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Fraunhofer Diffraction:
Single, multiple slit(s) & Circular aperture
Fri. Nov. 22, 2002
1
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,
2
d
or ,
Fraunhofer or far field limit
area of aperture
d
2
Fraunhofer, Fresnel limits
The near field, or Fresnel, limit is
2
d
See 10.1.2 of text
3
Fraunhofer diffraction
Typical arrangement (or use laser as a
source of plane waves)
Plane waves in, plane waves out
screen
S
f1 f2
4
Fraunhofer diffraction
1. Obliquity factor
Assume S on axis, so n r ' 1
ˆ ˆ
Assume small ( < 30o), so n r 1
ˆ ˆ
2. Assume uniform illumination over aperture
ikr '
r’ >> so
e is constant over the aperture
r'
3. Dimensions of aperture << r
r will not vary much in denominator for calculation of
amplitude at any point P
consider r = constant in denominator
5
Fraunhofer diffraction
Then the magnitude of the electric field at P is,
ikEo eikr '
EP
2rr ' eikr dS
aperture
6
Single slit Fraunhofer diffraction
P
y=b
dy r
y ro
r = ro - ysin
dA = L dy
where L ( very long slit)
7
Single slit Fraunhofer diffraction
I o bC
2
EP C eikr dA
_______________
b
EP C eikro eiky sin dy
sin 2
I Io
o
sin
EP bCeikro e i
2
where,
Fraunhofer single slit diffraction pattern
kb ikEo eikr '
sin C
2 2rr '
8
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
b
1
0
-1
-2
-3
1.0
0.8
0.6
0.4
0.2
0.0
I/Io
9
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()
10
Diffraction from an array of N slits, separated by a distance a and of width b
y=(N-1)a + b
y=(N-1)a
y=3a+b
P
y=3a
y=2a+b
y=2a
y=a+b
y=a
y=b
y=0
11
Diffraction from an array of N slits
It can be shown that,
2
sin sin N
2
I P Io
sin
where,
b a
k sin k sin
2 2
12
Diffraction and interference for N slits
The diffraction term
Minima for sin = 0 sin
= p = k(b/2)sin
or, sin = p(/b)
The interference term
Amplitude due to N coherent sin N
Io
sources sin
Can see this by adding N phasors
that are 2 out of phase. See
Hecht Problem 10.2
13
Interference term
Maxima 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
14
Interference term
Minima 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)
15
Interference term
Subsidiary or Secondary Maximum
Now between zeros must have secondary
maxima
Assume these are approximately midway
sin N 2N
Then first at [ m+3/(2N) ]
sin 3
Then it can be shown that
4 2
I N I o 0.045 I m ax
9 2
16
Single slit envelope
Now interference term or pattern is
modulated by the diffraction term
2
sin
which has zeros at =(b/)sin=p
or, sin = p/b
But, sin = m/a locate the principal
maxima of the interference pattern
17
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
18
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, _________________
19
Diffraction gratings
Resolution
Imagine trying to resolve two wavelengths
1 2
Assume resolved if principal maxima of
one falls on first minima of the other
See diagram___________
20
Diffraction gratings
m1 = a sin
m2 = a sin ’
But must have
a sin a sin ' 1
m m
1 2 N
Thus m(2 - 1 )= a (sin’ - sin) = (1/N)
Or mΔ =/N
Resolution, R = /Δ = mN
E.g.
21
Fraunhofer diffraction from a circular aperture
y
P
r
x
Lens plane
EP C e dxdy
ikr
22
Fraunhofer diffraction from a circular aperture
Path length is the same
Do x first – looking down Why?
for all rays = ro
R2 y 2
R y
2 2 EP C eikr 2 R 2 y 2 dy 23
Fraunhofer diffraction from a circular aperture
Do integration along y – looking from the side
P
+R
y=0
ro
-R
r = ro - ysin
24
Fraunhofer diffraction from a circular aperture
R
EP 2Ce ikro
R
e iky sin R 2 y 2 dy
y
Let
kR sin
R
Then ky sin k R
kR
(1)
R2 y 2 R 2 2 R 2 R 1 2 (2)
Rd dy (3)
25
Fraunhofer diffraction from a circular aperture
1
e 1 d
i
EP 2Ce ikro
R 2 2
1
1
J1
i
The integral
e 1 d 2
1
where J1() is the first order Bessell function of the first kind.
26
Fraunhofer diffraction from a circular aperture
These Bessell functions can be represented as
polynomials: 2k p
1k 2
J P
k 0 k!k p !
and in particular (for p = 1),
2 4 6
2 J1 2 2
1 2
2! 2!3! 3!4!
27
Fraunhofer diffraction from a circular aperture
Thus,
2 J1
2
I Io
where = kRsin and Io is the intensity when =0
28
Fraunhofer diffraction from a circular aperture
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 at
sin = 1.22/D, 2.23 /D, 3.24 /D, …
29
Fraunhofer diffraction from a circular aperture
2J 1
1.0
2 J 1
0.5 2
-10 -8 -6 -4 -2 0 2 4 6 8 10
30
The central Airy disc contains 85% of the light
Fraunhofer diffraction from a circular aperture
D
sin = 1.22/D
31
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 <
10
8
6
f
4
D
2
0
1.0
0.5
-2
-4
-6
32
8
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
10
become indistinguishable
8
10 6
8 4
6 2
S1
4 0
1.0
0.5
2 -2
0 -4
S2
1.0
0.5
-2 -6
-4 -8
33
-6-10
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,
1.22
m in
D
e.g. telescope D = 10 cm = 500 X 10-7 cm
5 X 10 5
m in 5 X 10 6 rad
10
e.g. eye D ~ 1mm min = 5 X 10-4 rad
34
