Astronomy 203/403, Fall 1999

1999 University of Rochester 1 All rights reserved

13. Lecture, 14 October 1999

13.1 Example: diffraction from a circular aperture

Most telescopes have circular apertures, so the application of the Kirchhoff integral to diffraction fromsuch apertures is of particular interest. Let us start with a plane wave incident on a circular hole withradius a in an otherwise opaque screen, and calculate the distribution of the intensity of light on a screen adistance R a>> away, as in Figure 13.1. The amplitude of the near field we consider to be constant, so

that E x y t E eN Ni t( , , )′ ′ = −

0ω . With the definitions

′ = ′ ′′ = ′ ′′ = ′ ′ ′

==

≅ =

≅

x r

y r

da r dr d

X q

Y q

Xr

qr

qr

x

y

cossin

cossin

cos

sin,

φφφ

κ κκ

κκ

ΦΦ

Φ

Φ(13.1)

x

y

z,Z

da’

r’

r

θ

Φ

φ

a

R

Y

X

q

Figure 13.1: geometry of circular aperture and distant screen for calculation of diffractionof plane wave incident normally on the aperture. The “screen” axes X and Y are parallelto the “aperture” axes x and y.

the far field, at point q,Φ on the screen, is

E te

rdr r d E e

i r qr

E er

dr r di r q

r

F x y

i r

Ni t

a

Ni r t a

κ κλ

φκ

φ φ

λφ

κφ

κω

π

κ ω π

, , exp cos cos sin sin

exp cos ,

e j b g

b ga f

= ′ ′ ′ −′

′ + ′FHG

IKJ

= ′ ′ ′ −′

′ −FHG

IKJ

−

−

zzzz0

0

2

0

0

0

2

0

Φ Φ

Φ

(13.2)

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where the trig identity cos cos cos sin sinα β α β α β− = +b g has been used in the last step. The aperture is

symmetrical about the z axis, so we expect that the answer will be independent of the “screen” azimuthalcoordinate Φ; without loss of generality, then, we can take Φ = 0. This leaves the following integral over

′φ :

I di r q

r= ′ −

′′F

HGIKJz φ

κφ

πexp cos

0

2

. (13.3)

Integrals of this form crop up frequently in physics, whenever there is axial symmetry. They cannot,however, be expressed in terms of the usual elementary functions such as the trig functions orexponentials, but instead are elementary functions themselves; this is a Bessel function.

The Bessel function of the first kind, of order m, has this integral representation:

J ui

e dvm

mi mv u va f a f=

−+z2

0

2

π

πcos . (13.4)

The first few of these are plotted in Figure 13.2. Note that the even-order Bessel functions are evenfunctions and the odd-order ones are odd functions; our integral, Equation 13.3, therefore resembles afirst-kind Bessel function of order zero:

J u J u e dv

I Jr qr

iu v0 0

0

2

0

12

2

− = =

=′FHGIKJ

za f a fπ

πκ

πcos ;

.

(13.5)

10 5 0 5 10

0.5

0

0.5

1

( )J u0

u

( )J u1

( )J u2

Figure 13.2: The first three of the first-kind Bessel functions.

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1999 University of Rochester 3 All rights reserved

Bessel functions are “oscillating” functions, but their zeros are not evenly spaced along the horizontalaxis; this keeps them from being expressible as linear combinations of just a few sines and cosines. Someof the zeroes are listed in Table 13.1. One of these is listed in boldface – it will turn out to be importantlater.

Table 13.1: the first five zeroes of the Bessel functions plotted in Figure 13.2.

We thus use Equation 13.5 in Equation 13.3, and in turn in Equation 13.2, to produce

E q tE e

rdr r J

qrr

E er

rq

vJ v dvFN

i r t aN

i r t aq r

, ./

b g a fa f a f

= ′ ′′F

HGIKJ =

FHGIKJ

− −z z2 200

0

02

00

πλ

κ πλ κ

κ ω κ ω κ

(13.6)

Integration of J0 is facilitated by a recurrence relation between Bessel functions of different order:

ddu

u J u u J u

u J u v J v dv

mm

mm

mm

mm

u

a f

a f a f

=

=

−

−z1

10

( ) ,

.or (13.7)

We can use the latter formula for m = 1 in Equation 13.6 to obtain

E q tE e

rrq

aqr

Jaqr

a E er

raq

Jaqr

E tE Ae

rJ a

a

FN

i r t

Ni r t

FN

i r t

,

,

, ,

b g

a f a f

a f

a f

a f

=FHGIKJFHGIKJFHGIKJ

= FHGIKJ

=

−

−

−

2

2

2

02

1

20

1

0 1

πλ κ

κ κ

πλ κ

κ

θλ

κ θκ θ

κ ω

κ ω

κ ωor

(13.8)

where we have substituted A a= π 2 and θ = q r/ . The intensity on the screen becomes

I ac

E t E tcE A

r

J aaF F F

Nκ θπ

θ θπλ

κ θκ θ

a f a f a f a f= =

LNM

OQP8 8

202 2

2 21

2* , , . (13.9)

in cgs units. Let us rewrite this expression in terms of the intensity at κ θa = 0 , which as you may suspectwill turn out to be the peak intensity. Immediately we see a problem with evaluation of Equation 13.9 here;

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

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the last factor is indeterminate at κ θa = 0 . To proceed, note first that J0 0 1a f = and that J1 0 0a f = (seeFigure 13.2 and Table 13.1), and that the recurrence relation, Equation 13.7, can be written for the first twofirst-kind Bessel functions as

J ud

duJ u

J uu0 1

1a f a f a f= + . (13.10)

Evaluation of this expression at u = 0 is complicated by the indeterminacy of the last term; we must writeit as

1 0

01

2 0

10

1

10

1

1

= +

= +

=

→

→

dJdu

J uu

dJdu

dJdu

u

dJdu

u

u

a f a f

a fa f

a f

lim

lim

,

(l’Hôpital’s rule) (13.11)

ordJdu

1 012

a f = , (13.12)

which, replaced in the first of Equations 13.11, gives

lim .u

J uu→

=0

1 12

a f(13.13)

We can use this in Equation 13.9 to get

IcE A

r

cE A

rF

N N08

212 8

02 2

2 2

20

2 2

2 2a f = LNMOQP =

πλ πλ, (13.14)

and

I a IJ a

aF Fκ θκ θ

κ θa f a f a f=

LNM

OQP0

2 12

. (13.15)

Because J1 also has zeroes at finite values of κ θa , I aF κ θa f has a set of concentric rings for which theintensity is zero (dark rings) The first of these lies at κ θa 1 3 832= . (see Table 13.1), or

θκ π

λ λ1

3 832 3 8322

1 22= = =. .. ,

a a D(13.16)

where D is the diameter of the aperture – a familiar result. Equation 13.15 is plotted in various ways inFigure 13.3 and Figure 13.4.

George Airy, Astronomer Royal of England during the 1840s, first did the calculation we just completed;the intensity distribution of Equation 13.15 is usually called the Airy pattern, and the central maximumwithin the first dark ring is called the Airy disc.

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( )( )0IaI θκ

θκa

10 5 0 5 1010 4

103

0.01

0.1

1

10 5 0 5 100

0.2

0.4

0.6

0.8

1

Figure 13.3: plots of the intensity given by Equation 13.15, on linear (left) and logarithmic(right) scales.

Figure 13.4: grayscale images of the far-field intensity diffracted by a circular aperture.Left: linear scale, for emphasis of the structure of the central maximum. Right:logarithmic scale with the central maximum “saturated” for intensity greater than 10-1.3 ofthe peak value, for emphasis of the surrounding bright and dark rings. Figure 13.3consists of meridional cross sections of these images.

Homework problem 13.1. Gaussian beams stay Gaussian as they propagate. Show that a Gaussian near-fielddistribution with 1/e radius ρN,

E x y E ex y

Ni t

N( , ) exp′ ′ = −

′ + ′FHG

IKJ

−0

2 2

2ω

ρ , (13.17)

gives rise to a Gaussian far-field distribution,

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1999 University of Rochester 6 All rights reserved

E x y z tE

ze

x y

zF

N i z t N( , , , ) exp( )= −

+F

HGG

I

KJJ

−πρλ

π ρ

λκ ω

20

2 2 2 2

2 2

e j , (13.18)

that has 1/e radius λz/πρN.

Homework problem 13.2. Most telescope primary mirrors have central obscurations, in addition to beingcircular, so their diffraction patterns differ somewhat from Equations 13.14-15.

a. Derive an expression the far-field intensity as a function of κ θa for an annular aperture, with outerhalf-diameter a and inner half diameter ka (k < 1).

b. Plot the intensity divided by peak intensity, I k Iθ , ,a f a f0 against κ θa , on the same plot with I Iθa f a f0for the filled circular aperture, as given by Equations 13.14-15, for k = 0.1 – a rather typical value fortelescopes – and the more extreme k = 0.9. What are the major differences between the diffractionpatterns of filled circular apertures and annular apertures?