Transcript Propagation of waves - Dalhousie University

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,
d
area of
aperture

Fraunhofer or far field limit
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

S
screen

f1
f2
4
Fraunhofer diffraction
1.
2.
Obliquity factor
Assume S on axis, so nˆ  rˆ'  1
Assume  small ( < 30o), so nˆ  rˆ  1
Assume uniform illumination over aperture
ikr'
r’ >>  so
3.
e
r'
is constant over the aperture
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,
ikE o e ikr'
EP  
2rr '
ikr
e
 dS
aperture
6
Single slit Fraunhofer diffraction
P
y=b

dy
y
r
ro
r = ro - ysin
dA = L dy
where L   ( very long slit)
7
Single slit Fraunhofer diffraction
I o  bC
2
E P  C  e ikr dA
__________
_____
b
E P  C  e ikro e ikysin  dy
o
E P  bCeikro e i
sin 

where,
kb
  sin 
2
ikEo e ikr'
C
2rr '
I  Io
sin 
2

2
Fraunhofer single slit diffraction pattern
8
Single Slit Fraunhofer diffraction: Effect of slit width
3
2
0.0
0.2
0.4
0.6
-3
-2
-1
0


1
b
0.8

1.0

Minima for sin  = 0
 = p = k(b/2)sin  or, sin = p(/b)
First minima at sin  = /b
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
y=3a
y=2a+b
y=2a
y=a+b
y=a
y=b
y=0

P



11
Diffraction from an array of N slits

It can be shown that,
 sin 
I P  I o 
 

where,
b
  k sin 
2
2
  sin N 
 

  sin  
2
a
  k sin 
2
12
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 N
Io
sin 
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  0.045I max
2 
 9 
16
Single slit envelope

Now interference term or pattern is
modulated by the diffraction term
 sin 

 






2
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







m1 = a sin 
m2 = 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
x

P
r

E P  C  e dxdy
ikr
Lens plane
22
Fraunhofer diffraction from a circular aperture
Do x first – looking down
Path length is the same
for all rays = ro
Why?


R2  y2
 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

-R
ro

r = ro - ysin
24
Fraunhofer diffraction from a circular aperture
R
EP  2Ceikro
ikysin 
e

R 2  y 2 dy
R
Let
Then
y
 
R
  kR sin 
  
ky sin   k R  
 kR 
R2  y2 
R 2  2 R 2  R 1  2
Rd  dy
(1)
( 2)
(3)
25
Fraunhofer diffraction from a circular aperture
1
EP  2Ce
ikro
R
2
e
i
1  d
2
1
1
The integral
e

1
i
J1  
1  d 

2
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
 1k   2 


J P    
k!k  p !
k 0


and in particular (for p = 1),
  
  
  
2
2
2
2 J1  



 1




2!
2!3!
3!4!
2
4
6
27
Fraunhofer diffraction from a circular aperture

Thus,
 2 J 1   
I  Io 

 


2
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
-10
-8
-6
-4
-2
0
2
4
6
8
2
10
The central Airy disc contains 85% of the light

30
Fraunhofer diffraction from a circular aperture
D

sin = 1.22/D
31
Diffraction limited focussing
6
8
10
Cannot focus any wave to spot with dimensions < 
2
0
-2
-4


0.5
D
4
f
1.0


W = 2fsin  2f  = 2f(1.22/D) = 2.4 f/D
W = 2.4(f#) > 
f# > 1
-6

sin = 1.22/D
The width of the Airy disc
8

32
Fraunhofer diffraction and spatial resolution

If S1, S2 are too close together the Airy patterns will overlap and
become indistinguishable
4 0
-6-10
-4 -8
-2 -6
0 -4
2 -2
0.5
0.5
S2

1.0
S1
1.0
6 2
8 4
10 6

8



Suppose two point sources or objects are far away (e.g.
two stars)
Imaged with some optical system
Two Airy patterns
10

33
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,
 min

e.g. telescope D = 10 cm  = 500 X 10-7 cm
 min

1.22 

D
5 X 105

 5 X 106 rad
10
e.g. eye D ~ 1mm min = 5 X 10-4 rad
34