ch 6-optics of ani so tropic media

Document info 6.

EM Wave Propagation in Anisotropic Media

Chapter 6Physics 208, Electro-optics

Peter Beyersdorf

1

Thursday, September 2, 2010

6.

Class Outline

The dielectric tensor

Plane wave propagation in anisotropic media

The index ellipsoid

Phase velocity, group velocity and energy

Crystal types

Propagation in uniaxial crystals

Propagation in biaxial crystals

Optical activity and Faraday rotation

Coupled mode analysis of wave propagation2


6.

Definitions

Isotropic material

A material that has optical properties that are independent of the orientation of the material

(glass is an example of an isotropic material)

Anisotropic material

A material that has optical properties that are dependent of the orientation of the material

(many crystals are anisotropic…why?)

3


6.

The Dielectric Tensor

ε relates the electric field to the electric displacement by

In anisotropic materials the polarization may not be in the same direction as the driving electric field. Why?

�D = � �E = �0 �E + �P

4


6.

Anisotropic Media

Charges in a material are the source of polarization. They are bound to neighboring nuclei like masses on springs.

In an anisotropic material the stiffness of the springs is different depending on the orientation.

The wells of the electrostatic potential that the charges sit in are not symmetric and therefore the material response (material polarization) is not necessarily in the direction of the driving field.

5


6.

Analogies

What every-day phenomena can have a response in a direction different than the driving force?

6


6.

The Dielectric Tensor

The susceptibility tensor χ relates the polarization of the material to the driving electric field by

or

Thus the material permittivity ε in

is also a tensor

Px

Py

Pz

= �0

χ11 χ12 χ13

χ21 χ22 χ23

χ31 χ32 χ33

Ex

Ey

Ez

�P = �0χ �E

� = �0�I + χ

�

�D = � �E = �0 �E + �P

7


6.

Tensor Notation

For brevity, we often express tensor equations such as

in the form

where summation over repeated indices is assumed, so that this is equivalent to

8

�D = �0�I + χ

��E

Di = �0(δij + χij)Ej

Di =�

j

�0(δij + χij)Ej


6.

Dielectric Tensor Properties

The dielectric tensor is Hermetian such that

In a lossless material all elements of ε are real so the tensor is symmetric!! ! ! ! ! and can be described by 6 (rather than 9) elements

�ij = �∗ji

�11 �12 �13�12 �22 �23�13 �23 �33

�ij = �ji

9


6.

Plane Wave Propagation in Anisotropic Media

For a given propagation direction in a crystal (or other anisotropic material) the potential wells for the charges will be ellipsoidal, and so there will be two directions for the driving field for which the material response is in the same direction.

These two polarization states each have a (different) phase velocity associated with them. Light polarized at either of these angles will propagate through the material with the polarization unchanged.

10


6.

Wave Equation in Anisotropic Materials

From Maxwell’s equations in a dielectric (ρ=J=0), using d/dt→iω and ∇→ik we have→

�∇× �E +∂ �B

∂t= 0

i�k × �E + iω �B = 0

�∇× �H − ∂ �D

∂t= �J

i�k × �H − iω �D = �J

�k × �k × �E + ω2µ� �E = 0Which is the wave equation for a plane wave, most easily analyzed in the principle coordinate system where ε is diagonal (i.e. a coordinate system aligned to the crystal axes) 11

→

�k × �E = ωµ �H �k × �H = −ω� �E


6.


�k × �k × �E + ω2µ� �E = 0

for which the determinant of the matrix must be zero for solutions (other than k=ω=0) to exist

ω2µ�x − k2

y − k2z kxky kxkz

kykx ω2µ�y − k2x − k2

z kykz

kzkx kzky ω2µ�z − k2x − k2

y

Ex

Ey

Ez

= 0

or

��

ω2µ�x − k2y − k2

z kxky kxkz


z kykz


y

��= 0

This defines two surfaces in k-space called the normal shells.12


6.

Normal Shells

In a given direction, going out from the origin, a line intersect this surface at two points, corresponding to the magnitude of the two k-vectors (and hence the two values of the phase velocity) a wave of frequency ω can have in a material with permittivity ε can have in this direction.

kx

ky

kz

13

The two directions where the surfaces meet are called the optical axes. Waves propagating in these directions will have only one possible phase velocity.


6.


�k × �k × �E + ω2µ� �E = 0

ω2µ�x − k2

y − k2z kxky kxkz


z kykz


y

Ex

Ey

Ez

= 0

and in the principle coordinate system

where

Ex

Ey

Ez

= E0

kxk2−ω2µ�x

ky

k2−ω2µ�ykz

k2−ω2µ�z

E20 ≡ E2

x + E2y + E2

z 14


6.

Example

Find the possible phase velocities (vp=ω/k) for a wave propagating along the x-axis in a crystal, and the associated polarization directions.

15


6.


�k × �k × �E + ω2µ� �E = 0

ω2µ�x − k2

y − k2z kxky kxkz


z kykz


y

Ex

Ey

Ez

= 0

or

16

X X X XX

XXX X

X�ω2µ�x

� �ω2µ�y − k2

� �ω2µ�z − k2

�= 0

ω = 0,ω

k=

1√

µ�y, or

ω

k=

1√

µ�z

So either have a DC field (ω=0) or a wave with

vp =1

√µ�y

vp =1

√µ�z

or

and kx→k, ky→0, kz→0Starting with


6.


Ex

Ey

Ez

= E0

kxk2−ω2µ�x

ky

k2−ω2µ�ykz

k2−ω2µ�z

17

Eigenpolarizations are found from

Writing kx=kcosφ, ky=ksinφsinθ, kz=ksinφcosθ the eigenpolarization can be evaluated using L’Hopital’s rule with variable φ. For!! ! ! ! ! ! we get eigenpolarization! , for!! ! ! ! ! ! we get .

x

y

z

kφθ

y zvp = 1/

√µ�y

vp = 1/√

µ�z


6.

Gauss’ lawin a dielectric (ρ=0) can be written as

implying the propagation direction is orthogonal to the displacement vector .

But the Poynting vectorwhich gives the direction of energy flow is orthogonal to the electric field , not the displacement vector. Thus energy does not flow in the direction of the wave’s propagation if the polarization of the wave ( ) is not an eigenvector of the material’s dielectric tensor!

Wave Propagation in Anisotropic Materials

i�k · �D = 0

�∇ · �D = ρ

�S = �E × �H

k�D

�E

E18


6.

Spatial Walkoff

19


6.

Eigenstates of a material

For any direction of propagation there exists two polarization directions in which the displacement vector is parallel to the transverse component of the electric field. These are the eigenstates of the system.

Requiring D and E be parallel in the matrix equation

is equivalent to finding the eigenvalues and eigenvectors of ε satisfying

�D = � �E

λi �ui = ��ui

→ →

20


6.

Index Ellipsoid

Index of refraction as a function of polarization angle

The polarization directions (of the electric displacement) that have a max and min index of refraction form the major and minor axes of an ellipse defining n(θ), the index for a wave with the electric displacement vector at an angle of θ in the transverse plane.

�k • n1

• n2

21

�D1

�D2


6.

Index EllipsoidThe union of all index ellipses plotted in three dimensions is the index ellipsoid, defined by

The optical axis is the direction of propagationthat has an index of refraction independentof polarization angle.(i.e. a circular index ellipse)

x2

n2x

+y2

n2y

+z2

n2z

= 1

22

^.

propagation (k)

Direction of Direction of

→.polarization (D)


6.

Example

Find the angle of the direction of the optical axis for Lithium Niobate

(LiNb03 uniaxial negative nx=ny=n0=2.300, nz=ne=2.208)

23


6.

Example

Consider topaz (nx=1.619 ny=1.620, nz=1.627)

What plane is the optical axis in?

How many optical axes are there?

Find the angle of the direction of the optical axis for Topaz

24


6.

Crystal Types

The index ellipsoid is determined by the three principle indices of refraction nx, ny and nz.

A crystal with nx≠ny≠nz will have ____ optical axes

A crystal with nx=ny≠nz will have ____ optical axes

A crystal with nx=ny=nz will have _____ optical axes

25


6.

Crystal PropertiesThe geometry of a crystal determines the form of the dielectric tensor

26


6.


Isotropic

� = �0

n2 0 00 n2 00 0 n2

27


6.


Unaxial

� = �0

n2

0 0 00 n2

0 00 0 n2

e

28


6.


Biaxial

� = �0

n2

x 0 00 n2

y 00 0 n2

z

29


6.

Biaxial Crystals

By convention nx<ny<nz so the optical axis is always in the xz plane

30


6.

Uniaxial Crystals

By convention no≡nx=ny and ne≡nz

The optical axis in a uniaxial crystal is always in the z-direction

A negative uniaxial crystal is one in which ne<no while a positive uniaxial crystal has ne>no.

31


6.

Example

A wave is propagating in the [1,1,1] direction in Mica, what are the two principle indices of refraction and in which polarization directions do these correspond to?

(Mica: nx=1.552, ny=1,582, nz=1.588)

32


6.

Example

A wave is propagating in the [1,1,1] direction in Mica, what are the two principle indices of refraction and in which polarization directions do these correspond to?

Since D is orthogonal to k, the plane of polarization is given by r.k=0 so for k in the [1,1,1] direction x+y+z=0. The intersection of this plane and the index ellipsoid, given by

is,

Find extremes of r2=x2+y2+z2 subject to the preceding two constraints to find directions of transverse eigenpolarizations. Plug directions back into r=(x2+y2+z2)1/2 to find index for each polarization

�k · �r = 0

x2

n2x

+y2

n2y

+(x + y)2

n2z

= 1

33

x2

n2x

+y2

n2y

+z2

n2z

= 1

→ →


6.

Example Solution in Mathematica

34

Parameters for Mica

Constraints for solution: point on index ellipsod in

plane transverse to propagation

Solve 3 equations with 4 vaariables

(x,y,z,r) for distance (r) of point at intersection of

plane and ellipoid (as a function of z)

Find extremum of r(z) corresponding to

major and minor axes of index

ellipse

Find x and y values corresponding to

the z values at the extermum of r(z), i.e. x and z for the

elipse axes

List index (r) and polarization (x,y,z) for the ellipse axes


6.

Index of extraordinary ray

The index of refraction of the ordinary ray is always equal to the index seen when the ray propagates along the optical axis

The extra-ordinary ray sees an index of refraction that is a function of the ray propagation angle from the optical axis. In a uniaxial crystal the index ellipse in a plane containing the optical axis (call it the x-z plane) is described by

where for a wave propagating at an angle θ to the optical axis! ! ! ! ! ! ! ! ! ! ! ! giving

35

�kD

xne(θ

)z

θ

θ

x = ne(θ) cos θ, z = ne(θ) sin θ

1n2

e(θ)=

cos2 θ

n2o

+sin2 θ

n2e

�x

no

�2

+�

z

ne

�2

= 1


6.

kisinθi knsinθn

θn

θi

n

kisinθi knsinθn

θo

θi

no

θene

Double Refraction

Finding the angle of refraction into an anisotropic material using Snell’s law !! ! ! ! ! ! is possible but n2 depends on θ2. Recognizing that Snell’s law requires the tangential component of the k-vector be continuos across a boundary a graphical solution is possible using the normal shells.

36

n1 sin θ1 = n2 sin θ2


6.

Conical Refraction

The normal surface is a surface of constant ω in k-space. Group velocity vg=∇kω is perpendicular to the normal surfaces, but in a biaxial crystal the normal surfaces are singular along the optical axis.

The surrounding region is conical, thus a wave along the optical axis will spread out conically

37

→


6.

Optical Activity

Quartz and other materials with a helical molecular structure exhibit “Optical Activity”, a rotation of the plane of polarization when passing through the crystal.

38


6.

B

Optical Activity

Optically active materials can be thought of as birefringent, having a different index of refraction for right-circular polarization and for left-circular polarization.

The specific rotary power describes how much rotation there is per unit length

The source of this optical activity is the induced dipole moment of the molecule formed by changing magnetic flux through the helical molecule

ρ =π

λ(nl − nr)

39


6.

The displacement vector thus has an additional contribution in a direction perpendicular to the electric field

where !! ! ! ! ! ! ! is called the gyration vector and

so we can define an effective dielectric tensor

so that

allowing the wave equation to be solved for eigenpolarizations of propagation and two corresponding indices of refraction that depend on

Optical Activity

�D = � �E + i�0 �G× �E

�D = �� E

40

�G = (gijkikj/k20)k0

�G× �E =

i j k

Gx Gy Gz

Ex Ey Ez

=

0 −Gz Gy

Gz 0 −Gx

−Gy Gx 0

Ex

Ey

Ez

= G �E

�� = � + i�0G

G


6.

Faraday Rotation

A form of optical activity induced by an external magnetic field.

where V is called the Verdet constant. The sense of rotation depends on the direction of propagation relative to the direction of the magnetic field

ρ = V B

41


6.

The motion of the charges in a material driven by the electric field feel a Lorentz force! ! ! ! ! resulting in an induced dipole with a term proportional to!! ! ! ! such that

where γ=-Vn0λ0/π.

For Faraday rotation the imaginary term is proportional to B, while for optical activity it is proportional to k.

Thus a wave double passing (in opposite directions) a material will net zero polarization rotation from optical activity, but twice the one-way rotation due to Faraday rotation.

Faraday Rotation

42

�D = � �E + i�0γ �B × �E

�B × �Eq�v × �B


6.

ExerciseGiven an isotropic material that is perturbed such that

where Δε is

find the eigenpolarizations for the electric field and associated indices of refraction. If a linearly polarized plane wave were to propagate along the z-direction through a thickness d of such material, what would the output wave look like?

�D = � �E + i∆� �E

43

∆� = �o

0 −Gz Gy

Gz 0 −Gx

−Gy Gx 0


6.

Exercise

44


6.

Coupled Mode Analysis

When the eigenstates of an unperturbed system are known, a small perturbation can be treated as a coupling between these states.

Rather than rediagonalizing the matrix for ε one could solve the wave equation with the perturbed matrix ε+Δε

E = A1ei(k1z−ωt)e1 + A2e

i(k2z−ωt)e2

E = A1(z)ei(k1z−ωt)e1 + A2(z)ei(k2z−ωt)e2

orE = A3e

i(k3z−ωt)e3 + A4ei(k4z−ωt)e4

with ε

with ε+Δε45


6.

ExerciseGiven an isotropic material that is perturbed such that

where Δε is

find the coupled mode expression for the electric field for a linearly polarized plane wave propagating along the z-direction.

∆� =

0 −Gz Gy

Gz 0 −Gx

−Gy Gx 0

�D = � �E + i∆� �E

46


6.

Fromfind solutions of the form

where ! ! ! i.e. we are finding slowly varyingamplitudes Ex(z) and Ey(z) to the unperturbed solutions.Since E varies only along z Use

! ! ! ! ! ! ! ! ! ! is ! the transverse component offor! ! ! ! (assume an isotropic medium)

47

The 1-dimensional wave equationin isotropic media

�∇× �∇× �E − ω2µ(� + ∆�) �E = 0

�E = Ex(z)ei(kz−ωt)i + Ey(z)ei(kz−ωt)j

ω

k=

1√

µ�

�∇ → kd

dz

�∇× �∇× �E = −∇2E + �∇(�∇ · �E)

− d2

dz2�E + k

d

dz

�k

d

dz· �E

�− ω2µ(� + ∆�) �E = 0

� �� Exi + Ey j + Ez k

�− kEz

�E · �k = 0 �Et = �E

�Et�E

d2

dz2�E − ω2µ(� + ∆�) �E = 0

− d2

dz2

��E − k(k · �E)

�

� �� −ω2µ(� + ∆�) �E = 0


6.48

This is from wave equation solutions to the unperturbed material

− d2

dz2

�Ex(z)ei(kz−ωt)

�i− ω2µ(� + ∆�)Ex(z)ei(kz−ωt)i

− d2

dz2

�Ey(z)ei(kz−ωt)

�j − ω2µ(� + ∆�)Ey(z)ei(kz−ωt)j = 0

Using the chain rule to evaluate terms like− d2

dz2

�Ex(z)ei(kz−ωt)

�= − d

dz

�Ex(z)ikei(kz−ωt) + ei(kz−ωt) d

dzEx(z)

�

= Ex(z)k2ei(kz−ωt) − ikei(kz−ωt) d

dzEx(z)

−ikei(kz−ωt) d

dzEx(z)− ei(kz−ωt) d2

dz2Ex(z)

=�

d2

dz2Ex(z)− 2ik

d

dzEx(z) + k2Ex(z)

�ei(kz−ωt)

and using k2=ω2με we get�

d2

dz2Ex(z)− 2ik

d

dzEx(z) + k2Ex(z)

�i

�d2

dz2Ey(z)− 2ik

d

dzEy(z) + k2Ey(z)

�j = 0


6.49

Envelope varies slowly compared to optical frequency

These two coupled first-order differential equations reduce to

uncoupled second order differential equaitons

Solving (1) for Ey gives

plugging into (2) gives

which can be solved to give

(1)

(2)

d2Ex

dz2= −

�ω2µ�0G

2k

�2

Ex

For slow variations! ! ! ! ! ! we can neglect!! ! terms d2E

dz2� k

dE

dzd2E

dz2

we have

Gx = Gy = 0; Gz = Gfor

and∆� = i�0

0 −G 0G 0 00 0 0

2ikdEx

dz+ ω2µ(i�0G)Ey(z)i

+ 2ikdEy

dz− ω2µ(i�0G)Ex(z)j = 0

Ey = − 2k

ω2µ�0G

dEx

dz

Ex(z) = A cos�

ω2µ�0Gz

2k

�+ B sin

�ω2µ�0Gz

2k

�


6.

Assuming the inputfield is linearly polarized along î and has amplitude E0

requiring A=E0

plugging into

gives

and with Ey(0)=0 we have B=0, and substituting ω2με0=k0 we can write

50

This is a field with a plane of polarization direction that is rotating CW as the wave advances along z

Ey =2k

ω2µ�0G

dEx

dz

Ex(z) = E0 cos�

12

k0

kGz

�

Ey(z) = −E0 sin�

12

k0

kGz

�

Ex(z) = A cos�

ω2µ�0Gz

2k

�+ B sin

�ω2µ�0Gz

2k

�

Ex(0) = A cos�

ω2µ�0Gz

2k

�+ B sin

�ω2µ�0Gz

2k

�= E0

Ey =�−E0 sin

�ω2µ�0Gz

2k

�+ B cos

�ω2µ�0Gz

2k

��


6.

Diagonalization

Consider the same problem where

Find the normal modes of propagation by finding the eigenvectors of the dielectric tensor

51

� = �0

n2 −iG 0i G n2 00 0 n2


6.

Diagonalization

Consider the same problem where

Find the normal modes of propagation by finding the eigenvectors of the dielectric tensor

λ=n2, n2-G,n2+G

→u=[0,0,1], [i,1,0], [-i,1,0]

52

� = �0

n2 −iG 0i G n2 00 0 n2


6.

Describing Polarization Mathematically

Since polarization, like vectors, can be described by the value of two orthogonal components, we use vectors to represent polarization

Amplitude of the components can be complex to represent a time delay between the components of the waves

Jones calculus allows us to keep track of the polarization of waves as they propagate through a system

53


�E(z, t) = ei(kz−ωt)

�Ex

Ey

�

6.

Jones VectorsExpressing polarization of a wave in terms of two orthogonal states with complex amplitude (i.e. amplitude and phase) of each component expressed in vector form

54

.�01

�

..�10

�

. .1√2

�11

�

..�cos θsin θ

�

..1√2

�i1

�

..1√2

�1i

�

.

y (vertical)x (horizontal)linear at +45°linear at θ

right circularleft circular

Note

the

exp

ression

s fo

r RC

P an

d LC

P dif

fer

from

the

tex

tboo

k du

e to

the

diff

eren

t

conv

entio

n us

ed for

defi

ning

phas

ors


6.

Jones Matrices

An optical element that transforms one polarization state into another can be treated as a 2x2 matrix acting on a jones vector

The Jones Matrices for a series of optical elements can be multiplied together to find how the optical system transforms the polarization of an input beam

55

�Ex,out

Ey,out

�=

�M11 M12

M21 M22

� �Ex,in

Ey,in

�


6.

Retarders

Devices which delay one polarization component with respect to the other

For a birefringent material of thickness d

Thus the Jones matrix can be written (up to a common phase factor)

where the y-component is delayed by Δφ for a wave expressed as

56

∆φ = (knyd− knxd) =2π

λ∆nd

Mretarder =�

1 00 ei∆φ

�

Note

the

sign

of Δφ for

is o

ppos

ite tha

t of

the

text

book

due

to

the

diffe

rent

con

vent

ion

used

for

defi

ning

phas

ors


�Ex

Ey

�


6.

Quarter Wave Plate

For a retarder with Δφ=π/2 (i.e. a retardation of λ/4) and fast axis horizontal

And when a wave with linear polarization at ±45° passes through the retarder it gets converted to left (right) circular polarization

57

�1 00 i

�

1√2

�1i

�=

1√2

�1 00 i

� �11

�

1√2i

�i1

�=

1√2

�1 00 i

� �1−1

�


6.

Pseudo-Isolator Example

Use Jones Calculus to determine the polarization state of light that goes through a horizontal polarizer and then through a quarter wave plate with its fast axis at 45° from horizontal.

What happens to light that goes through this system and then gets retro-reflected?

58


6.

Example

59

�Ex,out

Ey,out

�=

�cos(45o) sin(45o)− sin(45o) cos(45o)

� �1 00 i

� �cos(45o) − sin(45o)sin(45o) cos(45o)

� �1 00 0

� �Ex,in

Ey,in

�

�Ex,out

Ey,out

�=


� �1 00 i


� �Ex,in

0

�

�Ex,out

Ey,out

�=

1√2


� �1 00 i

� �Ex,in

Ex,in

�

�Ex,out

Ey,out

�=

1√2


� �Ex,in

iEx,in

�

i.e. left circular polarized light

�Ex,out

Ey,out

�=

12

�1 + i−1 + i

�Ex,in

�Ex,out

Ey,out

�=

1√2

�eiπ/4

ei3π/4

�Ex,in

�Ex,out

Ey,out

�=

eiπ/4

√2

�1i

�Ex,in


6.

Example

Left circular polarization becomes right circular polarization on reflection, and after returning through optical system

or

giving Eout=0

60

�Ex,out

Ey,out

�=

1√2

�1 00 0


� �1 00 i

� �cos(45o) sin(45o)− sin(45o) cos(45o)

� �−1i

�Ex,in

�Ex,out

Ey,out

�=

�1 00 0

� �cos(45o) sin(45o)− sin(45o) cos(45o)

� �1 00 i

�2 �cos(45o) − sin(45o)sin(45o) cos(45o)

� �1 00 0

� �Ex,in

Ey,in

�


6.

Compensator

Variable waveplate with Jones matrix

61

Babinet compensator Berek compensator

�1 00 ei∆φ

�



�Ex

Ey

�

6.

Jones Matrices

62

�0 00 1

�

�1 00 0

�

�sin2 θ 1

2 sin 2θ12 sin 2θ cos2 θ

�

�1 00 i

�

�cos2 θ + i sin2 θ (i−1)

2 sin 2θ(i−1)

2 sin 2θ i cos2 θ + sin2 θ

�

�1 00 −1

�

�cos 2θ − sin 2θsin 2θ cos 2θ

�

�cos θ − sin θsin θ cos θ

�

Vertical PolarizerHorizontal Polarizer

Polarizer at θQWP (fast axis horizontal)

QWP at θHWP (fast axis horizontal)

HWP at θCoordinate system rotation by θ

For an input field amplitude described byup to a phase factor the Jones matrices are


6.

Amplitude Modulator Example

Consider a pair of crossed polarizers with a variable waveplate in between with fast axis at 45° with respect to the polarizers. What is the transmission of this device for light which passes the inital polarizer as a function of the waveplate retardation?

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6.

Amplitude Modulator Example

640 0.25! 0.5! 0.75! !

0.25

0.5

0.75

1

P out/P

in

The compound Jones matrix is found from P(0°) R(θ) VWP(φ) R(-θ) P(90°)

�Eout,x

Eout,y

�=

�1 00 0

� �cos θ − sin θsin θ cos θ

� �1 00 eiφ

� �cos θ sin θ− sin θ cos θ

� �0 00 1

� �Ein,x

Ein,y

�

=�ei(φ−π)/2 sin(φ/2) sin(2θ) 0

0 0

� �Ein,x

Ein,y

�

Iout = E∗out,xEout,x + E∗

out,yEout,y = sin2(φ/2) sin2(2θ)Iin,y

For θ=45° we get Iout

Iin,y= sin2(φ/2)

φThursday, September 2, 2010

6.

Liquid Crystals

A state of matter containing with oblong molecules having some crystal-like properties and some liquid like properties.

When the molecular orientation is ordered the liquid is birefringent like a crystal.

Electric fields can cause adiabatic changes to the orientation direction allowing light polarization to be controlled by the applied electric field.

65


6.

Liquid Crystals

3 types of liquid crystal

Nematic - ordered orientation, random positions

smectic - ordered orientation, and ordered position in 1 dimension

Cholestric - like Nematic but with a helical rotation of the alignmnet direction

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6.

Twisted Neamtic Liquid Crystals

Nematic liquid crystal orientation is determined by boundary conditions. When placed between polished glass windows, the orientation on either side will align to the polishing direction resulting in a twisted structure.

A twisted nematic liquid crystal can be defined by the retardation coefficient of the nematic liquid crystal β=(ne-no)k0, and the twist coefficient α=dθ/dz

If β>>α the liquid crystal will guide the polarization of light that propagating along the helical axis.

67


6.


Treating the liquid crystal of thickness d as a series of N thin layers of thickness Δz=d/N, birefringence βΔz, and the angle of the fast axis of the mth layer is θm=mΔθ, where Δθ=αΔz. The Jones matrix for the mth layer is

For the entire crystal the Jones matrix is

which can be written as

68

TM = R(−θm)�1 00 eiβ∆z

�R(θm)

T =1�

M=N

TM

T = R(−θ)��

1 00 eiβ∆z

�R(∆θ)

�NR(∆θ)

T = R(−θ)�1 00 eiβ∆z

�R(θ)R(−(N − 1)∆θ)

�1 00 eiβ∆z

�R((N − 1)∆θ) . . . R(−∆θ)

�1 00 eiβ∆z

�R(∆θ)


6.


And with Δθ=αΔz and θ=αd this is

for α<<β the number of layers N can be chosen such that R(αΔz)≈I while still retaining the effect of birefringence so that

thus the twisted nematic liquid crystal adds a phase delay of βd to the slow axis and rotates the polarization by αd.

69

T = R(−θ)��

1 00 eiβ∆z

�R(∆θ)

�N

T = R(−αd)��

1 00 eiβ∆z

�R(α∆z)

�N

T = R(−αd)�1 00 eiβd

�


6.

Summary

Material polarization is not necessarily parallel to electric field in anisotropic materials

permittivity of material (ε) is a tensor relating E to D

Direction of wave vector and energy flow can be different

Normal shells and index ellipsoid are geometrical constructions to visualize the propagation parameters of a wave

Jones calculus can be used to keep track of polarization through an optic or a series of optics

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6.

References

Saleh and Teich Ch 5.2B, 6.1B, 6.3-6.4

Yariv & Yeh “Optical Waves in Crystals” chapter 4

Hecht “Optics” section 8.4

Fowles “Modern Optics” section 6.7

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ch 6-optics of ani so tropic media

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