Today in Physics 218: diffraction by a circular aperture ...dmw/phy218/Lectures/Lect_72b.pdf ·...

$: Today in Physics 218: diffraction by a circular aperture ...dmw/phy218/Lectures/Lect_72b.pdf · Today in Physics 218: diffraction by a circular aperture or obstacle V773 Tau: AO off$
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


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−′ ′ =


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≅



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⎜ ⎟⎝ ⎠

∫ ∫

∫

∫

∫ ∫

¥



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


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 − −⎡ ⎤ = ⇔ =⎣ ⎦ ∫


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



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

− ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠



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 =



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

→ →

→

= + = +

=

= =



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


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


The Airy pattern (continued)

Linear scale Logarithmic scale


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=


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

∞−

∞−

−

−−

′⎛ ⎞′ ′= ⎜ ⎟⎝ ⎠

′⎛ ⎞′ ′= ⎜ ⎟⎝ ⎠

′⎛ ⎞′ ′− ⎜ ⎟⎝ ⎠

⎛ ⎞= − ⎜ ⎟⎝ ⎠

∫

∫

∫


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

= −

= −

⎛ ⎞≅ =⎜ ⎟⎜ ⎟

⎝ ⎠



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

⎡ ⎤⎛ ⎞⎢ ⎥= + −⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦



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=


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.


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)

Today in Physics 218: diffraction by a circular aperture ...dmw/phy218/Lectures/Lect_72b.pdf ·...

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