⎜ ⎟⎝ ⎠
∫
∫
∫
2 April 2004 Physics 218, Spring 2004 16
The opaque circular obstacle (continued)
and for the intensity,
( )
( ) ( )( )
222
0 1
21
2 22 22
0 1 1
,8
1 28 2
2
21 2 2 cos .8 2 2
F F F
N
ik r z ik r z
N
cI q t E E
kaqc ka rE Jr kaq r
kaqka r J e er kaq r
kaq kaq kqc ka r ka rE J Jr kaq r r kaq r r
π
π
π
∗
− − −
=
⎡ ⎛ ⎞⎛ ⎞⎢= + ⎜ ⎟⎜ ⎟⎜ ⎟⎢ ⎝ ⎠⎝ ⎠⎣⎤⎛ ⎞− + ⎥⎜ ⎟
⎝ ⎠ ⎥⎦⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞⎢ ⎥= + −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦
( )( )
22
2 cos 1
2 cos 1 cos
2 cos 2 cos2 2
kr z r
kr
kqkr
r
θ
θ
= −
= −
⎛ ⎞≅ =⎜ ⎟⎜ ⎟
⎝ ⎠
2 April 2004 Physics 218, Spring 2004 17
The opaque circular obstacle (continued)
This result has the curious property of not being zero inside the shadow of the obstacle. In fact, there’s a sharp peak exactly in the center, with peak intensity
And there are concentric bright and dark rings that also lie within the shadow, though generally it is much darker there than it is outside the shadow.
( )22 2
20
2, 1 .8 2F Nc ka kaI q t E
r rπ
⎡ ⎤⎛ ⎞⎢ ⎥= + −⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
2 April 2004 Physics 218, Spring 2004 18
The opaque circular obstacle (continued)
Diffraction patterns at seen 5 m
away from 0.09375 inch, 0.15625 inch, and 0.1875 inch diameter spheres (Ioan Feier, Horst Friedsam and Merrick Penicka, Argonne National Laboratory).
0.635 m,λ µ=
2 April 2004 Physics 218, Spring 2004 19
Poisson’s spot
Not all effects named after famous physicists are meant to honor their namesakes.
In 1818, the French Academy, led by neo-Newtonians like Laplace, Biot and Poisson, offered a prize for the best work on the theme of diffraction, expecting that the result would be a definitive refutation of the wave theory of light.Fresnel, supported by Ampère and Arago, offered a paper in which he developed the scalar theory of diffraction in much the same way we did, based on the wave theory. During Fresnel’s talk, Poisson pointed out that one of the consequences of Fresnel’s theory was the intensity peak in the center of circular shadows that we just found.
2 April 2004 Physics 218, Spring 2004 20
Poisson’s spot (continued)
Poisson did this, of course, because he thought such a result was ridiculous; he meant it as a fatal objection to Fresnel’s theory. But right after the talk, Arago went into his lab, observed the intensity peak and concentric rings in the shadow directly, and proceeded to demonstrate it to the judges.Thus Fresnel was awarded the prize, the corpuscular theory of light stood refuted (until Einstein and Planck came along), and the intensity peak has been known ever since as Poisson’s spot.
Nick Nicola (University of Melbourne)