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

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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  
2rr '          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                              2rr '

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
   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                                


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

    1k   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

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