Download (y).

Polarization
Jones vector & matrices
Matrix treatment of polarization
• Consider a light ray with an instantaneous E-vector as
shown

E k , t   iˆEx k , t   ˆjE y k , t 
y
E x  Eoxe
Ey
i  kz t  x 
x
Ex
E y  Eoy e

i kz t  y

Matrix treatment of polarization
• Combining the components

i kz t  y 
i  kz t  x 
ˆ
ˆ
E  i Eoxe
 jEoy e

i y
i x
ˆ
ˆ
E  i Eoxe  jEoy e ei kz t 
 ~ i kz t 
E  Eo e


• The terms in brackets represents the complex amplitude of the
plane wave
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 two-element matrix, the
Jones vector
~
i x




Eox 
E
e
E
~
ox
ox
i x 
Eo   ~   
e 
i 
i y 
E
e
 Eoy   Eoy e 
 oy 
Jones vector: Horizontally polarized light
• The electric field oscillations are only
along the x-axis
• The Jones vector is then written,
~

Eox   Eoxei x   A
~
Eo   ~   
 
 Eoy   0   0 
The arrows indicate the
sense of movement as
the beam approaches
you
y
1
A 
0 
where we have set the phase x = 0,
for convenience
x
The normalized form
is
1
 
0 
 
Jones vector: Vertically polarized light
• The electric field
oscillations are only along
the y-axis
• The Jones vector is then
written,
~

  0  0
E
~
Eo   ~ox   
i y     
 Eoy   Eoy e   A
• Where we have set the
phase y = 0, for
convenience
0 
A 
1
y
x
The normalized form
is
0 
1 
 
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 E
  1 ox
Ey
Eoy
• and the Jones vector is simply a line
inclined at an angle  = tan-1(Eoy/Eox)
since we can write
~
~  Eox 
m cos  
Eo   ~   A 1 

 Eoy 
 sin  
y

x
The normalized form is
Circular polarization
• 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
y
x
t=0, Ey = 0, Ex = +A
t=T/8, Ey = +Asin 45o, Ex = Acos45o
t=T/4, Ey = +A, Ex = 0
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

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 yvibration, relative to x-vibration
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.
• The Jones vector for this case – where Ex leads Ey is
i
~  Eoxe x   A 
Eo  
i  
i y   
2
 Eoy e   Ae 
• The normalized form is,
1
A 
i 
1 1

2 i 
• 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?
Replace /2 with -/2 to get
1
2
1
 i 
 
Elliptically polarized light
• If Eox  Eoy , e.g. if Eox=A and Eoy = B
• The Jones vector can be written
 A
iB 
 
Type of rotation?
 A 
 iB 


Type of rotation?
counterclockwise
clockwise
What determines the major or minor axes of the ellipse?
Here A>B
Jones vector and polarization
• In general, the Jones vector for the arbitrary case is
an ellipse ( m; (m+1/2))
y
A

~  Eox  
Eo  
i   

E
e


B
cos


i
sin


 oy  
Eoy
b
tan 2 
2 Eox Eoy cos 
E ox2  E oy2
a

x
Eox
Polarization and lissajous figures
• http://www.netzmedien.de/software/downlo
ad/java/lissajous/
• http://www.awlonline.com/ide/Media/JavaT
ools/funcliss.html
• http://fips-server.physik.unikl.de/software/java/lissajous/
Optical elements: Linear polarizer
• Selectively removes all or most of the E-vibrations
except in a given direction
TA
y
x
Linear polarizer
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.
Jones matrix for a linear polarizer
• For a linear polarizer with a transmission axis at 
 cos 2 
M 
sin  cos
sin  cos 

2
sin  
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
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 
Eoy e
i y
int o
Eoy e
• It is easy to show by inspection that,

i  y  y

ei x
0 
M 
i y 
0
e


• Here x and y represent the advance in phase of the components
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 
Jones matrices: HWP
• For |Δ| = 
e i 2
M 
 0
ei 2
M 
 0
0
i 1
0 
2
e
0  1
i
2
e 


0
i 1
0 
2
e 

i
2
0

1
e



HWP, SA vertical
HWP, SA horizontal
Optical elements:
Quarter/Half wave plate
• When the net phase difference
Δ = /2 : Quarter-wave plate
Δ =  : Half-wave plate
/2

Optical elements: Rotator
• Rotates the direction of linearly polarized light by a
particular angle 
y

x
SA
Rotator
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 b cos
cos   

c


 


d   sin    sin    
• One then finds
cos 
M 
 sin 
 sin  
cos  
Polarization Representation
Polarization Ellipse
Jones Vector
1
 
0
0
 
1
1
2
1
2
1
 
1
 1 
 
 1
(,y)
 0, 0 
 0,  / 2 
 0,  / 4 
(f,)
Stokes
 0, 0 
1
 
1
0
 
0
 / 2, 0 
 1 



1


 0 


 0 
 / 4, 0 
1
 
0
1
 
0
 , / 4   / 4, 0
 1 


0


 1 


0


Polarization Representation
Polarization Ellipse
Jones Vector
(,y)
(f,)
1  1 


2  i 
  / 2, / 4
 0,  / 4 
1 1
 
2 i
 / 2, / 4
 0,  / 4
1  1


5  2i 
1  2 


5  i 
1 2i


10  2  i 
Stokes
 1 


 0 
 0 

 1 



1
 
0
0

1

 

 1 


 3 / 5 
 0 

 4/5 




1 1 


/
2,
tan


2

 1 


 3/5 
 0 

 4 / 5 





1 1 

/
2,

tan
 / 2, tan 2 

2

1

1 1 
0,
tan


2


 
1 4
1 1 

tan
,

/
4

/
4,
tan

 

3
2

 
 1 


 0 
 3/5 

 4 / 5 



Examples
Polarization State
Polarizer with transmission
axis oriented  to X-axis
 if   1
P0  e

0
E
Y-axis
0

0
f
X-axis
If polarizer is rotated by y about Z
1 1
 
2 i
P  R  y  P0 R y 
ignoring f’ polarizers transmitting
light with electric field vectors  to
x and y are:
0
0
 Py  
0
0
 cos f 


sin
f


1 1
 
2  i 
f’ is due to finite optical thickness
of polarizer.
1
Px  
0
Jones Vector
a
b
0

1
a
b
f
f
 a cos f  ib sin f 


 a sin f  ib cos f 
f
 a cos f  ib sin f 


a
sin
f

ib
cos
f


Examples
¼ Wave Plate

and

and the thickness
2
y  450

t
4 n
and incident beam is
1 1
W 
2  i
vertically polarized:
Polarization State
Y-axis
E
X-axis
f
1 1
W 
2 1
Incident
Jones Vector
1
2
1
 
i
1

1
Emerging Jones Vector
1
 
0
 cos f 


 sin f 
  i 

exp

0

1   4 
 1 1


1 
 i   2  1
0
exp   

 4 

i 

1
V 
V 
1 1

2  i
1 1

2  i
V 
i   1 
1 1

 
 
1 0
2  i 
i   cos f 
1  cos f  i sin f 





1   sin f 
2  sin f  i cos f 
1 1

2  i
i  1  1 
    
1 i  0
Birefringent Plates
45
45
Parallel polarizers


cos

 0 0
2
E'  

 0 1   i sin 

2



i sin 

2  1  0   1  cos 
2
 
  2 1
2 

cos 
 0 
2 
1
 1
  (ne  no )d 
I  cos 2    cos 2 

2

2 2


Cross polarizers


cos

 1 0
2
E'  

 0 0   i sin 

2


 
2  1  0   i  sin 
2
 
  2 1
2 

cos 
 0 
2 
i sin
1
 1
  ( ne  no ) d 
I  sin 2    sin 2 

2

2 2


Wave Plates
Jones Matrices
y
x
c-axis
c-axis
c-axis
In
general:
c-axis
450
y
 ei  / 2
W 
 0
0 

ei  / 2 
 ei  / 2
0 
W 
i / 2 
0
e




cos

2
W 
 i sin 


2

2

 
cos 
2 
i sin
Remember:

2
 ne  no  d

 ei  / 2
0 
W  R    
R 
i  / 2 
e
 0

 cos   sin    e i  / 2 0   cos  sin  
W 



ei  / 2    sin  cos  
 sin  cos    0
Polarizers
Jones Matrices
y
transmission
axis
x
transmission
axis
transmission
axis
450
In
general:
transmission
axis
y
1
W 
0
0

0
0
W 
0
0

1
 1/ 2 1/ 2 
W 

1/
2
1/
2


Remember:
 1 0
W  R    
R 
0
0


 cos   sin   1 0  cos  sin  
W 



 sin  cos   0 0   sin  cos  
 cos 2 
cos  sin  


2
sin

cos

sin


