HALF WAVE: it is a phase shift of

HALF WAVE: it is a phase shift of 

E_x0 cost

E_y0 cost E_x0 cos(t + ) E_y0 cost

with respect to polarization, the angle is doubled

Matrix treatment of polarization

• Consider a light ray with an instantaneous E-vector as shown

 ^{k t iE}  ^{k t jE}  ^{k t}

E ,  ˆ _x ,  ˆ _y ,

x y

E_x E_y

 

 

x

t kz i

t kz i ox

x

E e

E





















Matrix treatment of polarization

• Combining the components

• The terms in brackets represents the complex amplitude of the plane wave

 

 

 

 ^{t kz}

i o

kz t i i

oy i

ox

kz t i oy kz

t i ox

e E E

e e

E j e

E i E

e E

j e

E i E

x y

x y







 















 





 



~

ˆ ˆ

ˆ ˆ







Jones Vectors

• The state of polarization of light is determined by

– the relative amplitudes (E_ox, E_oy) and, – the relative phases ( = _y - _x )

of these components

• The complex amplitude is written as a two-element matrix, the Jones vector

 

 



 

 





 



 

 





 



 



i oy i ox

y i oy

i ox oy

ox

o

E e

e E e

E

e E

E

E E

x

~

~ ~

Jones vector: Horizontally polarized light

• The electric field oscillations are only along the x-axis

• The Jones vector is then written,

where we have set the phase _x = 0, for convenience



 



 



 



 



 



 











 

0 1 0 0

~

~ ~

A A e

E E

E E

i x

ox oy

ox o

 x

y

The arrows indicate the sense of movement as the beam approaches you

The normalized form is

 1 

x y

Jones vector: Vertically polarized light

• The electric field

oscillations are only along the y-axis

• The Jones vector is then written,

• Where we have set the phase _y = 0, for



 



 



 



 



 



 











 

1 0 0 0

~

~ ~

A A e

E E

E E _{i y}

oy oy ox

o 

The normalized form is

 

  0

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,

and the Jones vector is simply a line inclined at an angle  = tan^-1(E_oy/E_ox)

since we can write

 

oy m ox

y x

E E E

E  1

~

x y



The normalized form is



cos

Circular polarization

• Suppose E_ox = E_oy = A and E_x leads E_y by

90^o=/2

• At the instant E_x reaches its maximum

displacement (+A), E_y is zero

• A fourth of a period later, E_x is zero and E_y=+A

x y

t=0, E_y = 0, E_x = +A

t=T/4, E_y = +A, E_x = 0

t=T/8, E_y = +Asin 45^o, E_x = Acos45^o

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 positive

 





x

kz t i oy y

kz t i ox x

e E

E

e E

E





















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

• The negative sign before  indicates a lag in the y- vibration, relative to x-vibration

 

  







t i y

t i x

Ae E

Ae

E

Circular polarization

• To see this lag (in action), take the real parts

• We can then write

• Remembering that =2/T, the path travelled by the e-vector is easily derived

• Also, since E²=E_x² + E_y² = A²(cos²t + sin²t)= A²

• The tip of the arrow traces out a circle of radius A.

A= 1 for normalization.

t A

t A

E

t A

E

y x

 





2 sin cos

cos

 



 



 





Circular polarization

• The Jones vector for this case – where E_x leads E_y is

• 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?



 



 



 



 











 

A i Ae

A e

E e

E E i i

oy i ox

o _y

x 1

~

 2







 



 i 1 2 1





Elliptically polarized light

• If E_ox  E_oy , e.g. if E_ox=A and E_oy = B

• The Jones vector can be written



 









 





iB A iB

A Type of rotation?

Type of rotation?

counterclockwise

clockwise

Jones vector and polarization

• In general, the Jones vector for the arbitrary case is an ellipse ( m; (m+1/2))

 ^_^





 



 



 



 

sin cos

~

i B

A e

E

E E _i

oy ox o

a

b

E_ox E_oy

x y

2 2

cos 2 2

tan E E

E E_{ox oy}

  





Summary:

Given the expression for the field:

the components of the Jones matrix are:

Optical elements: Linear polarizer

• Selectively removes all or most of the E- vibrations except in a given direction

TA

x y

Jones matrix for a linear polarizer



 



 



 







 





1 0 1

0 d

c

b a



 0 0

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,



 



 



 







 





0 0 0

1 d

c

b a

Operating on horizontally polarized light,

Jones matrix for a linear polarizer

 

 



 













2 2

sin cos

sin

cos sin

M cos

• For a linear polarizer with a transmission axis at



Optical elements: Phase retarder

• Introduces a phase difference (∆) between orthogonal components

• The fast axis (FA) and slow axis (SA) are shown

FA

x y

SA

Jones matrix of a phase retarder

• We wish to find a matrix which will transform the elements as follows:

• It is easy to show by inspection that,

 

 _{y y} 

y

x x x

i oy i

oy

i ox i

ox

e E o

e E

e E o

e E

















int int



 



 

y x

i i

e

M e _



0

0

Jones matrix of a Quarter Wave Plate

• Consider a quarter wave plate for which |∆| = /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^ e^

M ⁱ

i ⁴ 0 ^ ₄ 1 0





Jones matrices: HWP

• For |∆| = 



 





 











 



 





 











 



 

1 0

0 1

0

0

1 0

0 1

0

0

2 2

2

2 2

2













i i

i

i i

i

e e

M e

e e

M e HWP, SA vertical

HWP, SA horizontal

Jones matrix of a Quarter Wave Plate

• Consider a quarter wave plate for which |∆| = /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^ e^

M ⁱ

i ⁴ 0 ^ ₄ 1 0





Optical elements: Rotator

• Rotates the direction of linearly polarized light by a particular angle 

x y

 SA

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 ( +  )

• One then finds

 

 ^_^







 



 







 

















sin cos sin

cos d

c b a



 



 

  





cos sin

sin M cos

HWP QWP

HWP QWP

Jones vector and polarization

• In general, the Jones vector for the arbitrary case is  an ellipse ( m; (m+1/2))

 ^_^





 



 



 



 

sin cos

~

i B

A e

E

E E _i

oy ox o

a

b

E_ox E_oy

x y

2 2

cos 2 2

tan E E

E E_{ox oy}

  



Problems 

Optical system

x

y



LASER



.

P t

Objective

filament

A

Spectral interferometry