Download Lecture 34

Polarization
Jones vector & matrices
Fri. Nov. 29, 2002
1
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  Eoxe
i  kz t  x 
Ex
E y  Eoy e

i kz t  y

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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
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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 
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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 e i 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 
 
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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 
 
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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 


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Circular polarization
y



Suppose Eox = Eoy = A
and Ex leads Ey by
90o=/2
At the instant Ex
reaches its maximum
displacement (+A), Ey
is zero
A fourth of a period
later, Ex is zero and
Ey=+A
x
t=0, Ey = 0, Ex = +A
t=T/8, Ey = +Asin 45o, Ex = Acos45o
t=T/4, Ey = +A, Ex = 0
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Circular polarization


For these cases it is
necessary to make y
>x . Why?
This is because we
have chosen our
phase such that the
time dependent term
(t) is negative
E x  Eoxe
i  kz t  x 
E y  Eoy e

i kz t  y

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Circular polarization



In order to clarify this, consider the wave at z=0
Choose x=0 and y=, so that x > y
Then our E-fields are
E x  Ae
 i t 
E y  Ae

i t  
The negative sign before  indicates a lag in
the y-vibration, relative to x-vibration
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Circular polarization


To see this lag (in action), take the real parts
We can then write E x  A cos t


E y  A cos t    A sin t
2




Remembering that =2/T, the path travelled by the evector is easily derived
Also, since E2=Ex2 + Ey2 = A2(cos2t + sin2t)= A2
The tip of the arrow traces out a circle of radius A.
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Circular polarization

The Jones vector for this case – where Ex leads Ey is
i
~  Eoxe x   A 
Eo  
i  
i y   
2
 Eoy e   Ae 
1
A 
i 

The normalized form is,

This vector represents circularly polarized light, where
E rotates counterclockwise, viewed head-on
This mode is called left-circularly polarized light
What is the corresponding vector for right-circularly
polarized light?


1
2
1
i 
 
Replace /2 with -/2 to get
1
2
1
 i 
 
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Elliptically polarized light


If Eox  Eoy , e.g. if Eox=A and Eoy = B
The Jones vector can be written
 A
iB 
 
 A 
  iB 


Type of rotation?
Type of rotation?
counterclockwise
clockwise
What determines the major or minor axes of the ellipse?
Here A>B
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Jones vector and polarization

In general, the Jones vector for the arbitrary
case is an ellipse ( m; (m+1/2))
y
Eoy
 Eox  
A

~
Eo  

i 
 Bcos   i sin  
E
e
oy


 
b
a
tan 2 
2 Eox Eoy cos 
E ox2  E oy2

x
Eox
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Polarization and lissajous figures

http://www.netzmedien.de/software/download/java/lissajous/
http://www.awlonline.com/ide/Media/JavaTools/funcliss.html

http://fips-server.physik.uni-kl.de/software/java/lissajous/

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Optical elements: Linear polarizer

Selectively removes all or most of the Evibrations except in a given direction
TA
y
x
Linear polarizer
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Jones matrix for a linear polarizer
Consider a linear polarizer with transmission axis along the
vertical (y). Let a 2X2 matrix represent the polarizer
operating on vertically polarized light.
The transmitted light must also be vertically polarized. Thus,
 a b  0   0 
 c d  1  1

   
Operating on horizontally polarized light,
a b  1 0
 c d  0    0 

   
Thus,
0 0
M 

0
1


Linear polarizer with TA
vertical.
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Jones matrix for a linear polarizer

For a linear polarizer with a transmission
axis at 
 cos 
M 
sin  cos
2
sin  cos 

2
sin  
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Optical elements: Phase retarder


Introduces a phase difference (Δ) between
orthogonal components
The fast axis(FA) and slow axis (SA) are shown
FA
y
x
SA
Retardation plate
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Jones matrix of a phase retarder

We wish to find a matrix which will transform the
elements as follows: Eoxei x int o Eoxei  x  x 
i y
i  y  y 
Eoy e
int o Eoy e

It is easy to show by inspection that,
ei x
M 
 0

0 
i y 
e 
Here x and y represent the advance in phase of the
components
20
Jones matrix of a Quarter Wave Plate




Consider a quarter wave plate for which |Δ| =
/2
For y - x = /2 (Slow axis vertical)
Let x = -/4 and y = /4
The matrix representing a Quarter wave plate,
with its slow axis vertical is,
e i 4
M 
 0
i 1
0 
4

e


0
i
4
e 

0
i 
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Jones matrices: HWP

For |Δ| = 
e i 2
M 
 0
e i 2
M 
 0
i 1
0 
2

e


0
i
e 2 

i 1
0 
2

e


0
i
2
e


0
 1
0
 1
HWP, SA vertical
HWP, SA horizontal
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Optical elements:
Quarter/Half wave plate

When the net phase difference
Δ = /2 : Quarter-wave plate
Δ =  : Half-wave plate

/2
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Optical elements: Rotator

Rotates the direction of linearly polarized
light by a particular angle 
y

x
SA
Rotator
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Jones matrix for a rotator


An E-vector oscillating linearly at  is rotated by
an angle 
Thus, the light must be converted to one that
oscillates linearly at ( +  )
a
c

b  cos  cos   

d   sin    sin    
cos 
 One then finds M  
 sin 
 sin  
cos  
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