Fraunhofer Diffraction: Circular aperture

$Page 1: Fraunhofer Diffraction: Circular aperture$
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Fraunhofer Diffraction: Circular aperture

Wed. Nov. 27, 2002

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Fraunhofer diffraction from a circular aperture

x

y

P

Lens plane

r

dxdyeCE ikrP

3


22 yR

22 yR

Do x first – looking downPath length is the same for all rays = ro

dyyReCE ikrP

222

Why?

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Do integration along y – looking from the side

-R

+R

y=0

ro

r = ro - ysin

P

5


R

R

ikyikrP dyyReCeE o 22sin2

sinkRR

y

)1(sin

kRRkky

)2(1 222222 RRRyR

)3(dyRd

Let

Then

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1

1

22 12 deRCeE iikrP

o

The integral 1

1

1

21J

de i

where J1() is the first order Bessell function of the first kind.

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These Bessell functions can be represented as polynomials:

and in particular (for p = 1),

0

2

!!

21

k

pkk

P pkkJ

!4!3

2

!3!2

2

!2

21

2

642

1

J

8


Thus,

where = kRsin and Io is the intensity when =0

2

12

J

II o

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Now the zeros of J1() occur at, = 0, 3.832, 7.016, 10.173, … = 0, 1.22, 2.23, 3.24, … =kR sin = (2/) sin

• Thus zero atsin = 1.22/D, 2.23 /D, 3.24 /D, …

10

-10 -8 -6 -4 -2 0 2 4 6 8 10

0.5

1.0

-10 -8 -6 -4 -2 0 2 4 6 8 10

0.5

1.0


12J

2

12

J

The central Airy disc contains 85% of the light

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D

sin = 1.22/D

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Diffraction limited focussing

sin = 1.22/D The width of the Airy disc

W = 2fsin 2f = 2f(1.22/D) = 2.4 f/D

W = 2.4(f#) > f# > 1

Cannot focus any wave to spot with dimensions <

D

f

-10

-8-6

-4-2

02

46

810

0.5

1.0

13

-10

-8-6

-4-2

02

46

810

0.5

1.0

Fraunhofer diffraction and spatial resolution

Suppose two point sources or objects are far away (e.g. two stars)

Imaged with some optical system Two Airy patterns

If S1, S2 are too close together the Airy patterns will overlap and become indistinguishable

-10

-8-6

-4-2

02

46

810

0.5

1.0

S1

S2

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Fraunhofer diffraction and spatial resolution

Assume S1, S2 can just be resolved when maximum of one pattern just falls on minimum (first) of the other

Then the angular separation at lens,

e.g. telescope D = 10 cm = 500 X 10-7 cm

e.g. eye D ~ 1mm min = 5 X 10-4 rad

D

22.1min

radXX 6

5

min 10510

105

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Polarization

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Matrix treatment of polarization

Consider a light ray with an instantaneous E-vector as shown

tkEjtkEitkE yx ,ˆ,ˆ,

x

y

Ex

Ey

y

x

tkzioyy

tkzioxx

eEE

eEE

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Matrix treatment of polarization Combining the components

The terms in brackets represents the complex amplitude of the plane wave

tkzio

tkziioy

iox

tkzioy

tkziox

eEE

eeEjeEiE

eEjeEiE

yx

yx

~

ˆˆ

ˆˆ

18

Jones Vectors The state of polarization of light is determined by

the relative amplitudes (Eox, Eoy) and,

the relative phases ( = y - x )

of these components

The complex amplitude is written as a two-element matrix, the Jones vector

ioy

oxiyi

oy

iox

oy

oxo eE

Ee

eE

eE

E

EE x

x

~

~~

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Jones vector: Horizontally polarized light

The electric field oscillations are only along the x-axis

The Jones vector is then written,

where we have set the phase x = 0, for convenience

0

1

00~

~~

AAeE

E

EE

xiox

oy

oxo

x

y

The arrows indicate the sense of movement as the beam approaches you

The normalized formis

0

1

20

x

y

Jones vector: Vertically polarized light

The electric field oscillations are only along the y-axis

The Jones vector is then written,

Where we have set the phase y = 0, for convenience

1

000~

~~

AAeEE

EE

yioyoy

oxo

The normalized formis

1

0

21

Jones vector: Linearly polarized light at an arbitrary angle If the phases are such that = m for

m = 0, 1, 2, 3, … Then we must have,

and the Jones vector is simply a line inclined at an angle = tan-1(Eoy/Eox)

since we can write

oy

oxm

y

x

E

E

E

E1

sin

cos1~

~~ m

oy

oxo A

E

EE

x

y

The normalized form is

22

Jones vector and polarization

In general, the Jones vector for the arbitrary case

is an ellipse

i

oy

oxo eE

EE~

a

b

Eox

Eoy

x

y

22

cos22tan

oyoxEE

EE oyox

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Fraunhofer Diffraction: Circular aperture