Download Jones vector for horizontally polarized light The electric field

Chapters 14 & 18: Matrix methods
Welcome to the Matrix
http://www.youtube.com/watch?v=kXxzkSl0XHk
Let’s start with polarization…
y
light is a 3-D vector field
z
x
linear polarization
circular polarization
Plane waves with k along z direction
oscillating (vibrating) electric field
y
y
x
x
Any polarization state can be described as linear combination of these two:

i kz t  y 
i  kz t  x 
ˆ
E  E0 x e
x  E0 y e
yˆ



i
E  E0 x ei x xˆ  E0 y e y yˆ ei kz t 
“complex amplitude”
contains all polarization info
Vector representation



i
E  E0 x ei x xˆ  E0 y e y yˆ ei kz t 
• The state of polarization of light is completely determined by
the relative amplitudes (E0x, E0y) and
the relative phases (D = y  x ) of these components.
• The complex amplitude is written as a two-element matrix, or
Jones vector 
~
E  E for complex field components
~ ~ i kz t 
E  E0 e
~


E
~
0x
E0   ~ 
 E0 y 
i x


E
e
~
0x
E0  
i y 
 E0 y e 
Jones calculus (1941)
Indiana Jones
Ohio Jones
Jones vector for horizontally polarized light
• The electric field oscillations are only
along the x-axis
• The Jones vector is then written,
~
i x



  A
E
E
e
1
~
0x
0x
E0   ~   
     A 
 E0 y   0   0 
0 
where we have set the phase x = 0, for
convenience
• Normalized form:
1 
0 
 
y
x
Jones vector for vertically polarized light
• The electric field oscillations are only
along the y-axis
• The Jones vector is then written,
~

  0  0
E
0 
~
0x
E0   ~   
i y      A  
 E0 y   E0 y e   A
1
where we have set the phase y = 0, for
convenience
• Normalized form:
0 
1 
 
y
x
Jones vector for linearly polarized light at
an arbitrary angle
y
• The electric field oscillations make an
angle  with respect to the x-axis
If  = 0°, horizontally polarized
If  = 90°, vertically polarized
• Relative phase must equal 0:
x = y = 0
• Perpendicular component amplitudes:
E0x = A cos 
E0y = A sin 
• Jones vector:
 E0 x ei x   A cos  
~
E0  

i y   

 E0 y e   A sin  
cos  
A

sin




x
Circular polarization
• Suppose E0x = E0y = 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
• The vector traces a circular
path, rotating counterclockwise
Circular polarization
• The Jones vector for this case – where Ex leads Ey is
i
~  E0 x e x   A 
E0  
i  
i y   
2
 E0 y e   Ae 
• The normalized form is
1
2
1
i 

1
A 
i 
LCP
-This vector represents circularly polarized light, where
E rotates counterclockwise, viewed head-on
-This mode is called left-circularly polarized (LCP) light
• Corresponding vector for RCP light:
Replace /2 with -/2 to get
1
2
1
 i 
 
RCP
Elliptical polarization
If E0x  E0y ,
e.g. if E0x = A and E0y = B ,
the Jone vectors can be written as
 A
iB 
 
 A 
 iB 


Direction of rotation?
counterclockwise
clockwise
General case
resultant vibration due to two perpendicular components
D   y   x
D  n , a line
if
else,
an ellipse
Lissajous figures
matrices
a b 
M

c d 
• various optical elements modify polarization
• 2 x 2 matrices describe their effect on light
• matrix elements a, b, c, and d determine the modification
to the polarization state of the light
Optical elements: 1. Linear polarizer
2. Phase retarder
3. Rotator
Optical elements: 1. Linear polarizer
Selectively removes all or most of the E-vibrations
except in a given direction
TA
y
x
z
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.
a(0)  b(1)  0
c(0)  d (1)  1
 a b  0   0 
 c d  1  1

   
b0
d 1
Operating on horizontally polarized light,
a b  1 0
 c d  0    0 

   
0 0 
M

0 1 
a(1)  b(0)  0
c(1)  d (0)  0
a0
c0
For a linear polarizer with TA vertical.
Jones matrix for a linear polarizer
For a linear polarizer with TA vertical
0 0 
M

0
1


For a linear polarizer with TA horizontal
1 0
M

0
0


For a linear polarizer with TA at 45°
1 1 1
M 
2 1 1
For a linear polarizer with TA at 
 cos 2 
M
sin  cos 
sin  cos  

2
sin  
Optical elements: 2. Phase retarder
FA
y
x
SA
z
Retardation plate
• Introduces a phase difference (Δ) between orthogonal
components
• The fast axis (FA) and slow axis (SA) are shown
D  /2:
quarter-wave plate
D  :
half-wave plate
when
¼ and ½ wave plates
/2

net phase difference:
Δ = /2 — quarter-wave plate
Δ =  — half-wave plate
Jones matrix for a phase retarder
• We wish to find a matrix which will transform the
elements as follows:
Eoxei x
Eoy e
i y
into
Eoxei  x  x 
into
Eoye

i  y  y

• 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
Jones matrix for 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
0
i 1
0 
4
e 

i
4
0
i
e 


M e
i
4
1 0 
0  i 


QWP, SA vertical
QWP, SA horizontal
Jones matrix for a half wave plate
For |Δ| = 
e i 2
M 
 0
ei 2
M 
 0
0
i 1
0 
2
e 

i
2
0

1
e 


0  i 2 1 0 
e 

i
e 2 
0  1
HWP, SA vertical
HWP, SA horizontal
Optical elements: 3. Rotator
rotates the direction of linearly polarized light by a 
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    


 

cos 
M 
 sin 
 sin  
cos  
Multiplying Jones matrices
To model the effects of more than one component on the
polarization state, just multiply the input polarization Jones
vector by all of the Jones matrices:
E1  A3 A 2 A1 E0
Remember to use the correct order!
A single Jones matrix (the product of the individual Jones
matrices) can describe the combination of several components.
Multiplying Jones matrices
Crossed polarizers:
E1  A y A x E0
E0
x
y
x-pol
E1
y-pol
 0 0  1 0   0 0 
Ay Ax  





0
1
0
0
0
0


 

so no light leaks
through.
rotated
x-pol
Uncrossed polarizers:
E0
0 0   1    0 0 
A y A x    





0
1

0

0


 

 Ex   0 0   Ex   0 
A y A x      
E   


E

E

0
y
y
   x
  
E1
y-pol
So Iout ≈ 2 Iin,x
z
Matrix methods for complex optical systems
In dealing with a system of lenses,
we simply chase the ray through
the succession of lens.
That is all there is to it.
Richard Feynman
Feynman Lectures in Physics
The ray vector
A light ray can be defined by two coordinates:

position (height), y
islope, 
y
optical axis
These parameters define a ray vector, which will change
with distance and as the ray propagates through optics:
 y
 
 
To “chase the ray,” use ray transfer matrices that characterize
translation
refraction
reflection
…etc.
Note to those using Hecht: vectors are formulated as
 
 y
 
System ray-transfer matrices
Optical system ↔ 2 x 2 ray matrix
 yin 
 
 in 
A
C

B
D 
 yout 
 
 out 
matrices
A B
M 

C D 
• 2 x 2 matrices describe the effect of many elements
• can also determine a composite ray-transfer matrix
• matrix elements A, B, C, and D determine useful properties
n0
Det M  AD  BC 
nf
Multiplying ray matrices
 yin 
 
 in 
O1
O2
O3

 yout 
 yin 
   yin  

   O3 O2  O1     O3O2O1  

 out 
 in 
   in  

Notice that the order looks
opposite to what it should be,
but it makes sense when you
think about it.
 yout 
 
 out 
Exercises
You are encouraged to
solve all problems in the
textbook (Pedrotti3).
The following may be
covered in the werkcollege
on 20 October 2010:
Chapter 14:
2, 5, 13, 15, 17
Chapter 18:
3, 8, 11, 12