Polarizers θ

Polarizers
Actual polarizers. Irradiance is
independent of the rotation angle θ
for the conversion of natural light
(unpolarized) to linear polarization.
In the figure below, only the
component E01cosθ is transmitted
I(θ) = I(0)cos2θ, which is known as
Malus’s Law.
If Iu = I (natural or unpolarized light),
then I(0) = Iu <cos2θ>t = Iu/2.
Dichroism –selective absorption
of one of two orthogonal E
components
Dichroic polarizers are not
suitable for high intensity light,
because of high joule heat due to
absorption
1. Absorption of E-field in the ydirection causes e’s to flow.
2. Re-radiation of waves that
cancel incident waves polarized in
the y-direction.
This results in transmission of
waves with E-fields perpendicular
to the wires (i.e., along the xdirection).
Polaroid sheet (H-Sheet), most commonly used linear polarizer.
Contains a molecular analogue of the wire grid.
1. Sheet of clear polyvinyl alcohol is heated and stretched.
2. Then it is dipped in an ink solution rich in Iodine.
3. Iodine is incorporated into straight long-chain polymeric molecules
allowing electron conduction along the chain, simulating a metal wire.
HN-50 is the designation of a hypothetical, ideal H-sheet that transmits 50%
of the incident natural light while absorbing the other 50%. In practice,
about 4% of the light will be reflected at each surface leaving a maximum
transmittance of 92% for linearly polarized light incident on the sheet.
Thus, HN-46 would transmit 46% of incident natural light, and might be the
optimal polarizer. In general, for HN-x, the irradiance of polarized light
transmitted would be I=Io(x/50), where Io is the irradiance for the ideal case.
In practice, it is possible to purchase HN-38, HN-32, and HN-22 in large
quantities for reasonable prices, each differing in the amount of iodine
present.
Tourmaline
Crystal System: Hexagonal (trigonal)
Habit: As well-formed, elongate, trigonal prisms, with
smaller, second order prism faces on the corners. Prism
faces are often striated parallel to direction of elongation
(c-axis). The rounded triangular cross-sectional shape of
tourmaline crystals is diagnostic; no other gem mineral has
such a shape.
Hardness: 7-7.5
Cleavage: none
High birefringence (Two differenct indices of
refraction)
Strong Dichroism
Any transparent gem having a mean R.I. of 1.63 and a
birefringence of 0.015-0.020 is tourmaline.
Tourmaline is widespread in metamorphic, igneous and
sedimentary rocks. Gem Elbaite is, however, nearly
restricted to pegmatites. Literally thousands of tourmalinebearing pegmatites are known; only a few hundred
apparently contain gem quality material in mineable
quantities.
Found in Brazil, Sri Lanka, U.S., Southern California
Tourmaline Boron Silicates)
Chemical
Formula
Specific
Gravity
XY Al B Si OH
(X
Y
Na or Ca)
Hardness
Mg, Li, Al or Fe
Refractive
Index
-
1) There is a specific direction within
the crystal known as the principal or
optic axis.
2) The E-field component of an
incident wave that is perpendicular to
the optic axis is strongly absorbed.
3) The thicker the crystal the more
complete will be the absorption.
4) A plate cut from a tourmaline
crystal parallel to its principle axis
and several mm thick will serve as a
linear polarizer.
5) Absorption depends on λ.
6) Advantages over H-sheet
polarizers with regard to maximum
irradiance permitted and can be used
with high power lasers.
Birefringence: Anisotropies in the index of refraction
Arrangement of atoms in a crystal can lead to both a structural asymmetry and
an anisotropy in the optical properties, as we just saw with tourmaline, which is
a dichroic crystal. The speed of the E-M wave depends on the refractive index n,
and therefore on the difference between the frequency of the E-field (ω) and the
natural frequency (ωo) of the atom. ωo depends on the binding forces attracting
the valence electrons to the atom.
2
qe N
1
n (ω ) = 1 +
ε o me ωo2 − ω 2
2
Consider a mechanical model for the atom in
which the spring stiffness, k, is directiondependent, and such that kx > ky = kz, where F
= kix. The resonant frequencies for oscillation
depend on the direction of polarization:
E || yˆ
ω0 y = k y / me
E || xˆ
ω0 x = k x / me
kx > k y
ω0 x > ω0 y
As a result of the anisotropy in ωo, the refractive Light with frequency ωd will appear
index will depend on the polarization direction. in the absorption band for the y
polarization and will be transparent
for the x polarization. This is an
(a)
example of a dichroic crystal which
will allow transmission of light E || xˆ
while strongly absorbing light with.
E || yˆ
2
E || yˆ
q N
1
n 2 (ω ) = 1 + e
ε o me ωo2 − ω 2
(b)
ωox > ωoy
E || xˆ
n y (ωb ) > nx (ω b )
The birefringent case is for ω = ωb in
which absorption is negligible for both
polarizations, but the index of refraction
depends on the polarization.
In the simple mechanical model, the spring stiffness
is such that cx > cy = cz and thus the x-axis is the
direction of the optic axis.
Optic axis
x
Optic axis
E
z
y
Plane ⊥ optic axis
Notice that for polarizations with E in a
plane normal to the optic axis, the index
of refraction n is independent of the
polarization orientation.
Let’s consider now calcite (calcium
carbonate) CaCO3 which is an
important birefringent material. It is
also the most common material for
making linear polarizers for high
power lasers.
NaCl Sodium Chloride
fcc – Face-centered Cubic
Calcite is the third most common mineral in
Earth’s crust (behind feldspar and quartz).
Because of its abundance, calcite can be found
in many rock types.
A Calcite in Sedimentary Rocks
Calcite or Calcium carbonate
(CaCO3) --- Rhombohedral unit cell
Unit Cell
The sedimentary rock limestone makes up
approximately 10 percent of all sedimentary
rock. Limestone consists almost entirely of
calcite. It forms because many marine
organisms make shells and other hard body
parts out of calcium carbonate. When the
organisms die, their shells or hard body parts
settle to the ocean floor. Through time, they
can accumulate into great thicknesses of
calcite mud. This mud, if turned into rock,
becomes limestone. Chalk, a form of
limestone, consists of the calcium carbonate
remains of innumerable microscopic floating
marine organisms, such as foraminifera and
coccolithophores.
From MSN Encarta
Calcite or Calcium carbonate (CaCO3)
3-fold symmetry
CO3 carbonate
groups are all in
planes normal to
the Optic axis.
Large
Birefringence
The birefringent property of calcite leads to
the formation of two images as shown in
these examples. The images are related to the
existence of ordinary rays (o-rays) and
extraordinary rays (e-rays). An analysis of
these rays shows that both these rays are
linearly polarized. We will describe the
nature of these rays in detail in the coming
slides.
Colorless Calcite Rhombohedron
with a long edge of ~12 cm.
It is possible to cleave calcite and form sharp faces that create a cleavage form
(rhombohedron) as shown below possessing faces of a parallelogram with
angles of 78.08° and 101.92°. There are only two blunt (not sharp) corners
(labeled A and B) where the surface planes meet. A line passing through the
vertex of each of these blunt corners and oriented so that it makes equal angles
with each face (45.5°) and each edge (63.8°) is clearly an axis of 3-fold
symmetry.
The 3-fold axis is related to the
3-fold symmetry of the CO3
carbonate groups shown
previously and the line
representing this axis must be
then parallel to the optic axis of
the crystal, as shown.
B
A
Principal Plane - Contains the optic axis. There are many planes that
can be drawn in this manner, as shown.
Principal Section – A principal plane that is normal to the pair of
opposite surfaces of the cleavage form (rhombohedron).
E
E
E||
When E lies in the principal
section, we can express the sum in
terms of ⊥ and || components, with
respect to the optic axis.
Optic Axis
The principal section is also a
parallelogram, with angles as shown.
Note that the ray direction is altered
(despite the normal incidence on the
section) when the electric field contains
non-zero components that are both
parallel and perpendicular to the optic
axis. This phenomenon is a result of
different propagation velocities, v|| and
v⊥, associated with each electric field
component. We will see that we can
express the propagation velocities, as
follows:
v|| = c/ne
ne = 1.486
v = c/no no = 1.658
The o-wave, with its
perpendicular polarization,
exhibits a single propagation
velocity, v⊥. The wave stimulates
numerable atoms at the surface
producing a source of radiating
spherical wavelets, the summation
of which leads to a plane wave
propagation as in the case of an
isotropic medium like glass.
Rays in the center calcite crystal contain
both || and ⊥ E-field components, leading to
the characteristic double image of the line.
KCl and NaCl are isotropic cubic crystals
and are thus do not exhibit birefringence.
Light emitting from a point source will be
isotropically emitted in all directions leading
to a spherical wave.
In the above figure, E lies in the principal
section, defining the e-wave, and E = E|| + E⊥,
where E|| || Optic-axis. Each component will
propagate with velocities, v|| and v⊥, respectively.
The result is that a point at the interface emits
waves that are elongated into an ellipsoid of
revolution rather than a spherical shape.
v|| = c/ne
ne = 1.486
v = c/no no = 1.658
The anisotropic propagation v|| > v⊥ causes a distortion in the wavefront and
changes the propagation (Poynting vector) direction indicated by S in the figure
for the e-wave. Remember that S = v2εE×B, and represents the direction in
which the irradiance of the wave propagates and thus defines the ray direction.
Note that solutions to Maxwell’s equations give: k ⊥D and S⊥E. For e-waves,
S||k only in propagation directions || or ⊥ to the optic-axis. For o-waves, E||D
and S||k for all propagation directions.
All crystals having symmetries that are hexagonal, tetragonal, and trigonal are
optically anisotropic and will lead to birefringence. In such crystals, an optic
axis exists and about which the atoms are arranged symmetrically. Crystals
possessing only one such optic axis are known as uniaxial. As mentioned, cubic
crystals like NaCl and KCl are symmetric, do not possess an optic axis, and do
not exhibit birefringence.
The difference ∆n = ne – no is a measure of the birefringence. For calcite ∆n =
1.486 – 1.658 = -0.172, v|| > v⊥, and the crystal is referred to as negative
uniaxial.
Consider hypothetical point sources of natural light embedded within negative
and positive uniaxial crystals, as shown in the left and right figures.
The shape of the ellipsoids depends on sign of ∆n (+ or -) as shown.
Thin layer of balsam cement
with n = 1.55
Other Crystallographic systems: Orthorhombic,
monoclinic, and triclinic have two optic axes and
are biaxial.
For example, Mica KH2Al3(SiO4)3 has three
different indices n.
Birefringent devices – Separation
of the o- and e- rays.
The Nicol Prism, a Birefringent
Polarizer.
Thin layer of
balsam cement
with n = 1.55
For calcite, again, ne = 1.486, no = 1.658
A calcite crystal that is cut,
polished, and painted,
separates the o-ray and e-ray
via TIR (total internal
reflection). A thin layer of
balsam glues two halves of
the crystal. Balsam has an
index of refraction, nb,
which is between that of the
o- and e-rays, i.e., ne < nb <
no. Thus, the o-ray
experiences TIR at the
balsam interface and is
absorbed by the layer of
black paint on the side. The
e-ray refracts normally at the
balsam interface an leaves
the crystal at the bottom.
Therefore, the emitted ray
can be used as a fully
linearly polarized beam.
The Wollaston Prism – Polarizing Beam-splitter
A polarizing beam-splitter
passes both orthogonally
polarized components that
can be separated. The oray becomes the e-ray and
vice-versa as the rays
traverse from the first to
the second section.
Interfaces are polished
flat and optically smooth
to allow rays to refract at
the interfaces.
For calcite, again, ne = 1.486, no = 1.658
Birefringent Polarizer
(mainly for high-power
lasers) where we can’t
use polaroid sheet
since it will melt.
o-ray absorping paint
e-ray
no sin θ o = nair sin θ r
sin θ cr −o = 1 / no = 1 / 1.658
θ cr −o = 37.1°
sin θ cr −e = 1 / ne = 1 / 1.486
Air gap
θ cr −e = 42.3°
Figure 8.27 The Glan-Foucault prism
Calcite-air interface: θcr-o < θ < θcr-e in
order for the o-ray to experience TIR
The e-ray will not experience TIR.
This prism is similar to the
Nicol, prism but without the
use of balsam cement.
Polarization by Reflection
Fig. 8.32 At θp , the reflected beam is linearly polarized ⊥ to the plane of incidence.
The transmitted beam, however, is strongly polarized || to the plane of incidence and
weakly polarized ⊥ to the plane of incidence, i.e., it is partially polarized.
Polarization by reflection and derivation of Brewster’s angle is found
by examining the Fresnel equations:
sin(θ i − θ t )
r⊥ = −
sin(θ i + θ t )
2 sin θ t cos θ i
t⊥ =
sin(θ i + θ t )
and
tan(θ i − θ t )
,
r|| =
tan(θ i + θ t )
and
2 sin θ t cos θ i
t|| =
sin(θ i + θ t ) cos(θ i − θ t )
tan 2 (θ i − θ t )
R|| = r =
→0
2
tan (θ i + θ t )
2
||
cos 2 (θ i + θ t ) = 0
θ i + θ t = 90°
Thus, the reflected wave becomes entirely polarized ⊥ plane of incidence
(and || to the surface) for θi = θr = θp (polarization or Brewster’s angle) when
θi + θt = 90°.
θ p + θ t = 90° and ni sin θ p = nt sin θ t = nt sin (90° − θ p ) = nt cos θ p
tan θ p = nt / ni
Brewster’s law
Different approaches for using
Brewster’s law to obtain
polarized light by reflecting
from multiple surfaces or thin
films.
Polarizing sunglasses take advantage of the fact that in nature "glare" consists
mainly of light-having a horizontal vibration direction. The reason for this is
that sunlight comes downward and hits mostly horizontal surfaces (such as the
oceans). Reflection from such a surface results in polarization with a
predominant horizontal vibration (i.e, normal to the plane of incidence). The
sketch below shows that in polarizing sunglasses the transmission axis is
vertical and thus reduces the glare reflected from horizontal surface.
Transmission
axis is vertical
Consider a beam of unpolarized
natural light striking a surface. If
the incoming light is unpolarized,
then
I i|| = I i ⊥ = I i / 2
I r|| = R|| I i / 2 and
R=
I r || + I r ⊥
Ii
=
I r ⊥ = R⊥ I i / 2
R|| + R⊥
2
,
where the subscripts i and r refer
to incident and reflected beams.
For a beam containing varying amounts of polarized and unpolarized
components, we can introduce a concept called “Degree of polarization”
V:
Ip
I max − I min
=
, where I P = I max − I min
V=
I max + I min I p + I n
and
I n = 2 I min
For polarized light – V=1
For natural light – V=0
Imax and Imin are the maximum and minimum intensities obtained by rotating
the linear polarizer relative to the detector. Likewise Ip and In are the
polarized and unpolarized contributions, respectively, of the beam.
(Ip + In)
Mixed polarization
from a source
Rotatable linear polarizer
Light
detector
Retarders: Methods for changing the polarization state of a beam
Cut and polish a calcite crystal so that its optic axis will be ⊥ to both the front and
back surfaces, as shown below. o- and e-waves propagate without any relative
changes in phases, i.e., the components with all polarization orientations will
traverse the distance AB in the same amount of time and have the same speed.
Suppose now that we cut the crystal so that the optic axis is || to the front and back
surfaces, as shown below. Since no > ne we have v|| > v⊥ and the e-wave reaches the
opposite face before the o-wave.
Difference in optical path length, OPD, Λ = d| no – ne|; ∆ϕ = koΛ = (2π/λo) d| no – ne|
λo is the vacuum
wavelength.
The final polarization state
depends on the amplitudes
of the o- and e-waves and
on ∆ϕ.
Note that the e-wave will
have a higher speed in a
negative uniaxial material,
i.e., v|| > v⊥. The situation
is reversed in a positive
uniaxial material.
A half-wave plate introduces a relative phase shift ∆ϕ =
π between the o- and e-waves, as shown below.
Therefore this phase-shift condition requires
∆ϕ =
2π
λo
d=
d no − ne = (2m + 1)π where m = 0, 1, 2, 3, ...
(2m + 1)λo
2 no − ne
Before striking the plate, we have a linearly
polarized beam of the form
E = (E0 x xˆ + E0 y yˆ )cos ωt
After reaching the second face:
E = E0 x cos(ωt + ∆ϕ ) xˆ + E0 y cos(ωt ) yˆ
Since ∆ϕ = (2m + 1)π
E = (− E0 x xˆ + E0 y yˆ )cos ωt
Therefore, the E-field vector is rotated
2θ about the y-axis.
θ = tan −1
E0 x
E0 y
Half-wave plates are sometimes called polarization rotaters. Note that these
plates will also change the handedness of circularly or elliptically polarized
light by introducing a ∆ϕ = π (relative phase) between the e- and o-waves:
[
]
(Right circularly polarized light)
E = E0 iˆ cos(kz − ωt ) + ˆj sin (kz − ωt )
E = E0 − iˆ cos(kz − ωt ) + ˆj sin (kz − ωt ) (Left circularly polarized light)
[
Note that elliptical light
will similarly be flipped
by an angle θ.
]
Quarter Wave Plate: Linearly polarized → Circularly polarized and vice versa
∆ϕ OPD Λ d no − ne
=
=
=
2π
λ0
λ0
λ0
∆ϕ =
2π
λ0
d no − ne = (4m + 1)
(4m + 1)λ0
d=
4 no −n e
R
π
2
m = 0, 1, 2, ..
Point of observation
is from behind the
quarter-wave plate
and so the eye sees a
Right-circular wave.
Since the angle is θ = 45°, Eox = Eoy.
for the linearly polarized beam
striking the quarter-wave plate.
Quarter-wave plate converts linearly polarized light to circularly polarized light
Right circular (R )
Linearly polarized light, oriented at 45° to the direction of the crystalline optic axis, is treated as
an x-component and a y-component, each equal to the other. Upon entering the birefringent
crystal, the y-component, aligned with the fast axis travels through the crystal faster than does
the x-component, which is aligned with the slow axis The two components are thrown 90° out of
phase with each other, such that upon emerging, one is always maximum while the other is
always zero and vice versa. The effect is to produce right circularly polarized light, as seen by an
observer looking into the crystal. The 90° phase shift is produced by a precise thickness d of the
birefringent crystal, which, because of the 90° shift, is referred to as a quarter wave or λ/4 plate.
Quarter Wave Plate: Linearly polarized → Circularly polarized and vice versa
Again we add the phase ∆ϕ to the x-component. If the fast axis is vertical along
y, then the x-component (o-wave) will lag behind the y-component (e-wave),
causing ∆ϕ = π/2.
v|| = c/ne ne = 1.486; v = c/no no = 1.658
E = E0 x cos(ωt − ∆ϕ ) xˆ + E0 y cos(ωt ) yˆ
and
E0 x = E0 y
Since ∆ϕ = (4m + 1)π / 2
E = E0 (xˆ sin(ωt ) + yˆ cos(ωt ) )
E0
If the initial linear polarization had
been rotated 90° (i.e., θ = -45°), then
the resulting polarization state would
have been left circular polarization.
Therefore, left and right circular can be
obtained by simply performing a 90°
rotation relative to the fast axis.
ω
Right circular polarization, i.e.
an R – state.
On the other side of the plate, we again examine the wave at a point where the fast-polarized
component is at maximum. At this point, the slow-polarized component will be passing
through zero, since it has been retarded by a quarter-wave or 90° in phase. If we move an
eighth wavelength farther, we will note that the two are the same magnitude, but the fast
component is decreasing and the slow component is increasing. Moving another eighth
wave, we find the slow component is at maximum and the fast component is zero. If we
trace the tip of the total electric vector, we find it traces out a helix, with a period of just one
wavelength. This describes circularly polarized light; right-circular light is shown in the
figure below. You may produce left-hand polarized light by rotating either the wave plate or
the plane of polarization of the incident light 90° in the figure below.
Right circular polarization
(R – state)
y
x
z
A quarter wave plate can similarly be used to transform circularly polarized
light into linearly polarized light. Again, care must be taken in order to
determine the resulting orientation of the linear polarization. The example
below shows how two quarter wave plates will act like a half-wave plate in
rotating the final polarization vector by 90° for linear polarization.
Unpolarized
source
Fast
axis
Left circular (L )
Rotation of 90°
Note that the
rotation of linear
polarizer D by 90°
will filter the light,
yielding zero
transmission. The
result is that the
combination of a
quarter-wave plate
and a linear
polarizer can be
used both as a
circular polarizer
(A+B) and
analyzer (C+D)
when used in
reversed
orientation.
90% In2O3-10% SnO2
!
"
!
!
"!
"
"
#
"
!
!
!
!
"
!
"
The polarization state of a light beam
As we saw earlier the degree of polarization is defined by:
Ip
I max − I min
=
, where I P = I max − I min
V=
I max + I min I p + I n
and
I n = 2 I min
V=1 For polarized light and V=0 for natural light
V can be measured by one linear polarizer.
The different states of polarizations can be analyzed by measuring
the beam intensity through several polarizers as will be described below:
The Stokes parameters, a description of the polarization state of light using
four parameters.
Begin with a set of four filters. Under natural illumination, each will transmit
half of the incident light. Use the following description for the filters, although
the choice is not unique:
Filter 1: Isotropic and passes all polarization states equally.
Filter 2: Ideal linear polarizer which transmits horizontal polarizations.
Filter 3: Ideal linear polarizer which transmits polarizations with θ = 45° (1st
and 3rd quadrants).
Filter 4: Circular polarizer filter that is opaque to left-circular light (see
previous slide).
Each of these filters is then placed alone in the path of a beam and the
transmitted irradiances I0, I1, I2, I3 are respectively measured with a detector for
each filter.
Define the Stokes parameters by the following relations:
S0 = 2I0,
S1 = 2I1 – 2I0,
S2 = 2I2 - 2I0,
and
S3 = 2I3 - 2I0
S0 – Reflects the incident irradiance, regardless of the beam polarization
S1 – Reflects the tendency for the polarization to be linear/horizontal (S1 > 0) or
linear/vertical (S1 < 0) or neither (S1 = 0).
S2 - Reflects the tendency for the polarization to be linear/with θ = 45° (S2 > 0)
or linear/with θ = -45°(S2 < 0) or neither (S2 = 0).
S3 – Reflects the tendency of the beam for being right circular (S3 > 0), left
circular (S3 < 0), or neither (S3 = 0).
Consider the expressions for quasi-monochromatic light:
[(
)
E x (t ) = iˆE0 x (t ) cos k z − ωt + ε x (t )
]
and
[(
]
)
E y (t ) = ˆjE0 y (t ) cos k z − ωt + ε y (t ) ,
where E (t ) = E x (t ) + E y (t )
The Stokes parameters can be expressed in terms of the E-M wave
parameters by time averages as
S 0 = E02x
T
+ E02y , S1 = E02x
T
T
− E02y , S 2 = 2 E0 x E0 y cos ε , and
S3 = 2 E0 x E0 y sin ε , where ε = ε y − ε x
T
T
T
(Note that we are dropping the constant
ε0c/2 for convenience.)
If the beam is unpolarized, then
E02x
T
= E02y
cos ε (t )
T
T
, S 0 = 2 E02x , S1 = S 2 = S3 = 0 because
= sin ε (t )
T
T
= 0.
Let us now normalize the Stokes parameters by dividing each one by S0.
Thus the set of Stokes parameters an incident beam of unit irradiance can
be expressed with 4 × 1 column vectors:
S0
1 S1
, which is
S0 S2
S3
1
0
0
0
for natural or unpolarized light
Representations of this vector for other polarization states are in the table.
For completely polarized light S 02 = S12 + S 22 + S 32 .
2
2
2
The degree of polarization is given by V = S1 + S 2 + S3 / S 0 .
If we have a superposition of two
quasi-monochromatic waves that
are incoherent, the resulting
Stokes parameters can be
determined by just adding the
vectors as follows:
S 0'
S 0''
S 0'+ S 0''
S1'
S1''
S1'+ S1''
S
S
'
2
'
3
+
S
S
'
'
2
'
'
3
=
S 2' + S 2''
S3'+ S3''
Suppose that a unit flux density vertical
P - state is added to an inchohrent L
- state with flux density 2, as follows:
1
−1
0
0
+
2
0
0
−2
=
3
−1
0
−2
The resultant wave is an ellipse of flux density 3, more vertical than
horizontal (S1 < 0), left-handed (S3 < 0) and has a degree of
polarization V = 1 + 4 / 3 = 5 / 3.
The Jones Vectors: Another representation of polarized light, being
applicable to only polarized waves. The Jones vector is written as
E x (t )
.
E=
E y (t )
In order to handle coherent waves, we will need to
preserve the phase information. We can write the
vector in complex form:
E0 x eiϕ x
~
E=
iϕ y .
E0 y e
Horizontal and vertical P - states are given by
E 0 x e iϕ x
~
Eh =
0
and
~
Ev =
0
iϕ
E0 y e y
~ ~ ~
with E = Eh + Ev
The sum of two coherent beams, as with the Stokes vector, is formed by the
sum of the corresponding components.
~
iϕ x 1
For example, if E0 x = E0 y , ϕ x = ϕ y , then E = E0 x e
1
which is a P - state at +45° (in the 1st quadrant), since the components have
equal amplitudes and ∆ϕ = 0.
In many applications it’s not necessary to know the exact amplitudes and
phases. Like the Stokes case, we can normalize the irradiance to unity,
which will result in a loss of phase information. This simplification leads to
1 1
E45 =
,
2 1
Eh =
1
0
, and
Ev =
0
1
.
Similarly, we can express right circular light as
~
ER =
1
E 0 x e iϕ x
iϕ x
iϕ x 1
= E0 x e
= E0 x e
− iπ / 2
i (ϕ x −π / 2 )
e
−i
E0 x e
Therefore, the normalized complex Jones vectors for circular polarization are
1 1
~
ER =
2 −i
1 1
~
.
EL =
2 i
and
Note that
1 1+1
2 1
~
~
ER + EL =
=
2 −i +i
2 0
The last sum is a horizontal P - state having an amplitude twice that of an Ex or
Ey component in either circular wave.
The orthogonality of two vectors, both real and complex, are defined as follows:
~ ~
A ⋅ B = 0 for real vectors and A ⋅ B * = 0 for complex vectors and
~ ~
~ ~
note that ER ⋅ EL* = (1)(1)* + (−i )(i )* / 2 = 0, and Eh ⋅ Ev* = (1)(0)* + (0)(1)* = 0 ,
~ ~
~ ~
also note that
ER ⋅ ER* = EL ⋅ EL* = 1.
[
]
[
]
Any polarization state can therefore be described as a linear combination of the
vectors in either one of the orthonormal sets. These ideas are analogous to those
developed in quantum mechanics.
Matrix methods for determination of the polarization state involving
transmission of waves through various optical elements
Let’s introduce 2 × 2 matrix A (the Jones matrix) to calculate the
transformation of an incident polarized beam on an optical element into a
final polarization state. That is we are going from polarization states i → t.
a11
~
~
Then Et = AEi , where A =
a21
a12
a22
The above table contains a brief listing of Jones matrices for various optical
elements. Suppose that we have a P - state at +45° passing through a quarter
wave plate whose fast axis is vertical (i.e., the y-direction). Then
0
~
iπ / 4 1
Et = e
0 −i
1 1 e iπ / 4 1 0 1 e iπ / 4 1
=
=
2 1
2 0 −i 1
2 −i
The resulting beam is right-circular (we can ignore here additional factors).
For a serious of optical elements, multiplication of the following matrices
gives
~
~
Then Et = An ... A3 A2 A1 Ei Note that the matrices do not commute
and order must be maintained.
Suppose that we have a P - state at +45° passing through two quarter wave
plates whose fast axis is vertical (i.e., the y-direction). Then
0 iπ / 4 1 0
~
iπ / 4 1
Et = e
e
0 −i
0 −i
1 1 e iπ / 2 1 0 1 e i π / 2 1
=
=
2 1
2 0 −1 1
2 −1
The transmitted beam is just a P - state at -45°, having been flipped 90° by a
two quarter wave plates acting as a half-wave plate. Note that just as we did for
lens matrices, we can calculate the 2 × 2 system matrix: A
= An ... A3 A2 A1
System
A similar method can be applied to beams that are best represented by the
Stokes vector. For example, there may be optical systems involving the
transmission of polarized and partially polarized light. This method
involves Mueller matrices, as listed in the table. Note that the Jones method
can deal with polarized waves whereas the Stokes/Mueller method deals
with polarized and unpolarized superposition of waves. The Mueller
matrices are used in much the same way as the Jones matrices.
For example, let’s pass a unit-irradiance unpolarized wave through a linear
horizontal polarizer. The Stokes vector of the transmitted wave is
1 1 0 0 1
1/ 2
1/ 2
1 1 1 0 0 0
St =
=
0
2 0 0 0 0 0
0 0 0 0 0
0
Thus, the transmitted wave has an irradiance of S0 = ½ and is linearly polarized
horizontally since S1 > 0.
As a second example, we take a partially polarized elliptical wave whose
Stokes parameters have been determined to be (S0, S1, S2, S3) = (4, 2, 0, 3). The
irradiance is 4. Since S1 > 0, it is more nearly horizontal than vertical. S3 > 0
implies that it is right handed. It has a degree of polarization V = (4+0+9)1/2/4 =
90%. Since S3 is not much less than S0, the ellipse resembles a circle.
Suppose that this beam now traverses a quarter-wave plate with a vertical
fast axis, then
1
0
St =
0
0
0
1
0
0
0 0
0 0
0 −1
1 0
4
4
2
2
=
−3
0
3
0
The transmitted beam has the same irradiance and degree of
polarization but is now partially linearly polarized since S2 < 0 and
reflects the tendency for the polarization to be linear with θ = -45°.
Again, using the 4 × 4 Mueller matrices, it is also possible to perform
calculations with a variety of optical elements leading to a system
matrix: A
= A ... A A A .
System
n
3
2
1