Chapters 14 & 18: Matrix methods Chapters 14 & 18: Matrix methods

http://www.youtube.com/watch?v=kXxzkSl0XHk

Welcome to the Matrix

Let’s start with polarization…light is a 3-D vector field

linear polarization circular polarization

y

x

z

y

x x

y

Plane waves with k along z directionoscillating (vibrating) electric field

Any polarization state can be described as linear combination of these two:

“complex amplitude” contains all polarization info

yeExeE yx

tkziy

tkzix ˆˆ 00

E

tkziiy

ix eyeExeE yx ˆˆ 00E

Vector representation

tkzie 0~~EE

EE~

tkziiy

ix eyeExeE yx ˆˆ 00E

for complex field components

y

x

E

E

0

00 ~

~~E

y

x

iy

ix

eE

eE

0

00

~E

• The state of polarization of light is completely determined bythe relative amplitudes (E0x, E0y) andthe relative phases (= y x ) of these components.

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

Jones calculus (1941)

Indiana Jones Ohio Jones

• 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

00~

~~ 0

0

00 A

AeE

E

EE

xix

y

x

x

y

• Normalized form:

0

1

Jones vector for horizontally 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 convenience

1

000~

~~

00

00 A

AeEE

EE

yiyy

x

x

y

• Normalized form:

1

0

Jones vector for vertically polarized light

sin

cos

sin

cos~

0

00 A

A

A

eE

eEy

x

iy

ixE

Jones vector for linearly polarized light at an arbitrary angle

• The electric field oscillations make an angle with respect to the x-axis

If = 0°, horizontally polarizedIf = 90°, vertically polarized

• Relative phase must equal 0:x = y = 0

• Perpendicular component amplitudes:

E0x = A cos E0y = A sin

• Jones vector:

x

y

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 counter-clockwise

iA

Ae

A

eE

eEE ii

y

ix

y

x 1~2

0

00

i

1

2

1

Replace /2 with -/2 to get

Circular polarization

• The Jones vector for this case – where Ex leads Ey 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 (LCP) light

• Corresponding vector for RCP light:

i1

2

1

LCP

RCP

iB

ADirection of rotation?

counterclockwise

clockwise

Elliptical polarization

If E0x E0y , e.g. if E0x = A and E0y = B ,

the Jone vectors can be written as

iBA

General caseresultant vibration due to two perpendicular components

if , a lineelse, an ellipse

xy n

Lissajous figures

dc

baM

• 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

matrices

Optical elements: 1. Linear polarizer

2. Phase retarder

3. Rotator

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

TA

x

y

z Linear polarizer

Optical elements: 1. Linear polarizer

1

0

1

0

dc

ba

10

00M

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.

For a linear polarizer with TA vertical.

Operating on horizontally polarized light,

Jones matrix for a linear polarizer

0

0

0

1

dc

ba

1)1()0(

0)1()0(

dc

ba

1

0

d

b

0)0()1(

0)0()1(

dc

ba

0

0

c

a

Jones matrix for a linear polarizer

10

00MFor a linear polarizer with TA vertical

For a linear polarizer with TA horizontal

00

01M

11

11

2

1M

2

2

sincossin

cossincosM

For a linear polarizer with TA at 45°

For a linear polarizer with TA at

• Introduces a phase difference (Δ) between orthogonal components

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

when /2: quarter-wave plate

: half-wave plate

FA

x

y

z Retardation plate

SA

Optical elements: 2. Phase retarder

/2

¼ and ½ wave plates

net phase difference: Δ = /2 — quarter-wave plateΔ = — half-wave plate

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

• It is easy to show by inspection that,

• Here x and y represent the advance in phase of

the components

yyy

xxx

ioy

ioy

iox

iox

eEeE

eEeE

into

into

y

x

i

i

e

eM

0

0

Jones matrix for a phase retarder

ie

e

eMi

i

i

0

01

0

0 4

4

4

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,

QWP, SA vertical

i

eMi

0

014

QWP, SA horizontal

10

01

0

0

10

01

0

0

2

2

2

2

2

2

i

i

i

i

i

i

ee

eM

ee

eM HWP, SA vertical

HWP, SA horizontal

Jones matrix for a half wave plate

For |Δ| =

x

y

Rotator

SA

Optical elements: 3. Rotatorrotates the direction of linearly polarized light by a

sin

cos

sin

cos

dc

ba

cossin

sincosM

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 ( + )

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:

1 3 2 1 0E EA A A

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:

0 0 1 0 0 0

0 1 0 0 0 0

y xA A

x

y z

1 0y xE EA A

0E1E

x-pol

y-pol

so no light leaks through.

Uncrossed polarizers:

0 0 1 0 0

0 1 0 0

y xA A

0E 1E

rotatedx-pol

y-pol

00 0

0x x

y y x

E E

E E E

y xA A So Iout ≈2 Iin,x

Multiplying Jones matrices

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 FeynmanFeynman Lectures in Physics

Matrix methods for complex optical systems

The ray vector

A light ray can be defined by two coordinates:

position (height), y

islope,

optical axis

optical ray

y

These parameters define a ray vector, which will change with distance and as the ray propagates through optics:

To “chase the ray,” use ray transfer matrices that characterize translation

refractionreflection

…etc.

y

Note to those using Hecht: vectors are formulated as

y

A B

C D

Optical system ↔ 2 x 2 ray matrix

System ray-transfer matrices

in

iny

out

outy

DC

BAM

• 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

matrices

fnn

BCADMDet 0

O1 O3O2

Multiplying ray matrices

Notice that the order looks opposite to what it should be, but it makes sense when you think about it.

in

iny

out

outy

in

in

in

in

out

out yOOO

yOOO

y

123123

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