Chapter 4 Electromagnetic Propagation in Anisotropic Media

Lecture 1 Light propagation in anisotropic media

Introduction:1) The optical properties (e.g., refractive indices and absorption) of an anisotropic

medium remarkably depend on the propagation direction and the polarization state of the incident light, as well as the external forces (electric, acoustic and mechanic) exerted on the materials.

2) Anisotropic media exhibit many interesting and important phenomena, including birefringence, double refraction, conic refraction, optical rotation, Faraday effect, and electro-optical phenomena. These effects are employed to design and fabricate various optical devices, including polarizers, filters, beam splitters, rotaters, and many electro-optical devices. A significant portion of laser technology, and nonlinear optics deal with the generation and control of light using the optical properties of anisotropic crystals.

3) We therefore conclude that studying the propagation of light in anisotropic media is inevitable in understanding the physics of lasers and their application in the lab.

1

4.1 The dielectric tensor of an anisotropic mediumCrystals are made up of regular periodical arrays of molecules. In an anisotropic crystal, the polarization induced by an electric field is in general not in the electric field direction. This can be seen if we consider that the electrons are anisotropically bonded in the crystal.

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We currently make the following assumptions on the media:1)Homogeneous. The medium is identical if translated internally.2)Nonabsorption. is real. Complex permittivity tensors will be considered later when we study the optical activity of crystals.3)Linear. does not depend on the strength of the external electric field. Nonlinear polarization is the basis of nonlinear optics, which we learn only a little in this course. 4)Nonmagnetic. Our results can be easily extended to magnetically isotropic media.

3

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The dielectric tensor ij is symmetric in a nonabsorption medium.

This is an intrinsic symmetry. It comes from the nature of the thermodynamic requirement that Ue is a state function of E, which takes all Ei as independent variables. This does not require the symmetry of the crystal, or the linearity of the response.This can be generalized as follows.

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4

Question: For a given wave normal direction s in the crystal, what are the eigen refractive indices and eigen polarizations?This is answered by solving the Maxwell’s equations with an anisotropic dielectric tensor.Suppose the phase of all the electromagnetic fields (E, H, D, and B) varies in the form:

itc

ni

c

ntiti

,Then . where,expexp s

k

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c

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t

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t

Relations between the directions of the vectors: 1) D, H, and s are mutually perpendicular.2) D, E, s, and E×H (energy flow) lie in the same plane.3) The Poynting vector S=E×H is generally not along s.

k (s)

E

H (B)

D

S=E×H (t)

4.2 Plane wave propagation in anisotropic mediaIn an anisotropic medium, the phase velocity of light depends on its polarization state and its propagation direction. For a given propagation direction, there exist in general two eigenwaves, each has its own eigen refractive index (or equivalently eigen phase velocity ) and eigen polarization. All light traveling in that direction can be decomposed onto the two eigenwaves.

5

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6

Lecture 2 Eigenwave equation

We continue to search for the eigenwaves propagating along a given direction in a crystal.

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This is the eigenwave equation for determining the eigen refractive indices (eigen values) and polarization (eigen states) of a plane wave propagating in a prescribed direction s. We now realize it in the matrix form.

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7

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Question: For a given wave normal direction s in the crystal, what are the eigen refractive indices and eigen polarizations?Answer:1)In general two refractive indices, n1 and n2, are given by solving the Fresnel’s equation of wave normals.

2)The polarization direction of E of each eigen state is given by

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One easier way to solve the eigenwave equation is given below:

9

4.4 Phase velocity, group velocity, and energy velocity

The phase velocity of a plane wave is

Therefore in general each of the two eigenwaves in a crystal has its own phase velocity.Group velocity is the velocity of energy flow of a laser pulse in a dispersive medium.We now generalize the concept of group velocity, where the light pulse moves in a bundle of directions centered at k0. We decompose the light in the k-space (momentum space).

.)(

with ,c

nk

kp

sv

.,,)( velocityGroup

)(exp)(

)()(exp)()(exp),(

)()()()(

)(exp)(),(

00

000

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titf

dtiAti

dtiAt

kv

rkkrk

kkrkkrkktrE

kkkkk

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k

kk

kkk

kk

k

k

k

In k-space, (k) = const is called a wave normal surface. Group velocity is the gradient of the wave normal surface in k-space, and is therefore always perpendicular to the wave normal surface.

10

The energy velocity is defined as

.

21

BHDE

HESv

Ue

For a non-absorptive medium, vg= ve. A neat proof is given in our textbook.

(k) = const

normal surface

group velocity kz

ky

kx

k-space

11

Lecture 3 Wave normal surface in the k-space

We now express the Fresnel’s equation of wave normals in terms of k=(kx, ky, kz), that is, in the k-space. This is not mathematically new compared to what we have done by using n and s to express k. However, the k-space gives us more convenience.

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12

This determinate can be simplified into

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x

This is an (k) = constant wave normal surface. The surface is composed by all the wave normals k that have the same frequency . Note that the relative permittivities also depends on frequency. For future use, we simplify the equation for some special cases.

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13

Wave normal surface in the k-space:The wave normal surface (k) = constant contains the end points of all the k vectors for a given frequency . Here are the general characteristics of the wave normal surface.1)The wave normal surface consists of two shells, with only four points in common.2)The two lines that go through the origin and the four common points are called the optic axes. When light is propagating in the direction of one of the two optic axes, there is only one possible k value, and thus only one refractive index.3)Any other light propagating direction intersects the two shells of the wave normal surface at two different points, giving two possible k values, and thus two refractive indices for the two eigenwaves.4)For each eigenwave, the energy flow (group velocity or energy velocity) of the light is perpendicular to the wave normal surface at its k point.

Once wave vector k is known, the polarization of the corresponding E field of the eigenwave is given by

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/ :

/ ::

cnk

k

cnk

k

cnk

kEEE

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z

y

y

x

xzyx

14

15

16

17

Spatial relations between the fields of the two eigenwaves:

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,,

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sed

e

s

)()()(0)()(

)()(0

2121211221

2121211221

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DDEEDEDE

UUU e

Orthogonality of the two eigenwaves:

k (s)

E1

H1 (B1)

D1

S1(t1)

D2

H2 (B2)

E2

S2(t2)

18

4.3 The index ellipsoidWhen light propagates in a crystal, the D vectors of the two eigenwaves and the wave normal direction s form a mutually perpendicular triad. It is therefore more convenient to present the field vectors in the D space. The electric energy density Ue is given by

The constant energy density surface in D space is then

If we denote then we have

This surface is called the index ellipsoid (or the optical indicatrix).

.2

1

2

1

2

12

2

2

2

2

2

0

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x

x

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y

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D

n

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n

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),,(2/ 0 zyxU e rD .12

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zyx n

z

n

y

n

x

Lecture 4 The index ellipsoid

19

The role of the index ellipsoid:For a given arbitrary wave normal direction s, the index ellipsoid can be used to1)Find the indices of refraction of the two eigenwaves.2)Find the corresponding directions of the D vectors of the two eigen waves.

The prescription is as follows:1)Draw a plane that is through the origin and is perpendicular to s. This plane intersects the index ellipsoid surface with a particular intersection ellipse.2)The lengths of the two semiaxes of the intersection ellipse, n1 and n2, are the two indices of refraction of the eigenwaves.3)The two axes of the intersection ellipse are each parallel to the D vectors of the eigenwaves.

s

D1

D2n2

n1

20

Proof of the prescription of using the index ellipsoid:

The intersection ellipse is given by

The lengths (squared) of the semiaxes of the ellipse is given by the extrema of subject to the above two auxiliary conditions. This can be solved by

the Lagrange undetermined multipliers method, construct

The extrema of is then given by solving

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12

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2

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zyx

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DrDr

That is, if we assume the extrema of the intersection ellipse be , then the resultant E fields will be on the right polarizations, and the resultant n1 and n2 will satisfy the Fresnel’s equation of wave normals.

Dr // and mm nr

22

Comparison of the wave normal surface method and the index ellipsoid method

We have studied the wave normal surface method and the index ellipsoid method on light propagating in anisotropic crystals. Let us compare them here.

1)The wave normal surface shows us the allowed two wave vectors, and thus the refractive indices, at a given light propagation direction. The index ellipsoid does similar things, plus it also shows us the allowed direction of the D vectors.

2)The index ellipsoid displays the refractive indices and the directions of the D vectors in a convenient visual way. It also involves easier mathematics. Because of this simplicity, it is often first considered in solving problems.

3)However, the knowledge in the wave normal surface is much more profound. For example, it displays the optic axes. It shows the wave vector variation in the k-space. It shows the energy transfer direction. It has more mathematical base, and therefore allows for deeper theoretical derivations.

4)We tentatively summarize that for solving problems of light propagating in anisotropic crystals, we may first try the index ellipsoid method. However, we need to keep in mind that we have a backup more powerful wave normal surface method. This is especially true when we are directly dealing with the wave vectors rather than merely the refractive indices in the problems.

Lecture 5 Light propagation in uniaxial crystals

4.5 Classification of anisotropic media (crystals)Crystals are optically classified into 3 groups, namely the isotropic (or cubic) , the uniaxial, and the biaxial crystals, according to the number of independent elements of their dielectric tensors in the principle axis systems.Our text shows many examples of isotropic and anisotropic crystals commonly used in making optical devices.

23

Isotropic Uniaxial Biaxial

Dielectric tensor

Positive crystal when z>x, negative crystal when z<x.

Usually choosex<y<z

Examples NaCl, diamond Quartz (positive)Calcite (negative)BBO (Beta-Barium Borate, negative)

MicaTopaz

z

y

x

00

00

00

z

x

x

00

00

00

00

00

00

24

4.6 Light propagation in uniaxial crystalsThe index ellipsoid:The equation of the index ellipsoid of a uniaxial crystal is s

Do

De

no

ne)

z

y

The index ellipsoid is rotationally symmetric around the z-axis. Let s be in the y-z plane with a polar angle . The two polarization directions of the D vectors are: Do is parallel to the x-axis, De is in the y-z plane and is perpendicular to s.The corresponding refractive indices are:

.sincos

)(1sin)(cos)(

,2/1

2

2

2

2

2

2

2

2

eoe

e

e

o

e

oo

nnn

n

n

n

n

nn

spheroid oblatecrystal uniaxial negative :

spheroid prolatecrystal uniaxial positive :

index refractiveary extraordin ,/

index refractiveordinary ,//

1

0

00

2

2

2

22

oe

oe

ze

yxo

eo

nn

nn

n

n

n

z

n

yx

When s is on the z direction, ne(0°) = no. Therefore the z-axis is the optic axis.

25

The wave normal surface:

.0

toreduced is surface

normal wave then the, , If

2

2

2

2

2

2

2

2

2

22

cn

k

cn

k

n

kk

nnnnn

oo

z

e

yx

ezoyx

The refractive indices are given by solving the two factors:

.sincos

)(

,2/1

2

2

2

2

eoe

oo

nnn

nn

s

Do

De()

no

ne)

ne

no ck y /

ckz /

s

Do

De()

no

ne)

ne

no

ck y /

ckz /

Positive uniaxial crystal

Negative uniaxial crystal

Notes to uniaxial crystals:1)At a given propagation direction, there are in general two eigen refractive indices, each has its own eigen polarization direction. This is called birefringence.2)The E field of the o-ray is always polarized perpendicular to the plane that contains s and the optic axis, while the e-ray is polarized parallel to that plane.

26

4.7 Double refraction at a boundaryUp to now we discussed the light propagation in an anisotropic crystal when it is already inside the crystal, but how is the light refracted into the crystal?We recall that the boundary condition requires1)The wave vectors of the incident, reflected, and refracted light ( ) lie in the plane of incidence.2)The tangential components of all three wave vectors on the boundary interface is the same:We know that in the crystal the length of kt depends on the angle of refraction t , as well as the orientation of the optic axes. Each t supports two kt because of the double shell structure of the wave normal surface.

.sinsinsin 00 constkkk ttrr

tr kkk and ,,0

002211

002211

sinsinsin

or ,sinsinsin

nnn

constkkk

k2

boundary

k1

k0

0

22

11

00

sin

sin

sin

k

k

k

wave normal surfaces

Lecture 6 Double refraction

Also because of the double shell structure of the wave normal surface, in general there are two refraction angles, 1 and 2, both satisfy

27

The refraction at a boundary can be explained on the intersection between the wave normal surface and the plane of incidence.At the boundary, the k0 beam is uniquely decomposed into the reflected beam and the two eigenstates of the refracted k1 and k2 beams, according to the boundary conditions. Each refracted beam then propagates separately. This is called double refraction.

Please note that in general the polarization of the k1 and k2 beams are not orthogonal since they are in different directions. The incident light still can be uniquely decomposed into the reflected beam and the two refracted beams according to the boundary conditions.

k2

boundary

k1

k0

0

22

11

00

sin

sin

sin

k

k

k

wave normal surfaces

Internal double reflection:When light is internally reflected from the surface of an anisotropic material, double reflection may occur. It can be discussed similarly on the intersection between the wave normal surface and the plane of incidence.

28

Double refraction at the boundary of a uniaxial crystal: ExamplesThe directions of the D vectors are shown. There exist an ordinary wave with the refractive index no, and an extraordinary wave whose refractive index ne depends on its direction of propagation.

ke

optic axis

ko

ki

ke

optic axis

ko

ki

ke

optic axis

ko

ki

ke

optic axis

ko

ki

ke

optic axis

ko

ki

ke

optic axis

ko

ki

Positiveuniaxial crystal

Negativeuniaxial crystal

29

o e

tanarctan then is and between angle The

tan/cos

/sin tan

. anglepolar a has Suppose

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)/sin,/cos,0()sin,cos,0(

. lar toperpendicu is and plane - in the is

)0,0,/()0,0,(

2

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tt

t

EsEt

ED

stEsEt

ED

tanarctan then is and between angle The

tan/cos

/sin tan

. anglepolar a has Suppose

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)/sin,/cos,0()sin,cos,0(

. lar toperpendicu is and plane - in the is

)0,0,/()0,0,(

2

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ED

stEsEt

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30

Deviation angle between the energy flow of the o-ray and e-raysSuppose the light is incident normally on the surface of a uniaxial crystal. Then the wave vectors koand ke are in the same direction. Let us see how much the energy flow of the e-ray is deviated away from that of the o-ray.

s (to)

Do(Eo)

De

no

ne)

z

y

te

Ee

Example:For KDP (KH2PO4) crystal, no=1.50737, ne=1.46685. At =45°, the angle between the e-ray and the o-ray is =1.56°.

./)( with 45at occurs

deviation maximum theand ,1tan1

tan2 have we, toclose very is

2

ooe

o

eoe

nnn

n

nnn

If

31

4.8 Light propagation in biaxial crystalsThe index ellipsoid:The equation of the index ellipsoid of a biaxial crystal is

) (suppose 12

2

2

2

2

2

zyxzyx

nnnn

z

n

y

n

x

s

D1

D2

ny

n()=ny

z

x

Generally for a wave normal direction s (,) there exist two allowed vectors D1 and D2 with their specific refractive indices n1 (,) and n2 (,).

If s is in the x-z plane, we have n1()= ny always, and n x<n2 ()< nz. Since n x<ny< nz,there exists a special angle which makes n2 ()= n1 ()= ny. That is, the intersection ellipse is a circle and there is no difference between the two refractive indices. This special direction is called the optic axis of the crystal.For a biaxial crystal, there are two such optic axes, both are on the x-z plane. They are located symmetrically on each side of the z-axis. Thus comes the name biaxial crystal.

It is not difficult to find that the polar angle of the two optic axes is given by

.1111

tan22

22

2222yz

xy

x

z

zyyx nn

nn

n

n

nnnn

Lecture 7 Light propagation in biaxial crystals

32

When an unpolarized light is propagating along the optic axis of a biaxial crystal, the allowed D vector is in any direction perpendicular to the s vector. This results in the allowed E vectors locating in one plane. The ray direction t is then found to form a cone on one side of the light propagation direction s. Any plane perpendicular to s intersects the cone with a circle. This phenomenon is called conical refraction.Conical refraction is predicted by William Hamilton in 1832, and was confirmed experimentally by Humphrey Lloyd the next year. Both were Irish scientists.

. and , , , of plane in the is

),0,(00

),0,(

EtDsEt

EsD

s

zzxxzzzxxxzzxx

zx

sssEsEsDsD

ss

D1

D2

E3(D3)

E1

E2

t3

t2

t1 s

y

We discuss the details of conical refraction later using the wave normal surfaces.

33

Wave normal surfaces:We already know that the wave normal surface of a biaxial crystal consists of two complex shells, with only four points in common. Let us consider the intersections between the wave normal surface and the three coordinate planes. If we set ky=0, then

02

2

2

2

2

2

2

2

2

22

cn

k

n

k

cn

kk

x

z

z

x

y

zx

This consists of a circle with the radius ny/c, and an ellipse with semiaxes nz/c and nx/c. The intersections with the other two coordinates planes are similar, each has a circle and an ellipse. The optic axes lie in the x-z plane.

ckx /

ck y /

xn

zn

ynzn

ckx /

ckz /

xn

yn

ynzn

ck y /

ckz /

xn

yn

xnzn

34

Conical refraction: The group velocity of light is the gradient of the wave normal surface in the k-space, and is thus perpendicular to the wave normal surface. However, on the optic axis, the two shells of the wave normal surface degenerate into a point, which is a singular point where the gradient is not well defined. The energy flow of light at this point is governed by the nature of the singularity. We then need to examine the shape of the wave normal surface at that point. The wave vector at the singular point is given by

22

22

22

22

0

0

22

22

0

0000

tan

cos

0

sin

with,),,(

yz

xy

x

z

xz

yzxyz

y

xz

xyzyx

zyx

nn

nn

n

n

nn

nn

c

n

cnk

k

nn

nn

c

n

cnk

kkk

k

ckx /

ckz /

xn

yn

ynzn

k0

Lecture 8 Conical refraction

35

We are interested in the neighborhood of this singular point. We therefore do a Taylor expansion of the wave normal surface at this point.

,Let

0

0

0

zkk

ykk

xkk

zz

yy

xx

,0111111

4

4

2

2

222

222

222222

22

2

22

2

22

2

ccnnk

nnk

nnkkkk

nn

k

nn

k

nn

k

yxz

zxy

zyxzyx

yx

z

zx

y

zy

x

substitute it into the equation of the wave normal surface

and keep up to the second power of x, y, and z, we have(I confirmed it by Mathematica.)

.0))(())(4( 222220

20

200 zyxyzzxxzx nnnnyzknxknzkxk

This is a cone with its vertex at k0. The cone is symmetric about the y =0 plane.

ckx /

ckz /

xn

yn

ynzn

212

The energy flow is everywhere perpendicular to this cone surface, which form a light cone of

for conical refraction.

36

22

2222

22

22

2

2

02

02

2

0

01

12

))((2tan

tan

tantan

tantan

2

zx

yzxy

yz

xy

x

z

z

x

zz

xx

z

x

nn

nnnn

nn

nn

n

n

n

n

kn

kn

k

k

The cone intersects the y=0 plane by two lines, which make an angle 2 given by

We then rotate the x-z axes by an angle of /2−+ to the x'-z' axes. The cone will be erect in the new coordinates ( I confirmed it by Mathematica):

22222 tan'')tan1(' xyz

2

22

22

tan

''

)tan1(

1'

xyz

ckx /

ckz /

xn

yn

ynzn

2

'z'x

k0

2

37

22222 tan'')tan1(' xyz The light cone due to conical refraction

has the following characters:1)The cone contains the optic axis as 2)All other lights on the cone are above the optic axis (if the z axis is upward). 3)The cone has an apex angle of 2 in the y'=0 plane.4)The cone intersects any plane that is perpendicular to the optic axis with a circle.

ckx /

ckz /

xn

yn

ynzn

2

'z'x

k0

2

.tan'' xz

Proof of 4): We rotate x'-z' axes by an angle of – to the x"-z" axes. The cone will be

It is a circle at any x"=const.

0"2tan""" 22 yzxz

Here (2) is the second order nonlinear optical susceptibility .According to the theory of radiation, the complex field for the second harmonic that is generated by the crystal inside dx but observed at L is

Here a is just a constant.

Second harmonic generationWhen an intense laser beam passes through a crystal, the molecules of the crystal can be nonlinearly polarized, which causes the crystal radiate at the doubled frequency of the incident light. This is called second harmonic generation.Suppose the incident light has a frequency of . It passes through a crystal with thickness L in the x direction. Suppose the E-field of the incident light (which is called the fundamental beam) isWe suppose the conversion to second harmonic is small so that the amplitude of the fundamental beam is almost constant. We take care of the polarization issue later.The second order nonlinear polarization of the material is

38

Lecture 9 Phase-matching in second harmonic generation

.)(exp)(),,( 0 txkiEtxE

).,,(),,(),; 2(),,2( )2()2( txEtxEtxP

dxxLiktxEtxaEdxtLxe

xLikdxtxPdxtLxe

))(2(exp),,(),,(),,,2(

))(2(exp),,2(),,,2( )2(

x

L

dx

39

The total field of the second harmonic at the output surface of the crystal is then

)()2(4

)(2)2( Here

2exp2

)2()(2exp2/

2/sin)(

))(2()(2exp2exp)(

))(2(exp),,(),,(),,2(

20

0

20

0

nnkkk

tiL

kkikL

kLaLE

dxxLikxkitiaE

dxxLiktxEtxEatLE

L

L

is the phase mismatch.

The intensity of the produced second harmonic is

.2/

2/sin)(),,2()2( 2

22222

kL

kLILatLEI

For efficient second harmonic generation:

1)Conversion efficiency

2)Phase-matching conditionThis requirement is critical due to the narrow shape of the sinc function.

).()(

)2(

II

I

).()2( is that ,0 nnk

(I believe it should be called the wave-vector mismatch, and kL should be called the phase mismatch. This mistake was made by somebody many years ago.)

40

BBO (Beta-Barium Borate, β-BaB2O4) is a negative uniaxial optical crystal, i.e., ne< no.

The index ellipsoid is

with the z-axis as the optic axis.

1)()( 2

2

2

22

eo n

z

n

yx

2

1

2

2

2

2

22

2

2

2

)(

sin

)(

cos),(

),(

1

)(

sin

)(

cos

eoe

eeo nnn

nnn

Let be the direction of the light wave, the refractive index of the e light, ne(, ), is then

Phase-matching in BBO crystal

Phase-matching condition requires that . This is impossible if both the fundamental and the second harmonic are o-rays, or both are e-rays, because of the dispersion of the material. However, since ne< no, and normally , we

can let be o-ray and 2 be e-ray, and hopefully to accomplish the phase-matching condition. That is o + o e, which is called type-I phase-matching. For contrast, type-II phase matching refers to o + e o or e.

)()2( nn

)()2( ,, eoeo nn

41

To achieve type-I phase-matching in BBO, the phase-matching angle m, where k=0, is given by

.)2()2(

)2()(arcsin

)2(

sin

)2(

cos),2(

)(),2(

22

22

2

1

2

2

2

2

oe

oom

e

m

o

mme

ome

nn

nn

nnn

nn

This is not hard to calculate because we have the analytic Sellmeier equations for the refractive indices as a function of the wavelength or frequency. The Sellmeier equations of BBO are (in m):

no2() = 2.7359+0.01878/( 2-0.01822)-0.01354 2

ne2() = 2.3753+0.01224/( 2-0.01667)-0.01516 2

We then find that the BBO crystal can achievethe phase-matching condition for inputwavelengths down to a little more than 400 nm.

0.4 0.6 0.8 1m

20

40

60

80

mdegreePhase-matching angle

in BBO crystal

42

Angular sensitivity of phase-matching:

)(),2(4

),( oe nnk

....),(),(

mm

kkkk m

meo

o nnn

k

m

2sin)2(

1

)2(

1)(

222

3

1

223

2

2

2sin)2(

1

)2(

1)(

783.22

783.2783.22

1

2/

2/sin

meo

o nnn

L

Lk

kLkL

kL

m

This is only 0.11°for a BBO crystal of 1-mm long with an input beam at 600 nm. Therefore the phase-matching condition is extremely angular-sensitive.In practice the experimental angular width of phase matching can be larger, because 1) The angle is measured outside the crystal. 2) The laser has a bandwidth. 3) Intensity saturation. 4) A Gaussian beam has a divergence in its propagation directions.Similarly we can also calculate the wavelength sensitivity for phase-matching. It is found to be about 1 nm for 1-mm thick BBO crystal, which is again quite sensitive.

kL

I (2)

0.5

2.783

1.0

43

0.599 0.6 0.601 0.602m40.3

40.4

40.5

40.6

40.7

m, , degreekL=0

kL=±2.783

Angular and wavelength sensitivity of phase-matching in BBO crystal

44

4.9 Optical activityOptical activity refers to the phenomena that when a linearly polarized light is passing through a medium, the polarization plane is rotated. It is thus also called optical rotation. When this happens, the medium is said to be optically active. The rotation of the polarization plane is proportional to the path length of the light, and can be measured by degree/centimeter. Common optically active media include quartz, sugar and syrup. It can be used to measure blood sugar concentration in diabetic people.

Optical activity occurs in a chiral molecule, where the molecule does not overlap with its mirror image, and the electron cloud takes some kind of helical shape. Left- and right-circularly polarized light is different in polarizing a chiral molecule, which causes different refractive indices. This is called circular birefringence (or circular double refraction). Most amino acids (the building blocks of life) are left-handed, which makes optical activity prominent in nature.

Lecture 10 Optical activity

If when observed facing the light the rotation of the polarization plane is counterclockwise, the substance is dextrorotatory (right-handed). When the rotation is clockwise, it is called levorotatory (left-handed).

45

Explaining optical activity using circular birefringence:Suppose light is linearly polarized on the x-axis, and is propagating along the z-axis. Suppose the refractive indices for circularly polarized light is nr and nl.

c

znnti

c

znn

c

znnE

tic

znnii

c

znni

c

znni

c

znnE

tizc

niz

c

niiz

c

niz

c

ni

E

tizc

niiz

c

nii

E

tizc

niz

c

ni

Ezz

tiE

tiEz

rlrlrl

rlrlrlrl

rllr

lr

lr

2

)(exp

2

)(sinˆ

2

)(cosˆ

exp2

)(exp

2

)(sin2ˆ

2

)(exp

2

)(cos2ˆ

2

expexpexpˆexpexpˆ2

expexpˆˆexpˆˆ2

expexpˆexpˆ2

)( : distanceAt

expˆˆ2

expˆ)0( :0At

0

0

0

0

0

00

yx

yx

yx

yxyx

LRE

LRxE

The specific rotatory power is

which is very sensitive in measuring circular birefringence.

),(2

)(rl

rl nnc

nn

Now our original eigenwave equation is revised to

46

Theory on optical activity:A helical molecule can be polarized along its axis by a circulating current driven by an circular electric field produced by a time-varying magnetic field:The constitutive equation of a medium driven by a plane wave is then revised to

Gs=G is called the gyration vector. The length G of the gyration vector describes the rotation power of the medium. It varies with the direction of the wave, and can be expressed by a gyration tensor g of the medium as

HEp

sEHEsED ii and , because ,G0

.gG jiij ssg ss

0 G 02

0 Esss ni

0 02 Ess n

0

)/1(

)/1(

)/1(

22222

2222

2

22222

z

y

x

zzxyzyxz

xzyyyzxy

yzxzyxxx

E

E

E

nnssn

Gisss

n

Giss

sn

Gissnnss

n

Giss

sn

Gisss

n

Gissnns

After realizing it into the matrix form, we have

1

0

0

0

ssss

s

xy

xz

yz

ss

ss

ss

47

For a nontrivial solution of E, the determinant of the matrix must be 0, which is simplified to

(*1)

12222222

2222222

222

2

22

2

22

2

zyx

zzyyxx

z

z

y

y

x

x

nnnnnnn

snsnsnG

nnn

s

nn

s

nn

s

The corresponding wave normal surface is

2

2

222

2

222

2

222

2222226

2

22

22

2

22

22

2

22

22

2

1

zyx

zzyyxx

z

z

y

y

x

x

nc

knc

knc

kk

knknknc

G

nc

k

k

nc

k

k

nc

k

k

I believe this surface is formidable for anyone attempting to draw using Mathematica.

Fortunately equation 1 can be simplified to , where n1 and n2 are the solutions of the equation with G=0, i.e., when there is no optical activity.On the optic axis, we have the eigen refractive indices are then

222

221

2 Gnnnn

small. is when2

and ,22 Gn

GnnGnn

,21 nnn

These are two circularly polarize waves, with a rotary power of .n

G

48

The eigen polarization state in an optically active medium:If we solve equation 1 for the two general eigen refractive indices, and then substitute into the eigenwave equation, we may have the eigen polarization states for light in an optically active medium. Surely it is too complicated. Our textbook did it in a simple way. We can rewrite the eigenwave equation into that for the D vector, and express it in the (D1, D2, s) coordinate system, thus reducing one dimension. The corresponding eigen polarization, when G is small, and in terms of D, is

22

21

2

22

21

2

22

21

22

21

11

4

111

2

1

nn

iGnn

G

nnnnJ

We conclude that1)The eigen polarization states are generally two orthogonal elliptically polarized waves, oriented along the D1 and D2 axes.2)In an isotropic medium, or when the light is propagating in the optic axis of an anisotropic medium, the eigen polarizations are left- and right- circularly polarized light.3)In an anisotropic medium, and when the light is not close to the optic axis, because G is usually much smaller than , the light is almost linearly polarized.

22

21 nn

D1

D2

49

4.10 Faraday rotationFaraday rotation refers to the phenomena that when a linearly polarized light is passing through a medium placed in an external magnetic field, which is along the light propagation direction, the polarization plane is rotated.The origin of the effect is as follows. The electrons are moving in a molecule, driven by the electric field of the light. The external magnetic field will displace the electrons laterally due to the Lorentz force qv×B. The induced dipole momentum then includes a term proportional to E×B. The constitutive equation is then

Hereis the megnetogyration coefficient, B is the gyration vector. The discusses followed should be similar to optical activity, with a specific rotation given by , where V is called the Verdet constant, can be measured by deg/Gauss·mm.

EBED 0i

VB

One distinct difference between optical activity and Faraday effect is the reversibility of the effect when the light is going back. Optical activity is reversible, while in Faraday effect when the light is reversed the rotation is doubled. This is ready to explain by the two constitutive equations:

effectFaraday ,

activity optical ,G

0

0

EBED

EsED

i

i

B

d

kBk

Bk

50

Optical isolator:An optical isolator, or optical diode, is an optical device that allows the transmission of light in only one direction. It is used to prevent unwanted light feedback into a laser cavity. The operation of the device depends on the non-reversibility of Faraday effect.

The optical isolator in the figure consists of three parts. An input polarizer allows only vertically polarized light to pass. A Faraday rotator rotates the polarization by 45°. An output polarizer (analyser) just let the 45° polarized light to pass by. If light is feed back from somewhere later in the path, the Faraday rotator will turn it into horizontally polarized light, which is then absorbed (or deflected) by the first polarizer.