Transcript Propagation of waves - Dalhousie University

Fraunhofer Diffraction:
Circular aperture
Wed. Nov. 27, 2002
1
Fraunhofer diffraction from a circular aperture

y
x

P
r

E P  C  e dxdy
ikr
Lens plane
2
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
3
Fraunhofer diffraction from a circular aperture
Do integration along y – looking from the side

P
+R

y=0

-R
ro

r = ro - ysin
4
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)
5
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.
6
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
7
Fraunhofer diffraction from a circular aperture

Thus,
 2 J 1   
I  Io 

 


2
where  = kRsin and Io is the intensity when =0
8
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, …
9
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

10
Fraunhofer diffraction from a circular aperture
D

sin = 1.22/D
11
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

12
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

13
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
14
Polarization
15
Matrix treatment of polarization

Consider a light ray with an instantaneous Evector as shown

E k , t   iˆEx k , t   ˆjE y k , t 
y
Ey
x
E x  Eox e
i  kz t  x 
Ex
E y  Eoy e

i kz t  y

16
Matrix treatment of polarization

Combining the components

i kz t  y 
i  kz t  x 
ˆ
ˆ
E  i Eox e
 jEoy e

i y
i x
ˆ
ˆ
E  i Eox e  jEoy e e i  kz t 
 ~ i  kz t 
E  Eo e



The terms in brackets represents the complex
amplitude of the plane wave
17
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 twoelement matrix, the Jones vector
~
i x



Eox 
Eox e 
Eox
~
i x 
Eo   ~   
e 
i 
iy 
 Eoy   Eoy e 
 Eoy e 
18
Jones vector: Horizontally polarized light


The electric field oscillations
are only along the x-axis
The Jones vector is then
written,
The arrows indicate
the sense of movement
as the beam
approaches you
y
~

Eox   Eox ei x   A
1
~
Eo   ~   
     A 
 Eoy   0   0 
0 
where we have set the phase
x = 0, for convenience
x
The normalized form
is
1 
0 
 
19
Jones vector: Vertically polarized light


The electric field
oscillations are only along
the y-axis
The Jones vector is then
written,
~

  0  0
E
0 
~
ox
Eo   ~   
i y      A 
 Eoy   Eoy e   A
1

Where we have set the
phase y = 0, for
convenience
y
x
The normalized form
is
0 
1 
 
20
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,
Ex
m Eox
  1
Ey
Eoy
y

x
and the Jones vector is simply a line
inclined at an angle  = tan-1(Eoy/Eox)
The normalized form is
since we can write
~


E
~
m cos 
ox
Eo   ~   A 1 

sin

E
 oy 


21
Jones vector and polarization

In general, the Jones vector for the arbitrary
y
case E~o   Eoxi 
 Eoy e 
Eoy
is an ellipse
tan 2 
b
2 Eox Eoy cos 
E ox2  E oy2
a

x
Eox
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