1 Fraunhofer Diffraction: Single, multiple slit(s) & Circular aperture Fri. Nov. 22, 2002.

$: 1 Fraunhofer Diffraction: Single, multiple slit(s) & Circular aperture Fri. Nov. 22, 2002.$
1

Fraunhofer Diffraction: Single, multiple slit(s) & Circular aperture

Fri. Nov. 22, 2002

2

Fraunhofer diffraction limit

If aperture is a square - X The same relation holds in azimuthal plane and 2

~ measure of the area of the aperture Then we have the Fraunhofer diffraction if,

apertureofaread

or

d

,

2

Fraunhofer or far field limit

3

Fraunhofer, Fresnel limits

The near field, or Fresnel, limit is

See 10.1.2 of text

2

d

4

Fraunhofer diffraction

Typical arrangement (or use laser as a source of plane waves)

Plane waves in, plane waves out

S

f1 f2

screen

5


1. Obliquity factorAssume S on axis, so Assume small ( < 30o), so

2. Assume uniform illumination over aperture

r’ >> so is constant over the aperture

3. Dimensions of aperture << rr will not vary much in denominator for calculation of amplitude at any point Pconsider r = constant in denominator

1'ˆˆ rn1ˆˆ rn

'

'

r

eikr

6


Then the magnitude of the electric field at P is,

aperture

ikrikr

oP dSe

rr

eikEE

'2

'

7

Single slit Fraunhofer diffraction

y = b

y

dy

P

ro

r

r = ro - ysin

dA = L dy

where L ( very long slit)

8

Single slit Fraunhofer diffraction

'2sin

2

,

sin

_______________

'

sin

rr

eikEC

kb

where

ebCeE

dyeeCE

dAeCE

ikro

iikrP

ikyb

o

ikrP

ikrP

o

o

2

2sin

oII

Fraunhofer single slit diffraction pattern

2bCIo

9

Single Slit Fraunhofer diffraction: Effect of slit width

Minima for sin = 0 = p = k(b/2)sin or, sin = p(/b) First minima at sin = /b

-3

-2

-1

01

23

0.0

0.2

0.4

0.6

0.8

1.0

I/Io

b

10

Single Slit Fraunhofer diffraction: Effect of slit width Width of central max 2 (/dimension of

aperture) This relation is characteristic of all Fraunhofer

diffraction If b is very large 0 and a point source is

imaged as a point If b is very small (~) /2 and light spreads

out across screen (diminishes at large angles for to F()

11

Diffraction from an array of N slits, separated by a distance a and of width b

y=0

y=a

y=a+b

y=2a

y=2a+b

y=3a

y=3a+b

y=(N-1)a

y=(N-1)a + b

y=b

P

12

Diffraction from an array of N slits

It can be shown that,

where,

22

sin

sinsin

N

II oP

sin2

bk sin

2

ak

13

Diffraction and interference for N slits

The diffraction term Minima for sin = 0 = p = k(b/2)sin or, sin = p(/b)

The interference term Amplitude due to N coherent

sources Can see this by adding N phasors

that are 2 out of phase. See Hecht Problem 10.2

sin

sin

sin NIo

14

Interference term

MaximaMaxima occur at = m (m = 0,1, 2, 3, ..) To see this use L’Hopital’s rule _______ Thus maxima occur at sin = m/a This is the same result we have derived for

Young’s double slit Intensity of principal maxima, I = N2Io

i.e. N times that due to one slit

15

Interference term

MinimaMinima occur for = /N, 2/N, … (N-1)/N and when we add m For example, _______________________ Thus principal maxima have a width determined by

zeros on each side Since = (/)a sin = /N The angular width is determined by

sin = /(Na) Thus peaks are N times narrower than in a single slit

pattern (also a > b)

16

Interference term

Subsidiary or Secondary MaximumSubsidiary or Secondary Maximum Now between zeros must have secondary

maxima Assume these are approximately midway Then first at [ m+3/(2N) ] Then it can be shown that

3

2

sin

sin NN

max2

2045.0

9

4IINI o

17

Single slit envelope Now interference term or pattern is modulated by

the diffraction term

which has zeros at =(b/)sin=p or, sin = p/b But, sin = m/a locate the principal maxima of

the interference pattern

2sin

18

Single slit envelope

Thus at a given angle a/b=m/p Then suppose a/b = integer For example, a = 3b Then m = 3, 6, 9, interference maxima are

missing

19

Diffraction gratings

Composed of systems with many slits per unit length – usually about 1000/mm

Also usually used in reflection Thus principal maxima vary sharp Width of peaks Δ = (2/N) As N gets large the peak gets very narrow For example, _________________

20


ResolutionResolution Imagine trying to resolve two wavelengths 1 2

Assume resolved if principal maxima of one falls on first minima of the other

See diagram___________

21


m1 = a sin m2 = a sin ’ But must have

Thus m(2 - 1 )= a (sin’ - sin) = (1/N) Or mΔ =/N Resolution, R = /Δ = mN E.g.

Nm

a 1'sin

2

m

a

1

sin

22

Fraunhofer diffraction from a circular aperture

x

y

P

Lens plane

r

dxdyeCE ikrP

23


22 yR

22 yR

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

dyyReCE ikrP

222

Why?

24


Do integration along y – looking from the side

-R

+R

y=0

ro

r = ro - ysin

P

25


R

R

ikyikrP dyyReCeE o 22sin2

sinkRR

y

)1(sin

kRRkky

)2(1 222222 RRRyR

)3(dyRd

Let

Then

26


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.

27


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

28


Thus,

where = kRsin and Io is the intensity when =0

2

12

J

II o

29


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, …

30

-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

32

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

33

-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

34

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

1 Fraunhofer Diffraction: Single, multiple slit(s) & Circular aperture Fri. Nov. 22, 2002.

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Transcript of 1 Fraunhofer Diffraction: Single, multiple slit(s) & Circular aperture Fri. Nov. 22, 2002.