Download - Ch 6-Optics of Ani So Tropic Media
Document info 6.
EM Wave Propagation in Anisotropic Media
Chapter 6Physics 208, Electro-optics
Peter Beyersdorf
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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
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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?)
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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
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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.
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6.
Analogies
What every-day phenomena can have a response in a direction different than the driving force?
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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
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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
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�D = �0�I + χ
��E
Di = �0(δij + χij)Ej
Di =�
j
�0(δij + χij)Ej
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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
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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.
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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
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6.
Wave Equation in Anisotropic Materials
�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
kykx ω2µ�y − k2x − k2
z kykz
kzkx kzky ω2µ�z − k2x − k2
y
������= 0
This defines two surfaces in k-space called the normal shells.12
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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.
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6.
Wave Equation in Anisotropic Materials
�k × �k × �E + ω2µ� �E = 0
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
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
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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.
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6.
Wave Equation in Anisotropic Materials
�k × �k × �E + ω2µ� �E = 0
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
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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
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6.
Wave Equation in Anisotropic Materials
Ex
Ey
Ez
= E0
kxk2−ω2µ�x
ky
k2−ω2µ�ykz
k2−ω2µ�z
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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
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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
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6.
Spatial Walkoff
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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
→ →
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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
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�D1
�D2
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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)
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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)
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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
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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
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6.
Crystal PropertiesThe geometry of a crystal determines the form of the dielectric tensor
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6.
Crystal PropertiesThe geometry of a crystal determines the form of the dielectric tensor
Isotropic
� = �0
n2 0 00 n2 00 0 n2
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6.
Crystal PropertiesThe geometry of a crystal determines the form of the dielectric tensor
Unaxial
� = �0
n2
0 0 00 n2
0 00 0 n2
e
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6.
Crystal PropertiesThe geometry of a crystal determines the form of the dielectric tensor
Biaxial
� = �0
n2
x 0 00 n2
y 00 0 n2
z
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6.
Biaxial Crystals
By convention nx<ny<nz so the optical axis is always in the xz plane
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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.
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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)
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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
→ →
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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
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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
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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.
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n1 sin θ1 = n2 sin θ2
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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
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→
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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.
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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)
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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
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�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
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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
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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
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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
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6.
Exercise
44
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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
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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
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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
Thursday, September 2, 2010
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
Thursday, September 2, 2010
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
�
Thursday, September 2, 2010
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
��
Thursday, September 2, 2010
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
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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
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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
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�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
Thursday, September 2, 2010
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
�
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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
�E(z, t) = ei(kz−ωt)
�Ex
Ey
�
Thursday, September 2, 2010
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
�
Thursday, September 2, 2010
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
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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
�=
�cos(45o) sin(45o)− sin(45o) cos(45o)
� �1 00 i
� �cos(45o) − sin(45o)sin(45o) cos(45o)
� �Ex,in
0
�
�Ex,out
Ey,out
�=
1√2
�cos(45o) sin(45o)− sin(45o) cos(45o)
� �1 00 i
� �Ex,in
Ex,in
�
�Ex,out
Ey,out
�=
1√2
�cos(45o) sin(45o)− sin(45o) cos(45o)
� �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
Thursday, September 2, 2010
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
� �cos(45o) − sin(45o)sin(45o) cos(45o)
� �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
�
Thursday, September 2, 2010
6.
Compensator
Variable waveplate with Jones matrix
61
Babinet compensator Berek compensator
�1 00 ei∆φ
�
Thursday, September 2, 2010
�E(z, t) = ei(kz−ωt)
�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
Thursday, September 2, 2010
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?
63
Thursday, September 2, 2010
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
Thursday, September 2, 2010
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
66
Thursday, September 2, 2010
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
Thursday, September 2, 2010
6.
Twisted Neamtic Liquid Crystals
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)
Thursday, September 2, 2010
6.
Twisted Neamtic Liquid Crystals
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
�
Thursday, September 2, 2010
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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Thursday, September 2, 2010
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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Thursday, September 2, 2010