Reference on polarization: Iizuka

Reference on polarization: Iizuka
Elements of Photonics, Volume I: In Free Space and Special Media. Keigo Iizuka
Copyright  2002 John Wiley & Sons, Inc.
ISBNs: 0-471-83938-8 (Hardback); 0-471-22107-4 (Electronic)
6
POLARIZATION OF LIGHT
Clever uses for polarized light are not restricted to just the field of photonics. Devices
for manipulating polarized light can be found in a wide range of settings, from advanced
research laboratories to the common household [1,2]. Perhaps one of the most familiar
household polarizers is a pair of sunglasses, a necessity for many car drivers on a
sunny day. Sunglasses filter the light specularly reflected from a flat paved surface.
The reflected light from the flat pavement is predominantly horizontally polarized, so
that a polarizer with a vertical transmission axis rejects the specular reflection.
Another example of polarizing glasses are those worn for viewing a stereoscopic
motion picture. In this case, the transmission axis of the polarizer covering the right
eye is orthogonal to that of the polarizer for the left eye. Likewise, the motion picture
scenes for the right and left eyes are projected using orthogonally polarized light. The
right eye polarizer passes the light for the right eye scene and rejects the left eye scene.
Similarly, the left eye polarizer passes the left eye scene and rejects the right eye scene.
The viewer enjoys a stereoscopic picture.
While the polarizing glasses of the previous examples are normally constructed with
linearly polarizing material, antiglare screens frequently employ circularly polarizing
sheets. Displays, such as radar screens, use these circularly polarizing sheets to suppress
glare. Light that enters the circular polarizer, and subsequently undergoes reflection at
some other surface, is blocked from reemerging from the circular polarizer because of
the reversal of handedness, while the light generated by the screen passes through.
Scientists in many disciplines use polarized light as a tool for their investigations.
Physicists are still trying to unfold the mysteries of the invariance of the state of polarization of a photon before and after collisions with high-speed particles. Polarization
also presents puzzles such as: “What happens when only one photon polarized at 45°
with respect to the birefringent crystal axis enters the crystal?” Is the photon, which
is considered the smallest unit of light, further split into horizontally and vertically
polarized half-photons?
While puzzles such as these make polarization itself an interesting study, applications
of polarization devices and phenomena to other disciplines are extensive. Spectroscopists
362
INTRODUCTION
363
High definition 3D television.
use the Lyot–Ohman filter [1] made out of a combination of polarizers and retarders for
their work. With this filter, resolving powers as high as 0.01 nm can be achieved.
Astrophysicists study the pattern of magnetic fields in nebulae by mapping the
pattern of perturbation of the state of polarization of light from a nebula. These perturbations result from the Faraday rotation caused by the magnetic field of the nebula.
Many organic materials rotate the direction of light polarization as light passes
through them. Chemists use this fact for analyzing the structure of new organic
molecules. One of the most familiar examples is the determination of the sugar content
of a sugar solution by measuring the rotation of the polarization.
Mechanical engineers use the strain birefringence pattern of a plastic model as an
aid to strain analysis. Colorful strain patterns in the plastic model can be viewed under
a polariscope.
Biologists are certainly beneficiaries of polarization microscopes [3] that enable
them to observe microbes that are transparent and invisible under normal light. The
polarization microscope sees the pattern of the retardance that the microbes create.
Biologists also know that the direction of polarization of the illuminating light controls
the direction of growth of some fungi is used for navigation by certain animals such
as bees and horseshoe crabs.
Principles of operation of many liquid crystal displays are based on the manipulation
of the polarized light as detailed in Chapter 5.
In the field of fiber-optic communication, many electrooptic devices are polarization dependent. Coherent optical communication systems detect the received light by
mixing it with local oscillator light. Fluctuations in the state of polarization of the
received light or the local oscillator light will cause the output power of the intermediate frequency IF signal to fluctuate. Countermeasures have to be exploited. The
concepts in this chapter establish the foundation for understanding polarization. In
Chapter 12, we will deal with issues such as countermeasures for polarization jitter in
coherent communication systems.
6.1
INTRODUCTION
The types of waves that have so far appeared in this book have been linearly polarized
waves. The E field component did not change direction as the wave propagated. As
shown in Fig. 6.1a, this type of wave is called a linearly polarized wave, and the
direction of the E field is called the direction of polarization.
364
POLARIZATION OF LIGHT
(a)
k
(b)
k
(c)
(d)
Figure 6.1 Various states of polarization (SOP). (a) Linearly (horizontally) polarized. (b) Right-handed
circularly polarized. (c) Left-handed circularly polarized. (d) Depolarized.
In this chapter, waves whose directions of polarization rotate as the waves propagate
will be described. The E vector rotates around the propagation direction k, as the wave
propagates, as shown in Figs. 6.1b and 6.1c. When the cross section of the helix is an
ellipse, the wave is said to be elliptically polarized. When the cross section is circular
as in Fig. 6.1b, it is naturally called a circularly polarized wave. If the E vector rotates
in a clockwise sense when observed at a distant location in the propagation path while
looking toward the light source, as in Fig. 6.1b, the handedness of the polarization is
right-handed rotation. Similarly, if the E vector rotates in a counterclockwise direction,
as in Fig. 6.1c, the rotation is left-handed. If there is no repetition in the pattern of
the E field as the wave propagates, as shown in Fig. 6.1d, the wave is said to be
unpolarized or depolarized.
For handedness to be meaningful, both the direction of observation and the direction
of propagation have to be specified. By convention, the handedness is specified by
looking into the source of light.
Information describing the pattern and orientation of the polarized light is called
the state of polarization. Any given state of polarization can be decomposed into
CIRCLE DIAGRAMS FOR GRAPHICAL SOLUTIONS
365
two linearly polarized component waves in perpendicular directions. The state of
polarization is determined by the relative amplitude and difference in phase between
the two component waves. This relative phase difference is termed retardance.
The three most basic optical components that are used for manipulating or measuring
the state of polarization are the (1) retarder, (2) linear polarizer, and (3) rotator.
In this chapter, the prime emphasis is placed on how to use these optical components.
The circle diagrams are predominantly used for explaining the operation. In the next
chapter, however, the Poincaré sphere will be used for explaining the operation.
6.2
CIRCLE DIAGRAMS FOR GRAPHICAL SOLUTIONS
Graphical and analytical methods for finding the state of polarization complement each
other. The graphical method is fail-safe and is often used to confirm the results obtained
by analytical methods. The graphical method helps visualize the state of polarization for
a given set of parameters and also makes it easier to visualize intermediate stages. On
the other hand, analytical methods provide higher accuracy and are easier to generalize.
This chapter begins with a look at graphical solutions to common polarization problems.
6.2.1
Linearly Polarized Light Through a Retarder
A retarder can be made from any birefringent material, that is, any material whose
refractive index depends on direction. As an example, let us take the uniaxial crystal
characterized by refractive indices ne and no as described in Chapter 4. The orthogonal linearly polarized component waves are the e-wave and the o-wave. It is further
assumed that the front and back surfaces of the retarder are parallel to the optic axis
of the crystal, and the propagation direction of the incident light is normal to the
front surface of the retarder. In this situation, the directions of the component e-wave
and o-wave do not separate as they propagate through the retarder; rather, they emerge
Left-handed?
Right-handed?
Ha! Right-handed wave
Right- or left-handedness can only be determined after both the direction of propagation and the
direction of observation have been specified.
366
POLARIZATION OF LIGHT
together. Depending on which is smaller, ne or no , one of the component waves moves
through the retarder faster than the other. The relative phase difference is the retardance . The polarization direction of the faster component wave is called the fast
axis of the retarder, and the polarization direction of the slower component wave is
called the slow axis. The emergent state of polarization is the superposition of the two
component waves and will depend on the relative amplitudes of the two component
waves, as well as the retardance.
A circle diagram will be used to find the state of polarization as the incident linearly
polarized light transmits through the retarder. Figure 6.2a shows the configuration. A
linearly polarized wave at azimuth D 55° is incident onto a retarder with retardance
S
y
B
E
q = 55°
x
A
0
F
∆ = 60°
(a)
y
∆
2A
2
1
3
4
2
1
0
3
0
4
60°
5
2B
−wt + ∆
5
11
C2
10
6
7
6
10
9
8
8
9
9
7
8
10
7
11
C1
6
0
x
−wt
1
5
4
3
Figure 6.2
2
(b)
Graphical solution. (a) Geometry. (b) Circle diagram.
CIRCLE DIAGRAMS FOR GRAPHICAL SOLUTIONS
367
The difference in the usage of the words retardance and retardation is analogous to that
between transmittance and transmission. Retardance is a measurable quantity representing
the difference in phase angles.
 D 60° . The direction of the fast axis of the retarder is designated by an elongated
F and in this case is oriented in the x direction. The direction of the slow axis is
perpendicular to that of the fast axis and is taken as the y direction. The z direction is
the direction of propagation.
The incident light E is decomposed into the directions of the fast and slow axes,
that is, in the x and y directions. In complex notation, the component waves are
Ex D AejωtCˇz
Ey D Be
jωtCˇzC
6.1
6.2
with
E D Ex Oi C Ey Oj
A D jEj cos 55°
B D jEj sin 55°
and the corresponding real expressions are
Ex D A cosωt C ˇz
6.3
Ey D B cosωt C ˇz C 
6.4
The phasor circle C1 in Fig. 6.2 represents Eq. (6.1) and C2 represents Eq. (6.2).
As time progresses, both phasors rotate at the same angular velocity as ejωt (for now
a fixed z), or clockwise as indicated by 0, 1, 2, 3, . . . , 11. The phase of Ey , however,
lags by  D 60° because of the retarder. The projection from the circumference of
circle C1 onto the x axis represents Ex , and that from the C2 circle onto the y axis
represents Ey . It should be noted that the phase angle ωt in C1 is with respect to the
horizontal axis and ωt C  in C2 is with respect to the vertical axis.
By connecting the cross points of the projections from 0, 1, 2, 3, . . . , 11 on each
phasor circle, the desired vectorial sum of Ex and Ey is obtained. The emergent light
is elliptically polarized with left-handed or counterclockwise rotation.
Next, the case when the fast axis is not necessarily along the x axis will be treated.
For this example, a retarder with  D 90° will be used. Referring to the geometry
in Fig. 6.3a, the fast axis F is at azimuth  with respect to the x axis, and linearly
polarized light with field E is incident at azimuth . Aside from the new value of 
and the azimuth angles, the conditions are the same as the previous example.
The only difference in the procedure from that in the previous case is that the fast
axis is no longer in the x direction and the incident field has to be decomposed into
components parallel to the fast and slow axes, rather than into x and y components.
Figure 6.3b shows the circle diagram for this case.
368
POLARIZATION OF LIGHT
y
y
2
F
E
1
3
C2
4
F
E
S
S
Θ
q
0
x
0
x
1
4
(a)
C1
3
2
(b)
Figure 6.3 Circle diagram for linearly polarized light entering a retarder ( D 90° ), where the azimuth
angle of the retarder is given by . (a) Geometry. (b) Circle diagram.
It is important to recall that the only allowed directions of polarization inside the
crystal are along the fast and slow axes; no other directions in between the two axes
are allowed. This is the reason why the incident field is decomposed into components
along the fast and slow axes.
6.2.2
Sign Conventions
As mentioned in Chapter 1, this book has employed the convention of
ejωt
6.5
ejωt
6.6
rather than
which appears in some textbooks. This section attempts to clarify some of the confusion
surrounding signs and the choice of Eq. (6.5) or (6.6). Let us take the example of
the retarder in Fig. 6.2 as the basis for discussion. The expression for the state of
polarization depends critically on the difference between x and y of the Ex and Ey
component waves. With the convention of Eq. (6.5), the phases of Ex and Ey for the
CIRCLE DIAGRAMS FOR GRAPHICAL SOLUTIONS
369
positive z direction of propagation are
x D ˇz ωt
6.7
y D ˇz ωt C 
6.8
where  D 60° . We will now explain why y was expressed as ˇz ωt C  rather
than ˇz ωt . The example was defined such that the x direction is the direction
of the fast axis, which means Ex advances faster than Ey , and hence Ex leads Ey .
Let us examine Eqs. (6.7) and (6.8) more closely to see if it is indeed the case that
Ex leads Ey . For simplicity, the observation is made on the z = 0 plane. Both x and
y are becoming large negative values as time elapses. At the time when x D 0, y
is still a positive number, namely, y D 60° and y lags x by 60° /ω seconds in the
movement toward large negative values. Hence, one can say that Ey is lagging Ex by
60° or Ex is leading Ey by 60° . This confirms that y D ˇz ωt C  was the correct
choice to represent Ex leading Ey , for the convention of Eq. (6.5). When Ex leads Ey ,
a left-handed polarization results for  D 60° , as shown in Fig. 6.2.
Now, let us look at the other convention of using ejωt instead of ejωt . The same
example of the retarder in Fig. 6.2 will be used. When Ex and Ey are propagating in
the positive z direction, the signs of the ˇz and ωt terms are opposite (Chapter 1). Let
x0 and y0 denote the phases of the x and y components using the ejωt convention.
x0 D ˇz C ωt
y0 D ˇz C ωt 
6.9
6.10
where  D 60° . As the x direction was specified as the direction of the fast axis,
Eqs. (6.9) and (6.10) have to represent the case where Ex leads Ey by 60° . Let us
verify that this is true. Taking z D 0 as the plane of observation, both x0 and y0
become large positive numbers as time elapses. At the time x0 D 0, the phase of y0 is
a negative number and is behind x0 by 60° /ω seconds in the movement toward large
positive numbers. This is consistent with Ex leading Ey by 60° .
To conclude this example, if Ex leads Ey by 60° , the resulting polarization is lefthanded, regardless of the choice of convention of Eq. (6.5) or (6.6). However, for this
previous statement to be true, the sign of  does depend on the choice of convention.
In Problem 6.1, the same reasoning is applied to the geometry of Fig. 6.3.
Next, the state of polarization of the emergent wave will be investigated as a function
of retardance. For simplicity, the amplitudes of Ex and Ey are kept the same, that is,
B/A D 1, and the fast axis is kept along the x axis. The case of B/A 6D 1 is left for
Problem 6.2. A series of circle diagrams were drawn to obtain the states of polarization
with  as a parameter. The results are summarized in Fig. 6.4.
With  D 0° 360° , the state of polarization is linear with azimuth D 45° . As
 is increased from 0° to 90° , the shape becomes elliptical, growing fatter and fatter
while keeping the major axis always at 45° , and the rotation of polarization always
left-handed. When  reaches 90° , the wave becomes a circularly polarized wave, still
with left-handed rotation. As soon as  passes 90° , the radius in the 45° direction
starts to shrink while that in the 135° direction expands, and the state of polarization
becomes elliptically polarized with its major axis at 135° , but still with left-handed
rotation. This trend continues until  D 180° .
370
POLARIZATION OF LIGHT
y
x
∆ = q = 360°
0
180°
45°
315°
90°
135°
270°
225°
180°
Figure 6.4 Elliptical polarizations with A D B. The fast axis is in the x direction, and  is the
represents left-handedness,
represents right-handedness, and
represents
parameter.
linear polarization.
Table 6.1 Summary of states of polarization with fixed B=A = 1 for various
retardance values
Retardance
∆
0°
90°
180°
270°
360°
Shape
Inclination
q
45°
135°
135°
45°
Sign of sin ∆
+
−
Handedness
Left
Right
With  D 180° , the wave becomes linearly polarized again, but this time the direction
of polarization is at 135° . As soon as  exceeds 180° , the wave starts to become
elliptically polarized but with right-handed rotation. As  increases between 180° and
360° , the state of polarization changes from linear (135° ) to right-handed elliptical (major
axis at 135° ) to right-handed circular to right-handed elliptical (major axis at 45° ) to
linear (45° ). In this region of , the handedness is always right-handed. The results are
summarized in Table 6.1.
CIRCLE DIAGRAMS FOR GRAPHICAL SOLUTIONS
371
90°
v
sin∆ > 0
Left-handed
∆
180°
u
−1
0
1
0°
sin∆ < 0
Right-handed
270°
Figure 6.5 Summary of elliptical polarization with A D B and  as a parameter on the Ey /Ex complex
plane.
Next, a method of classification other than Table 6.1 will be considered. As one
may have already realized, it is the combination of two numbers — B/A and  — that
determines the state of polarization of the emergent light from the retarder. These two
numbers, however, are obtainable from the quotient of Eqs. (6.1) and (6.2), namely,
Ey
B j
D
e
6.11
Ex
A
Each point on the complex number plane of Ey /Ex corresponds to a state of polarization. As a matter of fact, this representation will be extensively used in the next
chapter. Figure 6.5 shows such a complex plane. For this example, B/A D 1 and the
states of polarization are drawn on a unit circle for various values of . Figure 6.5
summarizes the results in Fig. 6.4.
Example 6.1 A quarter-waveplate, commonly written as a /4 plate, is a retarder with
 D 90° . As shown in Fig. 6.6, horizontally linearly polarized light is intercepted by
a /4 plate whose orientation  is rotated. Draw the sequence of elliptically polarized
waves of the emergent light as the fast axis of the /4 plate is rotated at  D 0° , 22.5° ,
45° , 67.5° , 90° , 112.5° , 135° , 157.5° , 180° , and 202.5° .
Solution The series of circles is drawn in Fig. 6.7. As the correct numbering of the
circles C1 and C2 is crucial to the final result, a few tips are given here on how to
set up the numbering. The convention of ejωt is being used, so that numbering of
both circles C1 and C2 is in a clockwise sense. Refer to the drawings with  D 22.5°
372
POLARIZATION OF LIGHT
S
Θ
F
E
F
S
x
F
y
S
Θ
O
Figure 6.6
rotated.
E
x
The state of polarization of the emergent light is observed as the quarter-waveplate is
and  D 157.5° as examples. Point P on the horizontal axis represents the tip of
the incident light polarization at t D 0. P is decomposed into points P1 and P2 on
circles C1 and C2 , respectively, as shown in Fig. 6.7. If the retardance had been zero
 D 0° , then P1 and P2 would correspond to point 1 for each of the circles C1 and
C2 . Because of the retardance of the /4 plate, C2 is delayed 90° with respect to C1 ,
which corresponds to a rotation of C2 by 90° in the counterclockwise direction. P2
now lines up with point 2 of C2 . Observe in the case of the /4 plate, for all diagrams
in Fig. 6.7, the line drawn from point 1 of C1 and the line drawn from point 2 of C2
intersect along the horizontal axis at P. This is a good method for obtaining the correct
numbering.
With  D 0° , the radius of C2 becomes zero, and with  D 90° , that of C1 becomes
zero. The emergent light is identical to the incident light for these cases.
The results are summarized in Fig. 6.8. The major or minor axis is always along
the direction of the fast axis. This is a characteristic of a quarter-waveplate when
the incident light is linearly polarized. First, the major axis follows the fast axis, and
then the minor axis, and then the major axis. They alternate at every 45° . In the region
0 <  < /2, the emergent light is right-handed, while in the region /2 <  < the
emergent light is left-handed. It is worthwhile remembering that the handedness of the
emergent circularly polarized wave alternates every 90° of rotation of the retarder. At
With the case of D 157.5° , two circles C2 are drawn, one on each side. Either circle
C2 can be used, as long as one makes sure that point 1 on circle C1 as well as on circle C2
correspond to point P if the retardance is momentarily reduced to zero.
373
CIRCLE DIAGRAMS FOR GRAPHICAL SOLUTIONS
1
C2
2
4
P
1
4
2
3
3
C2
4
Θ = 0°
Θ = 90°
C1
1
1
3
4
3
2
C1
2
4
4
3
C2 2
1
3
1
1
C2
P2
4
3
2
P
2
Θ = 112.5°
4
1
C1
2
2
3
P1
C1
4
3
1
4
P
1
Θ = 22.5° 3
3
C
1
1
1
C2
2
1
2
C2
4
4
2
2
1
Θ = 135°
3
1
4
Θ = 45°
3
P
2
4
4
3
2
3
1
4
C1
3
P
2
4
3
1
4
C2
P2
2
3
C2
3
3
2
P
4
1
2
P1
2
P
1
Θ = 157.5°
P2
2
1
3
Θ = 67.5°
3
2
3
1
2
1
4
1
4
4
C1
C1
C2
4
Figure 6.7 Circle diagrams as the /4 plate is rotated. The incident light is horizontally polarized.
Note: Handedness changes at the azimuth of the retarder,  D 90° and  D 180° .
374
POLARIZATION OF LIGHT
Θ = 90°
4
67.5°
3
F
F
2
45°
F
2
1
1
4
F
0
0
Θ=0
0
Θ = 112.5°
135°
5
F
F
6
7
6
F
157.5°
7
Θ = 202.5° 9
5
8
F
8
F
180°
22.5°
F
3
9
Figure 6.8 Transitions of the state of polarization as the /4 plate is rotated. The incident light is
horizontally linearly polarized.
 D 45° , right-handed circular polarization is obtained, and at  D 135° left-handed
circular polarization is obtained. This is a quick convenient way of obtaining a right
circularly or left circularly polarized wave.
At every 90° , the emergent light becomes identical with the incident light and
is horizontally linearly polarized. Note that the orientation with  D 180° is identical to that of  D 0° , and the orientation with  D 202.5° is identical to that of
 D 22.5° .
6.2.3
Handedness
The question of how the direction of the handedness is determined will be resolved.
The instantaneous value t of the direction of polarization with respect to the x
axis is
1 Ey t
t, z D tan
6.12
Ex t
CIRCLE DIAGRAMS FOR GRAPHICAL SOLUTIONS
375
From Eqs. (6.3) and (6.4), t, z is expressed as
t, z D tan1
B
cos  C tanωt ˇz sin 
A
6.13
For sin  D 0, the azimuth t, z becomes independent of time and location, and a
linearly polarized wave results. With other values of sin , t, z depends on time and
location.
The direction of the movement of t, z0 at a fixed point z D z0 is found from the
derivative of Eq. (6.13):

dt, z0 

D

dt

1C
ω sec2 ωt ˇz
B
[cos  C tanωt ˇz0 sin ]
A


2  sin 

6.14
The factor in the large parentheses of Eq. (6.14) is always positive, and dt, z0 /dt
is the same sign as sin . When sin  is positive, the azimuth t, z0 increases with
time and if negative, t, z0 decreases. These results are summarized in Eq. (6.15):
Counter clockwise (left-handed)
Linearly polarized
Clockwise (right-handed)
when sin  > 0
when sin  D 0
when sin  < 0
6.15
For the case of circularly polarized emergent light, that is,
B/A D 1 and
sin  D š1
6.16
the derivative simplifies to
dt, z0 D šω
dt
6.17
The angular velocity of the rotation of the polarization is the same as that of the
component wave. The sense of rotation of dt, z0 /dt also matches the sign of sin .
It should be noted that Eq. (6.15) is true only when the incident wave is linearly
polarized.
6.2.4
Decomposition of Elliptically Polarized Light
The graphical method for constructing an elliptical polarization has been described. Up
to this point, the incident light has been decomposed into components along the fast and
slow axes of the retarder. In this section, the graphical method will be generalized to
allow decomposition of a given elliptical polarization into an arbitrary set of mutually
perpendicular component waves.
The values B0 /A0 and 0 of the newly decomposed waves, however, vary according
to the desired orientation of the decomposed component waves. An example will be
used for explaining the decomposition.
376
POLARIZATION OF LIGHT
Example 6.2 Graphically decompose the elliptically polarized wave with B/A D 1
and  D 45° shown in Fig. 6.4 into E0x polarized at 22.5° and E0y polarized at 112.5° and
then determine the values of B0 /A0 and 0 of the newly decomposed component waves.
Solution
Referring to Fig. 6.9, the decomposition is performed as follows:
1. Draw coordinates x 0 y 0 in the desired directions.
2. Determine the radius of circles C1 and C2 from the points of the extrema on the
ellipse in the x 0 and y 0 directions.
3. Extend a line downward parallel to the y 0 axis from point 1 on the ellipse to
point 1 on circle C1 . Similarly, extend a line to the right parallel to the x 0 axis
from the same point on the ellipse to intersect point 1 and point 10 on the circle
C2 (set the point 10 aside for now).
y′
Ey
′
2
y
1
y′
°
55 ′
∆
90°−∆′
C2
1
B′
C′ 2
A′
x′
1′
3
4
22.5°
x
O
−A
′
4
−B
3
′
y′
′
4
Ex
1
x′
C1
3
2
Figure 6.9 Decomposition of an elliptically polarized wave into component waves along arbitrary
orthogonal coordinates.
CIRCLE DIAGRAMS FOR GRAPHICAL SOLUTIONS
377
4. From point 1 on the ellipse O, follow the ellipse in the direction of rotation of
polarization to point 2. Point 2 is the intersection of the ellipse O with the y 0
axis. Draw the extension line from point 2 on the ellipse parallel to the x 0 axis
to make point 2 on circle C2 .
5. Find point 3, which is the tangent to the ellipse parallel to the y 0 axis. Draw an
extension line parallel to the x 0 axis from point 3 on the ellipse O to point 3 on
circle C2 .
6. Find point 4, which is the intersection of the ellipse O with the y 0 axes, and draw
an extension parallel to the x 0 axis to point 4 on circle C2 .
7. Now, B0 /A0 and 0 can be obtained from this graph. The ratio of the radii of C1
and C2 gives
B0 /A0 D 0.59
and the angle 0 on circle C2 gives
0 D 55°
The Ey phasor rotates clockwise from the y 0 axis, so that E0y is lagging by
0 D 55° from that of E0x , consistent with the left-handed rotation of the ellipse
given by Eq. (6.15). As a matter of fact, examination of Fig. 6.9 shows that 0
can be calculated directly from the intersections of B0 and C0 of the ellipse on
the y 0 axis:
cos90° 0 D
OC0
OB0
and
sin 0 D
OC0
OB0
8. Note that if the phasor on E0y starts from point 10 and both E0x and E0y rotate
in the clockwise sense (ωt), then the intersections will not form the original
ellipse, and point 10 has to be discarded.
6.2.5
Transmission of an Elliptically Polarized Wave Through a l=4 Plate
Previous sections dealt with the transmission of linearly polarized light through a
retarder. This section treats the more general case of transmission of an elliptically
polarized incident wave through a retarder.
As shown in Fig. 6.10, let the azimuth and ellipticity of the incident wave be and
, respectively. The retarder is again a /4 plate. The circle diagram method starts
with the decomposition of the incident elliptic field into the field parallel to the fast
axis of the /4 plate and that parallel to the slow axis. The former component field is
represented by phasor circle C1 and the latter by phasor circle C2 .
Next, the retardance is considered. The endpoint P of the phasor vector of the
incident wave is projected onto point 1 of circle C1 and projected onto point 0 of
circle C2 . Point 0 on circle C2 is delayed by 90° with respect to point 1 in order to
account for transmission through the /4 plate. The circumference of the phasor circles
378
POLARIZATION OF LIGHT
1
y
0
2
C2
4
3
F
1
S
P
4
Θ
q
2
x
4
3
1
C1
3
2
Figure 6.10 An elliptically polarized wave (solid line) is incident onto a /4 plate at  D 45° . The
emergent ellipse (dotted line) is obtained by circle diagrams.
is divided into four and numbered 1, 2, 3, and 4 sequentially. The intercepts of the
projections from each of the circles parallel to the fast and slow axes are numbered 1,
2, 3, and 4. The emergent ellipse, shown as a dotted line in Fig. 6.10, is completed by
connecting 1, 2, 3, 4 and the handedness is in the direction of 1, 2, 3, and 4.
6.3
VARIOUS TYPES OF RETARDERS
A waveplate is a retarder with a fixed retardance. Waveplates providing retardances of
360° , 180° , and 90° are called full-waveplates, half-waveplates, and quarter-waveplates,
respectively. They are also written simply as , /2, and /4 plates. Retarders with
adjustable retardance are called compensators.
VARIOUS TYPES OF RETARDERS
6.3.1
379
Waveplates
Waveplates can be fabricated either from a single piece of birefringent crystal or from
a combination of two pieces of crystal. The difficulty with fabricating a single crystal
waveplate is that the plate has to be made extremely thin. The thickness d for a /4
plate is calculated as
2
djne no j D
2
6.18
Taking D 0.63 µm, the values of d for typical birefringent crystals are:
For calcite, d D 0.92 µm (ne D 1.4864, no D 1.6584).
For quartz, d D 17.3 µm (ne D 1.5443, no D 1.5534).
For mica, d D 31.5 µm (ne D 1.594, no D 1.599).
The thickness is in the range of tens of microns.
Even though calcite has a cleavage plane and need not be polished, its brittleness
makes it hard to handle thin pieces. Quartz is not as brittle but requires polishing
because it does not have a cleavage plane. Mica has more favorable properties. It is
not only flexible, but also possesses cleavage planes; however, there is some difficulty
in cleaving at exactly the right thickness. The plate with the desired thickness is selected
among many cleaved pieces.
The stringent requirement of excessively small thicknesses can be avoided by taking
advantage of the rollover of the retardance at every 2 radians. The retardance of a
/4 plate, for instance, is designed as
2
jne no j d D 2N C
2
6.19
where the value of N is normally a few hundred.
With N D 100, the thickness of a quartz /4 plate is 7 mm. The drawbacks of a
retarder with a large N are a higher sensitivity to temperature and to the angle of
incidence. The increase in the path of a ray with incident angle compared to the ray
normal to the plate of thickness d is
d 2
1
1 d
cos 2
6.20
The value of that creates a retardance error of /2 radians is
2
2
jne no j d D
2
2
6.21
Inserting Eq. (6.19) into (6.21) gives
1
p
rad
2N
With N D 100, the cone of the allowed angle of incidence is narrower than 4° .
6.22
380
POLARIZATION OF LIGHT
Ex
Ey
E
d1
d2
Figure 6.11 Structure of a quarter-waveplate.
The second approach to alleviating the thinness requirement is the combination of
two plates with their optical axes perpendicular to each other. Figure 6.11 shows the
construction of a waveplate of this kind. The optical paths x and y for Ex and Ey are
2
d1 ne C d2 no 2
y D
d1 no C d2 ne x D
6.23
6.24
and the retardance  D y x is
D
2
d1 d2 no ne rad
6.25
What matters in Eq. (6.25) is the difference in thickness, rather than the total thickness
so that the thickness of each plate can be comfortably large to facilitate polishing.
6.3.2
Compensator
Figure 6.12a shows the structure of a Babinet compensator. Two wedge-shaped birefringent crystals are stacked such that their optical axes are perpendicular to each other.
The apex angle, however, is made small so that the separation of the o- and e-waves
is negligibly small. They can be slid against one another so that the difference in
thicknesses d1 d2 is adjustable.
Depending on the location of the incident ray, the retardance is varied. At the
location where d1 D d2 , regardless of the values of ne and no , both waves Ex and Ey
VARIOUS TYPES OF RETARDERS
381
Ey
O
Ex
E
ne
a
ne
d2
no
d1
(a)
Ez
Ex
E
y
d2
d1
(b)
Figure 6.12 Comparison between Babinet-type and Soleil-type compensators. (a) Babinet compensator. Depending on the horizontal locations of the incident light, the emergent state of polarization
varies. (b) Soleil compensator. For a given d1 and d2 , the emergent state of polarization does not
change as the incident light position is varied.
382
POLARIZATION OF LIGHT
go through the same amount of phase shift and the retardance is zero. However, at the
locations where d1 6D d2 , the two waves do not go through the same phase shift. If the
crystal is a positive uniaxial crystal and ne > no like quartz and if d1 < d2 , then the
phase of Ey lags behind that of the Ex wave. The amount of phase lag, or retardation,
can continuously be adjusted by either shifting the location of the incident light while
keeping the relative positions of the crystal stack fixed, or sliding one crystal against the
other by means of a micrometer while keeping the location of the incident light fixed.
The compensator can be used at any wavelength provided that the light at that
wavelength is not significantly absorbed by the crystal. The compensator also provides
a full range of retardation so that it can be used as a zero-wave, quarter-wave, halfwave, or full-wave retarder. The sense of circular polarization can also be changed by
changing plus /4 to minus /4 of retardation. In some applications, the compensator
is used to measure the retardation of a sample material. Light of a known state of
polarization enters the sample, which causes a change in the polarization of the emergent light. The emergent light is passed through the compensator and the thickness d2
is adjusted to regain the initial state of polarization. From this adjustment of d2 , which
is precalibrated in terms of retardance, the retardation of the sample can be determined.
Figure 6.12b shows the structure of a Soleil compensator. The difference between
the Babinet and Soleil compensators is that the Soleil compensator has another block
of crystal so that the thickness d2 is independent of the location of the incident wave.
The top two blocks consist of two wedge-shaped crystals. They can be slid by a
micrometer against one another so that the thickness d2 is adjustable. The thickness
d1 of the bottom block is fixed. With the Soleil compensator, the emergent state of
polarization is independent of the location of incidence.
As mentioned earlier, when N is large, there are stringent requirements on the
collimation of the light entering the retarder, as well as a larger temperature dependence.
However, for the Soleil or Babinet compensator, it is possible to reduce N to zero.
6.3.3
Fiber-Loop Retarder
Bending an optical fiber creates birefringence within the fiber. This is the basis of
the fiber-loop retarder whereby a controlled amount of bend-induced birefringence is
used to change the retardance. An advantage of the fiber-loop retarder controller over
conventional optical devices, such as quarter-waveplates and half-waveplates, is that
the polarization control is achieved without interrupting transmission of light in the
fiber.
When an optical fiber is bent, the fiber is compressed in the radial direction of
the bend and is expanded in the direction perpendicular to it, as shown in Fig. 6.13.
The refractive index of glass is lowered where it is compressed and raised where it
is expanded. In the coordinates shown in Fig. 6.13, the difference between the change
nx in the x direction before and after the bending, and the change ny in the y
direction before and after the bending is calculated as [4,5]
ny nx D 0.0439n3
r 2
R
6.26
where n is the index of refraction of the core, 2r is the outer diameter of the fiber
(diameter of the cladding), and R is the radius of curvature of the loop as indicated in
VARIOUS TYPES OF RETARDERS
383
y
x
z
r
y
x
z
R
O
Figure 6.13
Bending of a fiber to create birefringence.
Fig. 6.14. The coefficient 0.0439 was calculated from Poisson’s ratio and the strainoptical coefficients for a silica glass fiber.
Figure 6.15 shows the structure of a fiber-loop polarization controller based on the
bend-induced birefringence in the fiber [6]. It combines both a /4 and /2 fiber-loop
retarder and a polarizer loop. The retarders are made of ordinary single-mode fibers.
The radius of the left-hand side retarder spool is designed such that the phase of the
y-polarized wave is /2 radians behind that of the x-polarized wave. This spool is the
fiber equivalent of a “quarter-waveplate” and it will be referred to as the /4 loop. The
right-hand side retarder spool is designed to create a -radian phase shift, analogous to
a “half-waveplate”, and will be referred to as the /2 loop. Usually, the two spools have
the same radius, and the /2 spool has twice the number of fiber turns as the /4 spool.
The orientation of the fiber loops can be changed as indicated by the arrows in
Fig. 6.15. When conversion of elliptic to linear polarization is desired, the /4 loop
is oriented so that the elliptically polarized wave going into the loop is converted into
a linearly polarized wave upon exiting the loop (see Section 6.4.3.2). The /2 loop is
then oriented to rotate the direction of the linear polarization to the desired direction.
Example 6.3 Find the loop radius of a fiber-loop polarization controller for a
D 1.30 µm wavelength system. The number of turns is N D 4 for the /4 loop
and N D 8 for the /2 loop. The diameter of the fiber is 125 µm, and the index of
refraction of the core is n D 1.55.
Solution The radius R of the /4 loop is first calculated. The difference ˇ in
propagation constants in the x and y directions is
ˇ D k0 n C nx k0 n C ny 6.27
where k0 is the free-space propagation constant. Combining Eq. (6.26) with (6.27),
ˇ D 0.0439 k0 n3
r 2
R
rad/m
6.28
384
POLARIZATION OF LIGHT
x
y
Ey
Ex
O
R
Ex
2r
y
S
Figure 6.14
x
Fiber-loop-type retarder using a single-mode fiber.
Photo
detector
Analyzer
Loop
l/2 Loop
l/4 Loop
Figure 6.15 Fiber-loop polarization controller.
HOW TO USE WAVEPLATES
385
The /4 condition requires
2RN Ð ˇ D
2
6.29
Equations (6.28) and (6.29) give
R D 0.176 k0 n3 r 2 N
6.30
With the parameters provided, the loop radius is
R D 4.95 cm
A choice of radius smaller than R D 1.5 cm is not recommended, because the
transmission loss (as will be mentioned in Section 6.5.3) becomes increasingly
significant as R is decreased.
6.4
HOW TO USE WAVEPLATES
The waveplate is one of the most versatile optical components for manipulating the
state of polarization. Various applications of the waveplate will be summarized from
the viewpoint of laboratory users.
6.4.1
How to Use a Full-Waveplate
A full-wave plate ( plate) combined with an analyzer (polarizer) functions like a
wavelength filter. The retardance of any thick waveplate critically depends on the
wavelength of light. This can be seen from Eq. (6.19), which gives the expression for
the retardance of a thick quarter-waveplate. The expression for the retardance of the
thick full-waveplate can be obtained by substituting 2 for /2 on the right-hand side
of Eq. (6.19).
The operation of the wavelength filter is described as follows. Multiwavelength
linearly polarized light is incident onto the full-waveplate. The azimuth of the incident
light is chosen so that both fast and slow component waves propagate through the fullwaveplate. Only those wavelengths that satisfy the full-wave retardance condition will
emerge as linearly polarized waves. The transmission axis of the analyzer is oriented
in the same direction as the azimuth of the incident light, so that the emergent linearly
polarized waves experience the least amount of attenuation on passing through the
analyzer. For all other wavelengths, the state of polarization is changed by going
through the full-waveplates, and these will be attenuated on passing through the
analyzer.
By adding additional full-waveplate and analyzer pairs, a narrower linewidth
M
wavelength filter is obtainable. Such filters are the Lyot–Ohman and Solic
filters [1,13].
6.4.2
How to Use a Half-Waveplate
The half-waveplate can be used to change the orientation and/or the handedness of a
polarized wave. The case of a linearly polarized incident wave is first considered. Let
386
POLARIZATION OF LIGHT
the vector OP of the incident E wave be in the vertical direction, and let the direction
of propagation be into the page. The plane of the half-waveplate is in the plane of
the page. Let the fast axis of the half-waveplate (/2 plate) be oriented at an angle from the direction of polarization, as shown in Fig. 6.16a. The incident vector OP is
decomposed into OQ and QP in the directions of the fast and slow axes, respectively.
After transmission through the /2 plate, the direction of the vector QP is reversed to
0
QP , while the vector OQ remains unchanged. Alternatively, OQ is reversed and PQ
0
is unchanged. The resultant vector for the transmitted wave becomes the vector OP .
0
The emergent vector OP is a mirror image of the incident vector OP with respect to
the fast axis. Another way of looking at the emergent wave is to say that the incident
vector has been rotated toward the fast axis by 2. For instance, a vertically polarized
incident wave can be converted into a horizontally polarized wave by inserting the /2
plate at an azimuth angle of 45° .
Next, the case of elliptically polarized incident light is considered. Let us say that
the direction of the major axis is vertical and the handedness is right. The direction of
the fast axis is degrees from the major axis of the ellipse. As illustrated in Fig. 6.16b,
the incident elliptically polarized wave is decomposed into components parallel to the
fast and slow axes, represented by circles C1 and C2 , respectively. The points 1,2,3,4
on the circle C1 correspond to points 1,2,3,4 on the circle C2 .
After transmission through the /2 plate, the points on circle C2 (slow axis) lag
the points on circle C1 (fast axis) by 180° . This is shown in Fig. 6.16b by moving the
points 1,2,3,4 on C2 diametrically opposite, as indicated by 10 , 20 , 30 , and 40 , while the
points 1,2,3, and 4 on C1 remain unchanged. The emergent wave is drawn with the
new combination. The emergent wave is elliptically polarized with the same shape but
rotated from that of the incident wave. The emergent wave looks like a mirror image
with respect to the fast axis of the /2 plate. The handedness is reversed. Alternatively,
the emergent ellipse also looks as if it were made by rotating the incident ellipse toward
the fast axis by 2, then reversing the handedness.
The /2 plate does not change the shape of the ellipse, only the orientation and
handedness. When just a simple change of handedness is desired, one of the axes of
the ellipse is made to coincide with the fast axis of the /2 plate.
6.4.3
How to Use a Quarter-Waveplate
The quarter-waveplate is probably the most popular of the waveplates. It has a
variety of uses including polarization converter, handedness interrogator, and retardance
measuring tool.
6.4.3.1 Conversion from Linear to Circular Polarization
by Means of a l=4 Plate
Figure 6.17 shows three configurations for converting a linearly polarized wave into
either a circularly or elliptically polarized wave by means of a quarter-waveplate.
In the figure, the incident light is vertically polarized and propagating from left to
right. In Fig. 6.17a, looking into the source of light, the direction of the polarization
of the incident light is 45° to the left of the fast axis of the quarter-waveplate.
With this configuration, the emergent light is a left-handed circularly polarized wave
(Problem 6.3).
387
HOW TO USE WAVEPLATES
P
Q
P′
Incident E
2f
Emergent E
f
O
(a)
1
2
nt
ide
Inc
1
nt ′
rge 1
e
Em
C1
2f
f
4
2,4′
3
2′,4
3′
3
3′
(b)
2′
4
1
C2
2
4′
1′ 3
Figure 6.16 Usage of a /2 plate. (a) Incident wave is linearly polarized. (b) Incident wave is elliptically
polarized.
388
POLARIZATION OF LIGHT
y
y
y
F
S′
Ey
45°
x
x
O
O
O
x
(a)
y
y
y
135°
Ey
x
x
O
x
O
O
(b)
y
Ey
x
x
O
y
y
x
O
(c)
Figure 6.17 Converting a linearly polarized wave by means of a /4 plate. (a) Linear to left-handed
circular polarization. (b) Linear to right-handed circular polarization. (c) Linear to elliptical polarization.
HOW TO USE WAVEPLATES
389
Figure 6.17b is similar to Fig. 6.17a except that the direction of polarization of the
incident wave is 45° to the right of the fast axis of the quarter-waveplate. The emerging
light is a right-handed circularly polarized wave.
In summary, looking toward the direction of the source, if the incident light
polarization is oriented 45° to the left of the fast axis, the handedness is also lefthanded. On the other hand, if the incident light polarization is oriented 45° to the right
of the fast axis, the handedness becomes also right-handed. Thus, for the same linearly
polarized incident wave, a change in the handedness of the emergent light is achieved
by just rotating the quarter-waveplate by 90° in either direction.
An elliptically polarized wave is generated by orienting the fast axis at an azimuth
angle other than 45° with respect to the direction of the incident light polarization, as
shown in Fig. 6.17c.
Inspection of the results shown in Fig. 6.17 and the answers to Problem 6.3 reveals
a shortcut method to drawing the emergent wave from a /4 plate when the incident
light is linearly polarized. Figure. 6.18 illustrates this shortcut method, and the steps
are explained below.
Step 1. Draw in the azimuth of the input light and  of the fast axis of the /4
plate.
Step 2. Draw the line ef perpendicular to the fast axis of the /4 plate.
Step 3. Complete the rectangle efgh. The center of the rectangle coincides with the
origin.
Step 4. The ellipse that is tangent to this rectangle represents the polarization of the
emergent light. If E is to the left of the fast axis, the emergent wave has left-handed
elliptical polarization.
6.4.3.2 Converting Light with an Unknown State of Polarization
into Linearly Polarized Light by Means of a l=4 Plate
Figure 6.19 shows an arrangement for converting an elliptically polarized wave into
a linearly polarized wave. The incident light beam goes through a /4 plate, a /2
plate, and an analyzer and finally reaches the photodetector. If the fast axis of the
quarter-waveplate is aligned to the major (or minor) axis of the elliptically polarized
incident light, the output from the quarter-waveplate will be linearly polarized. This
fact is detailed in Fig. 6.20, where elliptically polarized light is decomposed into
two component fields perpendicular and parallel to the major axis. The phase of the
perpendicular component is delayed from that of the parallel component by 90° (see
also Problem 6.4). If the fast axis of the quarter-waveplate is aligned with the delayed
perpendicular component, the two component waves become in phase and the emergent
wave is a linearly polarized wave as indicated by the vector OP.
If the fast axis of the /4 plate is further rotated by 90° in either direction, again
0
the emergent wave is a linearly polarized wave as indicated by vector OP .
A method for aligning the /4 plate in the desired location will now be explained.
Besides the /4 plate, a /2 plate and an analyzer are added as shown in Fig. 6.19. The
transmission axis (major principal axis) of the analyzer is for now set in the vertical
direction. The output of the analyzer is monitored with a photodetector. The function
of the /2 plate is to rotate the direction of the light polarization emerging from the
/4 plate.
390
POLARIZATION OF LIGHT
∆=
E
p
2
Θ
q
O
Step 1. Draw E and F.
E e
f
O
Step 2. Draw ef perpendicular to the F axis
E e
f
O
Step 3. Complete the rectangle efgh.
h
g
E
Step 4. Draw the ellipse tangent to the rectangle
O
Figure 6.18 A shortcut method of finding the state of polarization when linearly polarized light is
incident onto a /4 plate.
HOW TO USE WAVEPLATES
391
Photodetector
Analyzer
l/2 Plate
l/4 Plate
Figure 6.19 Converting from elliptical to linear polarization.
1′ 2
3
4 ′1
Emergent
C1
4
3'
P
2
2′
1
Incident
P′
O
When fast and slow
axes are exchanged
4
3
4
1
C2
3
2
Figure 6.20 Converting an elliptically polarized wave into a linearly polarized wave by means of a
/4 plate.
392
POLARIZATION OF LIGHT
First, with an arbitrary orientation of the /4 plate, the /2 plate is rotated. As
the /2 plate is rotated, the value of the minimum output from the photodetector is
noted. This is the beginning of an iterative procedure aimed at producing a sharp
null in the photodetector output. As the next step, the /4 plate is rotated by a small
amount in one direction. The /2 plate is again rotated. The new minimum output
from the photodetector is compared to the previous minimum. If the new minimum is
smaller, the /4 plate was turned in the correct direction. The procedure of rotating
the /4 plate by a small amount followed by a rotation of the /2 plate is repeated
until the absolute minimum has been found. Only when the input to the /2 plate is
linearly polarized is its output linearly polarized, and the photodetector output shows
sharp nulls where the linearly polarized output from the /2 plate is perpendicularly
polarized to the transmission axis of the analyzer.
Once a linearly polarized wave and a sharp null are obtained, the direction of
polarization can be changed to the desired direction by rotating the /2 plate.
The fiber equivalent of Fig. 6.19 is shown in Fig. 6.15. The fiber-loop /4 plate can
be treated as a conventional /4 plate whose surface is perpendicular to the plane of
the fiber loop. The alignment procedure for the conventional waveplates in Fig. 6.19
applies equally to the fiber-loop waveplates in Fig. 6.15. The direction of the fast axis
of the /2 and /4 loops lies in the plane of the fiber loop as explained in Section 6.3.3.
6.4.3.3 Measuring the Retardance of a Sample
Figure 6.4 summarizes the sequential change in the state of polarization as the
retardance  is increased from 0° to 360° for a linearly polarized initial state with
B/A D 1. By the same token, if linearly polarized light with B/A D 1 is incident on a
birefringent sample of known orientation but unknown retardance, the retardance can
be determined from the state of polarization of the emergent light.
Figure 6.21 shows an arrangement of Senarmont’s method for measuring the
retardance of a sample. The incident light is linearly polarized at D 45° with respect
to the x –y axes and the amplitudes of the x and y components are the same, namely,
B/A D 1. Either the fast or slow axis of the crystal is aligned to the x axis (a method
for locating the fast or slow axis can be found in Problem 6.10). The emergent light is
an elliptically polarized wave with a 45° (or 135° ) azimuth angle. The ellipticity tan ˇ,
which is the ratio of the major axis to the minor axis, depends on the value of the
y
Ey
y
y
y
x
x
E
y
S
Ex
F
b 45°
135°
O
x
45° + b
x
O
Sample
l/4 Plate
Analyzer
Figure 6.21 Measurement of the retardance  of a crystal by Senarmont’s method.
HOW TO USE WAVEPLATES
393
retardance and, as obtained in Problem 6.11,
 D 12 ˇ
6.31
The elliptically polarized wave further enters a /4 plate whose fast axis orientation
is set at 45° or 135° . The emergent light from the /4 plate is linearly polarized
because the fast axis is aligned to the major or minor axis of the ellipse as mentioned
in Section 6.4.3.2. The azimuth angle of the light emergent from the /4 plate is
D 45° C ˇ. The value of ˇ can be found from the direction of the sharp null when
rotating the analyzer. Finally,  is found from ˇ by Eq. (6.31).
It is important to realize that the direction of the major or minor axis of the
elliptically polarized light incident to the /4 plate is always at D 45° or 135° if
B/A D 1, and the fast axis of the sample is along the x axis, regardless of . Once the
fast axis of the /4 plate is set to D 45° or 135° , the /4 plate need not be adjusted
during the measurement. The only adjustment needed is the direction of the analyzer.
This is a noble feature of Senarmont’s method.
6.4.3.4 Measurement of Retardance of an Incident Field
The previously mentioned Senarmont’s method used a priori knowledge of the azimuth
angle of 45° of the emergent light from the sample, but in this case, the retardance
between the x and y directions of an incident wave with an arbitrary state of polarization
will be measured.
The measurement consists of three steps using a polarizer, a /4 plate, and a
photodetector. Let an arbitrary incident wave be represented by Eqs. (6.1) and (6.2).
The arrangement is similar to the one shown in Fig. 6.21, but the sample is removed.
Step 1. First, only the polarizer, which is used as an analyzer, and the photodetector are
installed. With the transmission axis of the analyzer along the x axis, the transmitted
power is measured. The transmitted power Ix is expressed as
Ix D 12 jEx j2 D 12 A2
where an ideal analyzer is assumed (i.e., lossless transmission of the through
polarization and complete rejection of the cross polarization). Next, the transmission
axis of the analyzer is rotated along the y axis, and the transmitted power is
measured. The transmitted power Iy is expressed as
Iy D 12 jEy j2 D 12 B2
The total transmitted power I0 is
I0 D Ix C Iy
Next, the transmission axis of the analyzer is rotated at a 45° azimuth angle to the
x axis and the transmitted light power is measured. Both Ex and Ey contribute to
394
POLARIZATION OF LIGHT
the component along the analyzer transmission axis:
I1 D
2
1 1
1
p
p
C
E
E
x
y
2
2
2
D 14 jA C Bej j2
D 14 A C Bej A C Bej D 14 A2 C B2 C 2AB cos 
Step 2. The /4 plate is inserted in front of the analyzer. The fast axis of the /4
plate is parallel to the y axis, and the analyzer is kept with its transmission axis at
45° to the x axis. The transmitted power is
2
1 1
1
j90° I2 D p Ex C p Ey e
2
2
2
D 14 A2 C B2 C 2AB sin 
From these measured values of I0 , I1 , and I2 , the retardance is
 D tan1
6.5
2I2 I0
2I1 I0
LINEAR POLARIZERS
A linear polarizer favorably passes the component of light polarized parallel to the
transmission axis and suppresses the component polarized parallel to the extinction
axis. The extinction axis is perpendicular to the transmission axis.
The three major types of linear polarizers are the (1) dichroic polarizer,
(2) birefringence polarizer, and (3) polarizer based on Brewster’s angle and scattering.
Each type has merits and demerits and a choice has to be made considering such
parameters as transmission loss, power of extinction, wavelength bandwidth, bulkiness,
weight, durability, and cost.
6.5.1
Dichroic Polarizer
Figure 6.22 shows an oversimplified view of the molecular structure of a dichroic
sheet polarizer. It is analogous to a lacy curtain suspending an array of long slender
conducting molecules.
The dichroic sheet is quite thin and is normally laminated on a transparent substrate
for strength. Transmission through the dichroic sheet depends on the direction of
polarization of the incident wave [7,8].
When the axis of a conducting molecule is parallel to the E field, the situation is
similar to a linear dipole antenna receiving a radio signal. A current is induced in
the axial direction and can flow freely along the molecule except at both ends. At
the ends, the axial current has to be zero and the direction of the current has to be
LINEAR POLARIZERS
E
395
k
k
kr
(a)
E
Figure 6.22
k
(b)
Dichroic linear polarizer sheet. (a) Extinction. (b) Transmission.
The “dichroic” polarizer has to do with “two colors.” Historically [1], certain crystals
displaying polarizing properties were observed to change color when they were held up to
sunlight and were viewed with a polarizer. The color changes occur when the polarizer is
rotated.
Although somewhat of a misnomer, the term dichroic has persisted. At present, any sheet
whose absorption depends on the direction of polarization of the incident light is called a
dichroic sheet. Another example is the dichroic filter. It reflects at a specified wavelength
and transmits at another specified wavelength, while maintaining a nearly zero coefficient of
absorption for all wavelengths of interest.
reversed, resulting in a current standing wave Iz with a sinusoidal distribution along
the molecular axis as shown in Fig. 6.23a.
When, however, the direction of the E field is perpendicular to the axis of the
molecule, the current is induced in a diametrical direction. The current has to be zero
at the left and right edges of the molecule. The magnitude of the excited current cannot
be large because the zero current boundary conditions are located so close to each other.
The distribution of the current Ix has a quasitriangular shape with a short height. The
magnitude of the induced current for a perpendicular orientation of the E field is small
compared to that for a parallel orientation of the E field, as shown in Fig. 6.23b.
Regardless of the conductivity of the molecule, the transmitted light is attenuated
as long as the direction of the E field is parallel to the molecular axis. For resistivetype slender molecules, the induced current is converted into heat and there is no
reflected wave. For slender molecules that are conductors, the induced current sets up
a secondary cylindrical wave whose amplitude is identical to that of the incident wave
but whose phase is shifted by 180° in order that the resultant field on the surface of the
molecule vanishes, thereby satisfying the boundary condition of a perfect conductor.
In the region beyond the molecule, both the transmitted and the 180° out-of-phase
secondary wave propagate in the same direction but in opposite phase, and they cancel
each other. When the incident wave is from the left to the right, there is no emergent
wave in the region to the right of the molecule, as illustrated in Fig. 6.24. In the region
396
POLARIZATION OF LIGHT
Fleming’s shaking rope.
Sir John A. Fleming (1849–1946) used to explain electromagnetic wave phenomena by
making analogies with waggling a rope, and it is tempting to apply a rope analogy to the
case of the slender molecule polarizer. Imagine that a rope is stretched horizontally through
a set of vertical bars. If on one side of the bars, the rope is shaken up and down to produce
a wave propagating with its crests in the vertical direction, the wave will pass through the
bars unhindered. On the other hand, if the rope is shaken left and right so that its crests
are in the horizontal direction, the propagating wave is blocked by the bars. This analogy is
opposite to reality in the case of light transmission through the slender molecule polarizer.
The transmission axis is perpendicular to the bars (axis of slender molecules), and the
extinction axis is parallel to the bars. The shaking rope analogy is shaky in this case.
in front of the molecule, these two waves are propagating in opposite directions, and
there exists a standing wave in the region to the left of the molecule in Fig. 6.24.
As mentioned earlier, when the E field is perpendicular to the molecule axis, the
degree of excitation of the induced current Ix on the molecule is small and the wave
can propagate through the molecule curtain with minimum attenuation.
The quality of a polarizer is characterized by two parameters: the major principal
transmittance k1 and the minor principal transmittance k2 . k1 is the ratio of the intensity
of the transmitted light to that of the incident light when the polarizer is oriented to
maximize the transmission of linearly polarized incident light. k2 is the same ratio but
when the polarizer is oriented to minimize transmission. The values of k1 and k2 are
defined when the incident light direction is perpendicular to the surface of the polarizer.
The performance of the polarizer is optimum at this angle of incidence.
The value of k2 can be reduced by increasing the density of the slender molecules,
but always with a sacrifice of a reduction in k1 . Figure 6.25 shows the characteristic
curves of k1 and k2 for a typical dichroic sheet polarizer. Even though the transmission
ratio defined as Rt D k1 /k2 can be as large as 105 , it is hard to obtain the ideal value of
k1 D 1 with a dichroic polarizer sheet. On the other hand, the birefringent-type polarizer
can provide both a large transmission ratio and a value of k1 very close to unity.
The advantages of the dichroic sheet are that it is thin, lightweight, and lowcost, but the disadvantages are low k1 values (70% is common) and relatively low
LINEAR POLARIZERS
z
397
z
Ez
x
0
0
Iz
0
x
(a)
z
Ix
Ex
0
x
(b)
Figure 6.23 Difference in the excited current on the conducting molecule with parallel and
perpendicular E. (a) E is parallel to the axis. (b) E is perpendicular to the axis.
power handling capability due to absorption. The transmission ratio deteriorates in the
ultraviolet region, < 300 nm.
Example 6.4 Find the state of polarization of the emergent wave Ei for the following
combinations of incident field E0 and polarizer.
398
POLARIZATION OF LIGHT
Slender molecule
r
O
Primary field
Secondary wave
O
Standing wave
Figure 6.24
molecule.
r
Zero field
Top view of the fields incident onto and scattered from a slender perfectly conducting
(a) The incident light is linearly polarized,
and a poor-quality
polarizer is used with
p
p
major and minor transmittances k1 D 1 and k2 D 0.5, respectively.
(b) This situation is the same as case (a) but with an elliptically polarized incident
wave.
(c) The incident wave is elliptically polarized inpthe same waypas case (b) but the
polarizer has ideal characteristics, namely, k1 D 1 and k2 D 0. Draw the
locus of the major axis of the emergent light as the polarizer is rotated in its
plane.
Solution
The solutions are shown in Fig. 6.27.
(a) Figure 6.27a shows the configuration. The transmission axis of the polarizer is
shown by an extended T. The incident field E0 is decomposed into components E01
LINEAR POLARIZERS
399
1
k1
Transmittance
10−1
10−2
10−3
k2
10
−4
10−5
400
550
700
850
Wavelength (nm)
Figure 6.25 Characteristics of type HN38 Polaroid polarizer. (From Polaroid Corporation Catalog).
The concept of an absorption indicatrix is shown in Fig. 6.26. It is used to analyze the
transmission of a polarized wave through a bulk medium that possesses an absorboanisotropy
like a dichroic crystal. The method for using this indicatrix is similar to that used for the
refraction indicatrix presented in Section 4.5.2. Referring to Fig. 6.26, consider the case
when light propagating along the direction ON is incident onto an absorboanisotropic
crystal. The intercept of the plane containing the origin and perpendicular to ON with
the ellipsoid generates the “cross-sectional ellipse.” The lengths of the vectors a1 and
a2 of the major and minor axes represent absorbancies in these two directions of
polarization.
If the direction of polarization of the incident light is arbitrary, the E field of the incident
light is decomposed into components parallel to a1 and a2 , which suffer absorbancies a1
and a2 , where a1 and a2 are the major and minor axes of the ellipse. The amplitude of the
emergent light is the vectorial sum of these two components [9].
The shape of the ellipsoid of the absorption indicatrix is significantly more slender than
that of the refraction indicatrix in Section 4.5.2.
and E02 , which are parallel and perpendicular to the transmission axis of the polarizer.
Their phasor circles are C1 and C2 . The incident field being linear, the phasors are in
phase and points p
1, 2, 3, and 4 are numbered accordingly. The E02 component suffers
an attenuation of k2 D 0.5, which is represented in Fig. 6.27a by shrinking circle C2 .
On C2 , the points 1, 2, 3, and 4 shrink to 10 , 20 , 30 , and 40 . Successive intersections of
400
POLARIZATION OF LIGHT
z
N
O
a1
a2
y
x
Figure 6.26
crystal.
Absorption indicatrix used in finding the transmission through an absorboanisotropic
points 1, 2, 3, and 4 of circle C1 and points 10 , 20 , 30 , and 40 of the shrunken circle C2
produce the state of polarization of the emergent wave.
The emergent wave is linearly polarized but the azimuth angle is not the same as
that of the incident wave.
(b) Circles C1 and C2 are set up in a similar fashion to that of part (a), the only
difference being that points 1, 2, 3, and 4 of the linear incident light are replaced by
points 1, 2, 3, and 4 of the incident ellipse, as shown in Fig. 6.27b. The emergent light
polarization is formed from points 1, 2, 3, and 4 of C1 and points 10 , 20 , 30 , and 40 of
the shrunken C2 .
The azimuth angle of the emergent wave is closer to the azimuth of the polarizer
than the incident wave.
(c) Figure 6.27c explains the case with an ideal polarizer. Since k2 D 0, the radius
of the circle C2 shrinks to zero, and the amplitude of the emergent wave is determined
solely by the radius of circle C1 . The emergent light is linearly polarized and the
direction of the emergent E field is always along the direction of the transmission axis.
Referring to Fig. 6.27c, the solid line ellipse represents the incident ellipse. When the
401
LINEAR POLARIZERS
1
y
4
E 02
k1
2
1′
2′
k 2C
2
3′
4′
3
E0 1
Ei 1′
E 02 ′
4
E01
2′
x
3′
3
4
1
E01
C1
3
2
(a)
E 02 2
y
E0
Incident
1
Ei
2
2′
Emergent
O
3′
2′
C2
O 3′
k1
1 1′
k2
1′
y
3
4′
4
t
h
x
4′ 4
3
Θ
x
O
4
1
E01
C1
3
2
(b)
(c)
Figure 6.27 Waves with various states of polarization are incident onto a polarizer. (a) A linearly
polarized wave is incident onto a polarizer with k1 D 1.0 and k2 D 0.5. (b) An elliptically polarized wave
is incident onto a polarizer with k1 D 1.0 and k2 D 0.5. (c) Locus of the amplitude of the emergent
wave when an ideal polarizer is rotated.
azimuth angle of the polarizer is at , the direction of the emergent linear polarization
is also at D , and the amplitude is represented by Oh, which is perpendicular to
the tangent to the incident ellipse. The dashed line in the figure shows the locus of the
field vector as the polarizer is rotated. The position of h in Fig. 6.27c corresponds to
10 when k2 D 0 in Fig. 6.27b. It should be noted that the answer is not an ellipse.
402
POLARIZATION OF LIGHT
6.5.2
Birefringence Polarizer or Polarizing Prism
As mentioned in Chapter 4, there are only two possible directions of polarization
(e- and o-waves) inside a uniaxial crystal. No other directions of polarization are
allowed. A birefringence polarizer, which is sometimes called a prism polarizer, creates
a linearly polarized wave by eliminating one of the two waves. One type makes use
of the difference in the critical angles of total internal reflection for e- and o-waves.
Another type makes use of the difference in the angle of refraction for the two waves.
A significant advantage of the birefringence polarizer over a simple dichroic polymer
polarizing sheet is its high transmission coefficient of 90% to 95% or better compared
to 70% for the polarizing sheet. Moreover, the polarizing beamsplitter gives access to
beams of each polarization.
Figure 6.28 shows a cut-away view of the Nicol prism. A calcite crystal is sliced
diagonally and is cemented back together with Canada balsam cement whose index
of refraction is n D 1.55. Since the indices of refraction of calcite are ne D 1.486 and
n0 D 1.658 at D 0.58 µm, the e-wave does not encounter total internal reflection and
exits through the crystal to the right. The crystal is sliced at such an angle that total
internal reflection of the o-wave takes place at the interface between the calcite crystal
and the Canada balsam cement. The reflected o-wave is absorbed by the surrounding
dark coating.
Even though the Nicol prism is one of the best-known polarizers because of its long
history, the Nicol prism has the following disadvantages: the Canada balsam cement
absorbs in the ultraviolet region of the spectrum, the power handling capability is
limited by the deterioration of the cement, and the emergent beam is laterally displaced
from the position of the incident beam. A favorable characteristic of the Nicol prism
is its reasonable field of view of 28° .
The Glan–Foucault or Glan–Air polarizing prism is shown in Fig. 6.29a. This
type eliminates the use of Canada balsam cement so as to avoid absorption in the
Canada balsam cement n = 1.55
Arbitrarily
polarized
light
Calcite crystal no = 1.658
ne = 1.486
48°
71°
Direction of the
optic axis
Figure 6.28
e-Wave
o-Wave
Cutaway view of the Nicol prism.
LINEAR POLARIZERS
403
ultraviolet region and the limitation on the power handling capability of the prism.
As shown in Fig. 6.29a, the front surface of the prism is cut parallel to the optic axis
and perpendicular to the incident beam. The angle of the slanted airgap is chosen
such that total internal reflection takes place for the o-wave. A polarizing prism
normally discards the o-wave by the use of an absorptive coating. However, if desired,
the side surface of the polarizer can be polished to allow the o-wave to exit. The
o-wave from this polarizing beamsplitter can be used for monitoring purposes or
for providing an additional source with an orthogonal direction of polarization to
Arbitrarily
polarized light
Air gap
Direction of
optic axis
e-Wave
o-Wave
(a)
e-Wave
o-Wave
(b)
Figure 6.29 Glan–Foucault prism polarizer/beamsplitter and its modification. (a) Cutaway view of
the Glan–Foucault prism polarizer/beamsplitter. (b) Same as (a) but modified by Taylor for better
transmission.
404
POLARIZATION OF LIGHT
the e-wave. An additional merit of this prism is its short longitudinal dimension.
The Glan–Air prism, however, suffers from the demerits of a narrow acceptance
angle of 15° to 17° and multiple images caused by multiple reflections in the
airgap.
The Glan–Faucault prism was modified by Taylor who rotated the crystal axis by
90° , as illustrated in Fig. 6.29b. With this orientation Brewster’s angle can be used to
minimize the reflection. The value of k1 was significantly increased.
The Glan–Thomson polarizing prism has the same geometry as the Glan–Foucault
prism but uses Canada balsam cement in place of the airgap in order to increase the
viewing angle to 25° to 28° at the cost of the aforementioned drawbacks of Canada
balsam cement.
Several other types of polarizing prisms are similar and all are shown for comparison
in Fig. 6.30. The direction of refraction is determined by assuming that a negative
birefringent crystal ne < no like calcite is used. The angular separation of the oand e-waves is made by different arrangements of the optic axes of two pieces of
the same crystal material. The Rochon, Senarmont, and Ahrens polarizing prisms do
not deviate the direction of one of the transmitted lightwaves from the direction
of the incident light. With reference to Fig. 6.30, the deviated transmitted light
from the Rochon prism is vertically polarized while that of the Senarmont prism
is horizontally polarized. The Wollaston polarizing prism maximizes the angular
separation between the two beams because what is labeled the output o-wave is,
in fact, the e-wave in the first prism and both waves are refracted at the interface.
The geometry of the Cotton-type prism is almost the same as the bottom piece of
the Ahrens type, except for the larger apex angle of the prism for optimization of
operation.
Prisms based on refraction create aberrations when they are introduced in a
convergent beam. This is because the vertical geometry is not the same as the horizontal
geometry, and the angle of refraction from the boundaries in the vertical direction is
different from that of the horizontal direction just like a cylindrical lens.
6.5.3
Birefringence Fiber Polarizer
Next, the fiber-loop birefringence polarizer will be explained. When an optical fiber
is bent too tightly, the light in the core starts to leak out. The amount of leakage,
however, depends on the direction of polarization of the light because the change in the
refractive index caused by the bending is not isotropic. The fiber-loop-type polarizer
makes use of this property. Needless to say, the fiber-loop polarizer is especially
advantageous for use in fiber-optic communications because polarization control is
achieved without having to exit the fiber and transmission of light in the fiber is
uninterrupted.
When an optical fiber is bent, the fiber is compressed in the radial direction of
the bend and is expanded in the direction perpendicular to it, as shown at the top of
Fig. 6.13. The refractive index of glass is lowered where it is compressed and raised
where it is expanded. The differential stress creates anisotropy in the refractive indices
in the two aforementioned directions in the fiber.
Both single-mode and polarization-preserving fibers can be used for fabricating
a polarizer, but better results are obtained with polarization-preserving fibers which
already have birefringence even before bending the fibers. With ordinary single-mode
LINEAR POLARIZERS
405
e-Wave
o-Wave
(a)
e-Wave
o-Wave
(b)
o-wave
e-Wave
(c)
o-Wave
e-Wave
(d)
o-Wave
e-Wave
(e)
o-Wave
Figure 6.30 Various types of birefringence polarizers (polarizing prisms) using calcite. (a) Rochon.
(b) Wollaston. (c) Senarmont. (d) Ahrens. (e) Cotton.
406
POLARIZATION OF LIGHT
fibers, the fiber has to be bent over a much tighter radius to achieve the desired effect
and, consequently, is prone to breakage. Birefringence in the Panda-type polarizationpreserving fiber, shown in the inset in Fig. 6.31, is produced by contraction of the glass
with a higher thermal expansion coefficient in the “eyes” region when the drawn fiber
solidifies. The index of refraction nx seen by the wave polarized in the direction of
the “eyes” is raised due to the expansion of the core and that of ny seen by the wave
polarized in the direction of the “nose” is lowered due to the contraction of the core.
The slow axis is in the direction of the “eyes” and the fast axis is in the direction of
the “nose.”
nx = ny with no strain (normal fiber)
nx with bending
ny with bending
nx
2b
n x (x )
x
ny
n y (x )
2a
∆x
∆y
Core
"Eye"
Cladding
0
a
10
b
20
30 µm
(a)
x
y
nx
ny
n x (y )
n y (y )
∆x
∆y
nx
ny
Core
0
a
10
Cladding
20
30 µm
y
(b)
Figure 6.31 Refractive index profiles of a Panda fiber. (a) Profile along the x axis. (b) Profile along
the y axis. (After K. Okamoto, T. Hosaka, and J. Noda [10].)
LINEAR POLARIZERS
407
By comparing the inset in Fig. 6.31 with Fig. 6.13, one soon notices that the same
effect caused by the contraction of the “nose” can be generated by just bending the
fiber in the y direction. As a result, the birefringence of the Panda fiber is even more
enhanced when the Panda fiber is bent in the y direction.
Figure 6.31 shows the calculated profile of the indices of refraction when the Panda
fiber is bent [10]. The distribution along the “eyes” direction of the Panda fiber is
shown in Fig. 6.31a while that along the “nose” direction is shown in Fig. 6.31b.
As long as one stays on the line connecting the centers of the “eyes” (x axis) the
strains inside the core and the cladding are identical and the difference between nx
and ny , which is directly related to the strain, is the same in the core and cladding, as
shown in Fig. 6.31a.
Along the “nose” direction (y axis), however, the amount of strain varies
significantly with the distance from the center of the fiber, and the difference between
nx and ny also varies with the radius in the y direction. As shown in Fig. 6.31b, even
though ny is smaller than nx inside and on the periphery of the core, ny grows bigger
than nx in the region beyond 20 µm. The difference y between ny in the core and
ny in the cladding also decreases with y, whereas the difference x between nx in
the core and nx in the cladding stays the same with y. The evanescent wave exists
in these regions and a slight decrease in y significantly increases the bending loss
of the Ey component [10]. Thus, the emergent light is predominantly Ex polarized in
the direction of the “eye.” It is this anisotropic strain distribution that makes the fiber
polarizer work.
The tensile stress due to the Panda “eyes” can be enhanced further by increasing the
bending of the fiber in the y direction. Excessive bending, however, starts incurring the
transmission loss of the x-polarized wave. The amount of bending has to be determined
from a compromise between the transmittance k1 and the extinction ratio (defined
as the inverse of the transmission ratio) R. A transmission loss of 0.5 dB with an
extinction ratio R D 30 dB is obtainable for a wide wavelength range by 10 turns of
a 3-cm-diameter Panda fiber loop [11].
It is difficult to know the orientation of the Panda eyes unless its cross section is
examined under a microscope. The use of a special Panda fiber whose cross section is
oval shaped to indicate the orientation of the Panda “eyes” makes it easier to bend the
fiber into a coil while maintaining the right bending orientation [10].
6.5.4
Polarizers Based on Brewster’s Angle and Scattering
Brewster’s angle is another phenomenon that depends on the direction of polarization
of light and can be utilized to design a polarizer. Brewster’s angle of total transmission
exists only for a lightwave whose direction of polarization is in the plane of incidence.
Figure 6.32 shows a pile-of-plates polarizer that is based on Brewster’s angle.
Brewster’s condition is
n1
tan B D
n0
The wave polarized in the plane of incidence transmits through totally without
reflection. The wave polarized perpendicular to the plane of incidence also transmits
through, with some loss due to reflection. To be effective as a polarizer, several plates
are necessary in order to increase the loss due to reflection of the wave polarized
perpendicular to the plane of incidence.
408
POLARIZATION OF LIGHT
r
Ev
qB = 57°
Ev
r
Eh
.55
=1
n1
=1
n0
Figure 6.32 Pile-of-plates polarizer.
Light with high purity of linear polarization is obtainable from an external cavitytype gas laser such as shown in Fig. 14.1. This type of laser uses a Brewster window,
and in the case of the He–Ne laser, the light goes back and forth more than 2000
times before exiting the cavity. This is equivalent to a pile of 2000 plate polarizers,
and light with very pure linear polarization is obtained.
6.5.5
Polarization Based on Scattering
A rather unconventional polarizer makes use of the nature of Rayleigh scattering.
Scattering from a particle smaller than the wavelength of light creates polarized light.
Referring to Fig. 6.33, the light scattered in the direction normal to the incident ray
is linearly polarized. The vertically polarized component of the incident light cannot
N2 or CO2
Figure 6.33 Polarizer based on Rayleigh scattering.
CIRCULARLY POLARIZING SHEETS
409
be scattered in the vertical direction because the E field would become parallel to
the direction of propagation. The light scattered toward the vertical direction is highly
horizontally polarized light, and the light scattered toward the horizontal direction is
highly vertically polarized light.
A chamber filled with either N2 or CO2 molecules makes a polarizer. Even though
the amount of the scattered light is small, the purity of the polarization is good.
The direction of polarization is perpendicular to the plane containing the path of
the light from the source to the observer by way of the scatterer, as indicated in
Fig. 6.33.
6.6
CIRCULARLY POLARIZING SHEETS
A polarizer sheet laminated with a /4 plate sheet is sometimes marketed as a circularly
polarizing sheet. Its usages are presented here.
6.6.1
Antiglare Sheet
In this section, a method of preventing glare using a circularly polarizing sheet will be
described. Figure 6.34 shows a circularly polarizing sheet being used as an antiglare
cover for a radar screen. Figure 6.35 explains the function of the sheet, and for
purposes of the explanation, the polarizer and the /4 plate sheet are separated.
Figure 6.35a shows the state of polarization of the light incident on to the radar
surface, and Fig. 6.35b shows the state of polarization of the reflected wave from
the radar screen. In Fig. 6.35a, the direction of the polarization is 45° to the left
of the fast axis, and left-handed circularly polarized light is incident onto the radar
screen.
l/4 Retarder sheet
Polarizer sheet
Figure 6.34 A circularly polarizing sheet, which is a lamination of polarizer and /4 plate sheets, is
used for prevention of glare on a radar screen.
410
POLARIZATION OF LIGHT
B
A
O
A′
B
Left-handed
Polarizer
(transmission
axis vertical)
l/4 plate
Perfect
reflector
(a)
Right-handed
Perfect
reflector
l/4 Plate
Operator
Polarizer
(transmission
axis vertical)
(b)
Figure 6.35 Step-by-step explanation of the antiglare circularly polarizing sheet. (a) Incident wave.
(b) Reflected wave.
The very right top inset in the Figure shows what happens on reflection. If the
surface is assumed to be a perfect reflector, at the moment when the field vector of the
incident light points in the direction OA, the field vector of the induced field should
point in the opposite direction OA0 to satisfy the boundary condition that the resultant
tangential E field is zero on the surface of a perfect conductor. At the next moment,
when the incident vector moves to OB, the induced vector moves to OB0 . Although
CIRCULARLY POLARIZING SHEETS
411
the incident vector and the induced vector always point in opposite directions, they
always rotate in the same direction.
The reflected wave is the expansion of the induced wave. Figure 6.35b shows how
the reflected wave propagates toward the operator. Recall that the observer looks
toward the source of light, and the reflected wave is right-handed circularly polarized.
Likewise, the azimuth angle of the fast axis of the /4 plate now looks to the observer
like  D 135° .
The light transmitted through the /4 plate is found by the circle diagram to be
horizontally polarized. The light cannot go through the polarizer, and the light reflected
from the radar surface does not reach the radar operator. The blips originating from
the radar screen, which are randomly polarized, reach the operator’s eye with some
attenuation.
6.6.2
Monitoring the Reflected Light with Minimum Loss
A reflectometer gathers information from reflected light. One of the simplest ways to
sample the reflected light is to use a nonpolarizing beamsplitter (NPBS) in the manner
shown in Fig. 6.36a. With this configuration, however, the reflected as well as the
Lost
Lost
NPBS
Source
Detector
Target
(a)
y
45° x
Source
PBS
l/4 Plate
Detector
Target
(b)
Figure 6.36 Comparison between two types of reflectometers. (a) Using a nonpolarizing
beamsplitter. (b) Using a polarizing beamsplitter and a /4 plate.
412
POLARIZATION OF LIGHT
incident beam will be split by the splitter, resulting in light being lost to the system.
If a beamsplitter with reflectance R is used, the intensity of the light collected by the
system is Iin TR+, where Iin , T, and + are incident light intensity, the transmittance of
the beamsplitter, and the reflectivity of the target, respectively. Because of the constraint
T C R D 1, the optimum intensity of the collected light occurs when R D 0.5, and the
collected light intensity is at best 0.25 Iin +. Only one-quarter of the incident light
intensity is useful.
The reflected light is often weak, as, for instance, in a system for remotely analyzing
the gas contents from a smokestack. The system shown in Fig. 6.36b can be used
to maximize the sensitivity. This reflectometer uses the combination of a polarizing
beamsplitter (PBS) and a quarter-waveplate.
The arrangement is quite similar to that for preventing the glare explained in
Fig. 6.35, where the vertically polarized incident light is converted into a horizontally
polarized reflected light after passing through the /4 plate twice. In the reflectometer
of Fig. 6.36b, the vertically polarized light transmits through the PBS and is converted
into a left-handed circularly polarized wave by the /4 plate whose azimuth is 45° .
The light reflected from the target is a right-handed circularly polarized wave, which
in turn is converted into a horizontally polarized light by the same /4 plate. The
horizontally polarized wave is reflected by the PBS to the detector.
The power loss due to the transmission loss of the optical components is 103 –105 ,
depending on the quality of the components.
6.7
ROTATORS
When a linearly polarized light propagates in quartz along its optic axis, the direction of
polarization rotates as it propagates. Similar phenomena can be observed inside other
crystals like cinnabar (HgS) and sodium chlorate (NaClO3 ), as well as solutions like
sucrose (C12 H22 O11 ), turpentine (C10 H16 ), and cholesteric liquid crystals. Even some
biological substances like amino acids display this effect. This phenomenon of rotation
of the direction of polarization is called optical activity. A substance that displays
optical activity is called an optically active substance. Each optically active substance
has a particular sense of rotation. Media in which the rotation of polarization is righthanded looking toward the source are called dextrorotary (dextro in Latin means right).
Media in which the rotation of polarization is left-handed are called levorotary (levo
in Latin means left). There are both d- and l-rotary varieties of quartz.
Fresnel explained the mechanism of optical activity by decomposing a linearly
polarized wave into circularly polarized waves. As shown in Fig. 6.37, linearly
polarized incident light can be considered as a combination of right- and left-handed
circularly polarized waves with equal amplitudes. If these two oppositely rotating
circularly polarized waves rotate at the same speed, the direction of the polarization of
the resultant wave remains unchanged. However, if the rotation speeds are different,
the direction of polarization of the resultant wave rotates as the two waves propagate.
In explaining the difference in rotation speeds of the left and right circular
component waves, Fresnel attributed this to the rotational asymmetry of the molecular
structure of the optically active medium.
A birefringent material is a material that is characterized by two indices of refraction.
If, for example, the refractive indices are nx and ny , corresponding to x and y linearly
ROTATORS
P
L
413
P
L
R
O
R
O
(a)
(b)
Figure 6.37 Fresnel’s explanation of optical activity. (a) Incident light. (b) Inside an optically active
substance.
One may wonder about the validity of Fresnel’s explanation of optical activity for optically
active liquids because of the random orientation of the molecules. As shown in the figure, a
coil spring that looks right-handed is still right-handed even when it is flipped over.
Optically active
solution
Keep the eyes fixed and flip the spring. There is no change in handedness.
polarized component waves, the material is said to be linearly birefringent. Retarders
are examples of linearly birefringent devices. If the refractive indices are nl and nr ,
corresponding to left and right circularly polarized component waves, the material is
said to be circularly birefringent. Optically active substances are examples of circular
birefringence.
414
POLARIZATION OF LIGHT
Let us compare the emergent polarization for these two different types of
birefringence. In the case of linear birefringence, the shape and/or orientation of the
emergent polarization may differ from that of the incident light, as illustrated in the
retarder examples shown in Figs. 6.7 and 6.8. On the other hand, for a circularly
birefringent medium, the orientation of the emergent polarization changes, but the
shape remains the same. For example, linearly polarized light incident on an optically
active medium will remain linearly polarized, but the direction will rotate, as illustrated
in Fig. 6.37.
The angle of rotation in an optically active medium is proportional to the distance
of propagation in the medium. The angle of rotation per unit distance is called the
rotary power. The rotary power of quartz, for instance, is 27.71° /mm at the D line of
the sodium spectrum ( D 0.5893 µm) and at 20° C. In the case of a liquid substance
like natural sugar dissolved in water, the angle of rotation is proportional to both the
length of transmission and the concentration of the solute. The saccharimeter detailed
in Section 6.7.1 determines the concentration of an optically active sugar solution by
measuring the angle of rotation.
The rotary power depends on the wavelength of the light as well as the temperature
of the substance. If an optically active medium is placed between the orthogonally
oriented polarizer and analyzer shown in Fig. 6.38, and white light is used as the
incident light, then the optical spectrum is attenuated for wavelengths whose angles of
rotation are an integral multiple of radians and their complimentary colors appear.
A beautiful color pattern is observed. This phenomenon is called rotary dispersion.
The Faraday effect causes a substance to behave like an optically active medium
when an external magnetic field is applied. This induced optical activity exists only
when an external magnetic field is applied. The sense of rotation is solely determined
y′
y
45°
x′
y
k2
k1
k2
x
k1 x
Polarizer
E0
Analyzer
E1
E2
E3
Figure 6.38 In the polariscope, an optical component is inserted between two crossed polarizers for
inspection. The transmission axis for the polarizer on the left is in the y direction, and the transmission
axis for the polarizer on the right is in the x direction.
ROTATORS
415
by the direction of the magnetic field and does not depend on the direction of light
propagation. This is an important distinction between ordinary optical activity (a
reciprocal phenomenon) and the Faraday effect (a nonreciprocal phenomenon). In the
case of natural optical activity, if the direction of light propagation is reversed in an
l-rotary material, the rotation is still l-rotary. In the case of the Faraday effect, if a
material is l-rotary when the light propagates in the direction parallel to the magnetic
field, the material is d-rotary when the light propagates antiparallel to the magnetic
field.
Example 6.5 As shown in Fig. 6.38, the polariscope consists of two polarizer sheets
arranged with their transmission axes perpendicular to each other. Find the amplitude
E3 of the emergent light when the following components are inserted between the
polarizers. Assume k1 D 1 and k2 D 0.
(a)
(b)
(c)
(d)
(e)
(f)
No component.
A polarizer sheet with transmission axis along x 0 and an azimuth angle of 45° .
A /4 plate with an azimuth angle of 45° (fast axis along x 0 axis in Fig. 6.38).
A /2 plate with azimuth angle of 45° .
A full-waveplate with azimuth angle of 45° .
A 90° rotator.
Solution
This time, the solutions are found without resorting to circle diagrams.
(a) Nothing is inserted. E3 D 0.
(b) A polarizer is inserted at 45° . As shown in Fig. 6.39a, E1 is decomposed into
Ex0 and Ey 0 . Ey 0 is extinguished. Only the horizontal component of Ex0 is transmitted
through the analyzer.
1
E1
1
E3 D E1 ð p ð p D
2
2
2
(c) A /4 plate is inserted at 45° . There are two ways to solve this problem.
(1) Decompose E1 into components Ex0 parallel to the x 0 axis and Ey 0 parallel
to the y 0 axis. The components are
E1
Ex 0 D p
2
E1
and Ey 0 D p ej90°
2
These two waves are further decomposed into both horizontal and vertical (x
and y) components, but one needs to be concerned only with the x component
because only the horizontal component passes through the analyzer. The
horizontal component of the incident wave to the analyzer is
1 E1
1 E1
Ex D p p p p ej90°
2 2
2 2
416
POLARIZATION OF LIGHT
y
y′
x′
E1
Ey ′
Ex ′
x
0
1
°
E e j 90
2 1
1
E
2 1
(a)
y
y′
x′
E1
Ey ′
Ex ′
E3
x
0
−Ey′
(b)
Figure 6.39 Solutions of Example 6.5.
where the second term is from Ey 0 as shown in Fig. 6.39a and the first term
is from Ex0 , and
E3 D
E1
E1
1 ej90° D p ej45°
2
2
(2) Notep that emergent wave E2 is a circularly polarized wave with p
radius
E1 / 2. The magnitude of the horizontally polarized wave E3 is E1 / 2.
ROTATORS
417
(d) E1 is decomposed into Ex0 and Ey 0 . The vector Ey 0 is reversed in direction
because of the /2 retarder, as shown in Fig. 6.39b. The resultant of Ex0 and Ey 0
becomes E3 .
E1
E3 D 2 p cos 45°
2
D E1
Another way of obtaining the same result is to make use of the fact that a /2
plate rotates the polarization by 2. The vertical polarization becomes the horizontal
polarization.
(e) The full-waveplate does not disturb the state of polarization and the answer is
the same as (a):
E3 D 0
(f) If a vertical vector pointing toward the Cy direction is rotated by 90° , the result
is a horizontal vector pointing toward the x direction.
E3 D E1
6.7.1
Saccharimeter
As a sugar solution is an optically active substance, the concentration of sugar can be
determined by measuring the angle of rotation of the transmitted light polarization. The
Lausent-type saccharimeter such as shown in Fig. 6.40 is widely used to monitor the
sugar concentration of grapes in a vineyard. This is a pocketable outdoor type and uses
white light. The combination of a wavelength filter F and a polarizing beamsplitter
(PBS) converts the incident white light into quasimonochromatic linearly polarized
light. In the left half of the field, the light passes through a thin quartz rotator R, while
in the right half of the field, the light misses the rotator. Thus, a slight difference in
the direction of polarization is created between the light EL passing through in the left
field and ER in the right field. This slight difference in the direction of polarization is
for the purpose of increasing the accuracy of reading the azimuth of the analyzer A
through which the incident light is viewed.
L1
PBS
F
R C
C12H22O11
L2
A
P
Filter Polarizer
PBS
Rotator Intensity
compensator
but no
"rotator"
Test sample
Protractor
for analyzer
Figure 6.40 Lausent-type saccharimeter.
Analyzer
Telescope
418
POLARIZATION OF LIGHT
The first step is the calibration without solution. When the direction k2 of the
extinction axis is adjusted at exactly the midpoint of the angle between the two
directions of polarization, the contrast in the intensities between the left and right fields
diminishes. The azimuth angle 1 of the analyzer of diminishing contrast is noted.
Next, the solution under test is poured into the chamber. The directions of
polarization in both left and right fields will rotate by an amount that is proportional
to the concentration of the sugar.
The analyzer is again rotated so that the direction k2 of the extinction axis lies at
the midpoint of the rotated directions of polarizations, and the contrast between the
left and right field diminishes. This new azimuth angle 2 of the analyzer is noted.
The difference D 2 1 of the azimuth angles of the analyzer is the angle of
rotation caused by the optical activity of the sugar solution.
The explanation of the operation of the saccharimeter will be repeated referring to
Fig. 6.41. As shown in Fig. 6.41a, when the analyzer is not exactly adjusted such that
k2 is at the midpoint of EL and ER , a contrast between the left and right field intensities
can be seen (the right side is darker). As soon as k2 of the analyzer is adjusted to the
midpoint, as shown in Fig. 6.41b, the contrast disappears. The azimuth angle 1 is
noted. In the field, this calibration is performed prior to introducing the sample, as the
power of rotation of the quartz rotator R is temperature dependent.
As the second step, the test sample is introduced into the chamber. Both EL and
ER rotate by the same amount due to the optical activity of the sample, as shown in
Fig. 6.41c, and a contrast in field intensities appears again (the left side is darker).
The analyzer is rotated to find the azimuth 2 that diminishes the contrast, as shown
in Fig. 6.41d. The rotation is computed as D 2 1 .
The concentration P of sugar in grams per 100 cc of solution is given by the formula
D []t l
P
100
where []t is the specific rotary power of the substance at a temperature t° C and
°
a light wavelength µm. For sugar, []20
0.5893 µm D 66.5 (per length in decimeter ð
concentration in grams per 100 cc). The quantity l is the length of the chamber in
units of 10 cm.
The high-accuracy performance of this type of saccharimeter is attributed to the
following:
1. The contrast of two adjacent fields rather than the absolute value of the
transmitted light through the analyzer was used. The eyes are quite sensitive
to detecting differences in intensities between adjacent fields.
2. The region of minimum rather than maximum light transmission through the
analyzer was used. The sensitivity of the eyes to detecting a change in the
transmitted light is greater near the minimum of transmission.
Another approach is to eliminate both the quartz rotator and the intensity
compensator. The incident light to the sample is not divided. The direction of the
polarization of the emergent light is directly measured by a split-field polarizer. The
split-field polarizer is made up of two analyzers side by side with a 5° to 10° angle
between the extinction axes, as shown in Fig. 6.41e. When the split-field analyzer is
rotated so that the d0 –d0 axis aligns with the direction of polarization of the light, the
contrast between the left and right sections disappears. The split-field polarizer again
ROTATORS
k2
EL
k2
ER
EL
k1
O
ER
q1
O
(a)
419
k1
(b)
k2
EL
k2
EL
ER
ER
k1
O
O
q2
k1
(c)
(d)
d
Transmission axis
d′
(e)
Figure 6.41 Field view of Lausent-type saccharimeter. (a) Without sample and unadjusted.
(b) Without sample and adjusted to eliminate contrast. (c) With sample and unadjusted. (d) With
sample and adjusted to eliminate contrast. (e) Split-field polarizer.
uses the contrast of the fields near the minimum of transmission and the precision
reaches 0.001° .
6.7.2
Antiglare TV Camera
It is often difficult for a TV reporter to videotape a passenger inside a car due to the
light reflected from the surface of the car window.
The geometry of an antiglare camera [12] is shown in Fig. 6.42. When the incident
angle to the car window is in the vicinity of Brewster’s angle (56° for glass), reflection
of the p-polarized light is suppressed, but the s-polarized light is not, and thus the light
reflected from the car window is strongly linearly polarized. Removal of this particular
component of the light minimizes glare to the TV camera. One way of accomplishing
this is by means of a liquid crystal rotator such as the one shown in the display pannel
in Fig. 5.33, but without the input polarizer P1 .
420
POLARIZATION OF LIGHT
90° TN LC
45° TN LC
Polarizer
∆S
E
Figure 6.42 Operation of an antiglare TV camera.
(a) off
(b) on
(c) off
(d) on
Figure 6.43 Demonstration of antiglare TV camera. (Courtesy of H. Fujikake et al. [12].)
THE JONES VECTOR AND THE JONES MATRIX
421
The total amount of light into the camera is monitored and used as an electrical
servosignal. The amount of light for the same scene should be at a minimum when
the glare light has successfully been removed. The electrical servosignal rotates the
polarization direction of the incident glare light until it becomes blocked by polarizer
sheet P2 . The servosignal is minimized when the required amount of polarization
rotation is achieved. In order to construct a variable rotator, TN liquid crystal rotators
of 45° and 90° are combined. By selecting the appropriate combination of the applied
electric field to the two TN liquid crystal rotators, the amount of rotation can discretely
be varied from 45° to 135° at intervals of 45° . The photographs in Fig. 6.43 demonstrate
the effectiveness of the antiglare TV camera. The photographs on the left were taken
with an ordinary camera, while those on the right are the same scenes taken with the
antiglare camera. With the antiglare camera, the passengers in the car, and the fish in
the pond, are clearly visible.
6.8
THE JONES VECTOR AND THE JONES MATRIX
A method of analysis based on 2 ð 2 matrices was introduced by R. Clark Jones [13,14]
of Polaroid Corporation to describe the operation of optical systems. Each component
of the system has an associated Jones matrix, and the analysis of the system as a whole
is performed by multiplication of the 2 ð 2 component matrices. Moreover, the state
of polarization at each stage of the multiplication is easily known.
The state of polarization is described by the Jones vector whose vector components
are Ex and Ey . From Eqs. (6.1) and (6.2), the Jones vector is
Ex
A
D ejˇzωt
6.32
Bej
Ey
The common factor is generally of no importance and is omitted. Eliminating the
common factor ejˇzωt , Eq. (6.32) is written as
A
ED
6.33
Bej
Representative states of polarizations expressed by the Jones vector are shown in
Fig. 6.44.
If only the relative phase between Ex and Ey is important, the common factor ej/2
can be removed, and Eq. (6.33) becomes
j/2 Ae
ED
6.34
Bej/2
If one interprets an optical component as a converter of the state of polarization from
[Ex Ey ] into [E0x E0y ], then the function of the optical component is represented by
the 2 ð 2 matrix that transforms [Ex Ey ] into [E0x E0y ].
E0x
E0y
D [2 ð 2]
Such a 2 ð 2 matrix is called the Jones matrix.
Ex
Ey
6.35
422
POLARIZATION OF LIGHT
1
0
A
1
A
1
A
1
0
1
B
1
Be j ∆
j
Be −j ∆
−j
0
0
0
0
Figure 6.44
6.8.1
0
0
0
0
Jones vector and the state of polarization.
The Jones Matrix of a Polarizer
The Jones matrix of a polarizer whose major principal transmission axis is along the
x axis is
k
0
PD 1
6.36
0 k2
For an ideal polarizer, k1 D 1 and k2 D 0.
Next, the case when the polarizer is rotated in its plane will be considered. Let the
direction of the transmission axis be rotated by  from the x axis. The incident field
E is expressed in x –y coordinates.
In this case, the incident field E has to be decomposed into Ex1 , which is along the
major principal axis of the polarizer, and Ey1 , which is along the minor principal axis.
Referring Fig. 6.45,
Ex1 D E cos 
D E cos cos  E sin sin 
D Ex cos  Ey sin 
Similarly,
Ey1 D Ex sin  C Ey cos 
Ex1 and Ey1 can be rewritten in a matrix form as
Ex1
Ey1
D
cos  sin 
sin  cos 
Ex
Ey
6.37
The matrix in Eq. (6.37) is a rotation by  degrees from the original coordinates; the
incident field E is expressed in coordinates x1 and y1 that match the directions of the
principal axes of the polarizer.
423
THE JONES VECTOR AND THE JONES MATRIX
y
y1
x1
E y1
E
E x1
Ey
q
E ′y 1 = k 2E y1
E
= k 1 x1
Θ
Ex
0
Figure 6.45
E ′x 1
x
Finding the Jones matrix for a polarizer rotated by .
The light emergent from the polarizer is now
E0x1
E0y1
k
D 1
0
0
k2
Ex1
Ey1
6.38
The emergent wave, however, is in x1 and y1 coordinates, and needs to be expressed
in the original x and y coordinates.
Referring Fig. 6.45, the sum of the projections of E0x1 and E0y1 to the x axis provides
0
Ex . A similar projection to the y axis provides E0y .
E0x D E0x1 cos  E0y1 sin 
E0y D E0x1 sin  C E0y1 cos 
which again can be rewritten in a matrix form as
E0x
E0y
cos 
D
sin 
sin 
cos 
E0x1
E0y1
6.39
This is a rotation by  degrees from the x1 and y1 coordinates. The emergent wave
is expressed in the original x and y coordinates.
424
POLARIZATION OF LIGHT
Combining Eqs. (6.37) to (6.39), the Jones matrix P for a polarizer rotated by 
is given by
E0x
E0y
D P
Ex
Ey
6.40
cos  sin 
k1 0
cos 
sin 
P D
sin  cos 
0 k2
sin 
cos 
k cos2  C k2 sin2  k1 k2 sin  cos 
P D 1
k1 k2 sin  cos  k1 sin2  C k2 cos2 
6.41
If the polarizer is ideal and k1 D 1 and k2 D 0, Eq. (6.41) becomes
P D
6.8.2
cos2 
sin  cos 
sin  cos 
sin2 
6.42
The Jones Matrix of a Retarder
The Jones matrix of a retarder whose fast axis is oriented along the x axis is
1 0
RD
0 ej
j/2
e
0
RD
0
ej/2
6.43
6.44
The Jones matrix of the half-waveplate is
j 0
HD
0 j
6.45
and that of the quarter-waveplate is
ej/4
QD
0
0
6.46
ej/4
When a retarder is rotated, the treatment is similar to that of the rotated polarizer.
The Jones matrix whose fast axis is rotated by  from the x axis is
cos 
R D
sin 
sin 
cos 
ej/2
0
0
ej/2
cos 
sin 
sin 
cos 
6.47
Noting that Eq. (6.47) becomes the same as Eq. (6.40) if k1 and k2 are replaced by
ej/2 and ej/2 , respectively, the product of the matrix Eq. (6.47) is obtained as

ej/2 cos2  C ej/2 sin2 

R D

j2 sin sin  cos 
2


sin  cos 
2

2
j/2
j/2
2
sin  C e
cos 
e
j2 sin
6.48
THE JONES VECTOR AND THE JONES MATRIX
425
The Jones matrix of a retarder with retardance  rotated by š45° is, from Eq. (6.48),

Rš45°

2

D

Ýj sin
2
cos

Ýj sin
2 

cos
2
6.49
When a half-waveplate is rotated by  D š45° , the Jones matrix is
Hš45°
0 1
D
1 0
6.50
where a common factor of eÝj/2 which appears after inserting /2 D /2 and
 D š45° into Eq. (6.48), is suppressed.
When a quarter-waveplate is rotated by  D š45° , the Jones matrix is
Qš45°
6.8.3
1
1
Dp
Ýj
2
Ýj
1
6.51
The Jones Matrix of a Rotator
A rotator changes the azimuth angle without disturbing all other parameters of the state
of polarization.
Let the incident linearly polarized field E be converted into E0 by rotation as shown
in Fig. 6.46. Noting that jE0 j D jEj,
E0x D E cos0 C D E cos 0 cos E sin 0 sin E0y D E sin0 C D E cos 0 sin C E sin 0 cos 6.52
Since
Ex D E cos 0
Ey D E sin 0
Eq. (6.52) is equivalent to
E0x
E0y
cos D
sin sin cos Ex
Ey
6.53
which is the same expression as that for rotating the coordinates by .
While the above explanation dealt with the rotation of a linearly polarized incident
light, the same holds true for elliptically polarized incident light. For elliptical
polarization, each decomposed wave rotates by the same amount and the elliptical
shape does not change, but the azimuth of the axes rotates by .
Regardless of the orientation of a light wave incident onto a rotator, the amount of
rotation is the same.
Example 6.6 Find the answers to Example 6.5 using the Jones matrix.
426
POLARIZATION OF LIGHT
y
E'
Ey′
Ey
E
q
q0
Ex′
0
Figure 6.46
Ex
x
Rotation of a field vector by a rotator.
Solution
(a) No plate is inserted. From Eq. (6.42) with  D 90° and then 0° , the Jones matrix
expression is
E0x
E0y
1 0
0 0
Ex
D
0 0
0 1
Ey
0 0
Ex
D
0 0
Ey
There is no output.
(b) A polarizer is inserted at  D 45° . The Jones matrix expression from Eq. (6.42)
is
0
1 1 0
Ex
1 1
0 0
Ex
D
1 1
0 1
E0y
Ey
2 0 0
1 1 0
Ey
D
Ey
2 0 0
In the above expression, the vector just after the inserted polarizer is linearly polarized
at 45° . The advantage of the Jones matrix is that the state of polarization can be known
THE JONES VECTOR AND THE JONES MATRIX
427
at each stage of manipulation. Performing the final matrix multiplication gives
E0x
E0y
1 Ey
D
2 0
which is a linearly polarized wave along the x direction.
(c) A quarter-waveplate is inserted at  D 45° . From Eqs. (6.42) and (6.51), the
Jones matrix expression is
E0x
E0y
1 1 0
1 j
0
Dp
0
0
j
1
0
2
ej/2 1 0
Ey
D p
j/2
0
0
E
ye
2
0
1
Ex
Ey
The intermediate state of polarization after passing through the polarizer and quarterwaveplate is left-handed circular polarization from Fig. 6.44. The emergent wave is
E0x
E0y
ej/2
D p
2
Ey
0
(d) A half-waveplate is inserted at  D 45° . From Eqs. (6.42) and (6.50), the Jones
matrix expression is
E0x
E0y
1
D
0
1
D
0
0
0
0 1
1 0
0
Ey
0
0
0 0
0 1
Ex
Ey
The light leaving the half-wave plate is linearly polarized along the x direction. The
emergent wave is
0 Ex
Ey
D
E0y
0
(e) A full-waveplate is inserted with azimuth 45° . From Eqs. (6.42) and (6.49) with
 D 2, the Jones matrix expression is
E0x
E0y
1 0
0 0
1 0
0 1
1 0
0
D
0 0
Ey
0
D
0
D
0
0
0
1
Ex
Ey
428
POLARIZATION OF LIGHT
(f) A 90° rotator is inserted. From Eqs. (6.42) and (6.53), the Jones matrix expression
is
E0x
E0y
D
D
D
1
0
1
0
0
0
0 1
1 0
0
Ey
0
0
0 0
0 1
Ex
Ey
Ey
0
Example 6.7 Apply Jones matrices to Senarmont’s method for measuring the
retardance  of a crystal plate.
Solution As shown in Fig. 6.21, a linearly polarized wave inclined at 45° is incident
onto the crystal under test. The light emergent from the crystal further goes through a
quarter-waveplate at 45° , where the wave is converted into a linearly polarized wave
whose azimuth angle determines the retardance of the sample under test.
The output field E is from Eqs. (6.44) and (6.51)
j/2
1 1 j
Ex
e
0
1
Dp
j/2
j
1
0
e
1
Ey
2



p j45°  cos 2 C 45° 

D 2e


C 45°
sin
2
The emergent wave from the quarter-waveplate is linearly polarized with azimuth
angle /2 C 45° .
6.8.4
Eigenvectors of an Optical System
With most optical systems, if the state of polarization of the incident wave is varied,
the state of polarization of the emergent wave also varies. However, one may find a
particular state of polarization that does not differ between the incident and emergent
waves, except for a proportionality constant. The field vector that represents such an
incident wave is called an eigenvector and the value of the proportionality constant
is called an eigenvalue of the given optical system. For instance, a lasing light beam
(see Section 14.2.3) bouncing back and forth inside the laser cavity has to be in the
same state of polarization after each trip, over and above the matching of the phase,
so that the field is built up as the beam goes back and forth. When the laser system is
expressed in terms of the Jones matrix, the eigenvector of such a matrix provides the
lasing condition
and the eigenvalue, the gain or loss of the system. [15]
Ex
Let E be an eigenvector of the optical system, and let be its eigenvalue. The
y
relationship between incident and emergent waves in Jones matrix representation is
Ex
Ey
a
D 11
a21
a12
a22
Ex
Ey
6.54
THE JONES VECTOR AND THE JONES MATRIX
429
Equation (6.54) is rewritten as
a11 a12
a21
a22 Ex
Ey
D0
6.55
The eigenvalues and corresponding eigenvectors of the system will be found by
solving Eq. (6.55).
For nontrivial solutions for Ex and Ey to exist, the determinant of Eq. (6.55) has to
vanish:
a11 a22 a12 a21 D 0
6.56
Equation (6.56) is a quadratic equation in eigenvalue and the solution is
1 D
1
2
2 D
1
2
a22 C a11 C
2
4a12
a22 a11 2 C
2
4a12
a22 a11
2
a22 C a11 C
6.57
The convention of choosing 1 < 2 will become clear as the analysis progresses (see
the discussion surrounding Eq. (6.86)).
Next, the eigenvectors will be found. Inserting 1 into either the top or bottom row
of Eq. (6.55) gives
Ey D
a11 1
Ex
a12
6.58
Ey D
a21
Ex
a22 1
6.59
or
The equality of Eqs. (6.58) and (6.59) is verified from Eq. (6.56). One has to be
careful whenever a12 D 0 or a22 1 D 0, as explained in Example 6.6.
Eigenvector v1 , whose components Ex and Ey are related by either Eq. (6.58) or
(6.59), is rewritten as
Ex1
a12
v1 D
D
6.60
Ey1
a11 1
and similarly for 2
Ex2
v2 D
Ey2
a12
D
a11 2
6.61
By taking the inner product of the eigenvectors given by Eqs. (6.60) and (6.61), we
will find the condition that makes the eigenvectors orthogonal. Simplification of the
product using Eq. (6.57) leads to
[Ex1
Ex2
Ey1 ]
Ey2
D a12 a12 a21 6.62
430
POLARIZATION OF LIGHT
Thus, these vectors are orthogonal if
a12 D a21
6.63
meaning Eq. (6.54) is a symmetric matrix.
Example 6.8 Find the eigenvalues and eigenvectors of a quarter-waveplate with its
fast axis along the x axis.
Solution
The Jones matrix expression of a quarter-waveplate is, from Eq. (6.46),
j/4
Ex
e
0
Ex
D
6.64
0
ej/4
Ey
Ey
Comparing Eq. (6.64) with (6.54) gives
a11 D ej/4
a22 D ej/4
6.65
a12 D a21 D 0
Inserting Eq. (6.65) into (6.57) gives
1
1,2 D p 1 Ý j D eÝj/4
2
6.66
As mentioned earlier, if a12 or a22 1 is zero, one has to be careful. Here, the
original equation Eq. (6.55) is used,
a11 Ex C a12 Ey D 0
6.67
a21 Ex C a22 Ey D 0
and is combined with Eqs. (6.65) and (6.66) with D 1 to give
0Ex C 0Ey D 0
Ey D 0
0Ex C 2j sin
4
The above two equations are simultaneously satisfied if Ey D 0 and Ex is an arbitrary
number, meaning a horizontally polarized wave. The output is 1 Ex .
Similarly, with D 2 , Eq. (6.55) becomes
2j sin
Ex C 0Ey D 0
4
Ex C 0Ey D 0
which leads to Ex D 0 and Ey can be any number. The eigenvector is a vertically
polarized wave. The output is 2 Ey .
The magnitude of the transmitted light is j1,2 j D 1. If the phase of the output
is important, the phase factor (ej/2 ) that was discarded from Eq. (6.34) should be
retained in Eq. (6.64).
STATES OF POLARIZATION AND THEIR COMPONENT WAVES
6.9
431
STATES OF POLARIZATION AND THEIR COMPONENT WAVES
Relationships existing among ellipticity, azimuth of the major axes of the ellipse, Ex
and Ey component waves, and retardance will be derived. Such relationships will help
to convert the expression for an elliptically polarized wave into that of Ex and Ey
component waves.
6.9.1
Major and Minor Axes of an Elliptically Polarized Wave
The lengths of the major and minor axes will be found from the expressions for the
Ex and Ey component waves.
Letting 0 D ωt C ˇz, Eqs. (6.3) and (6.4) can be rewritten for convenience as
Ex
D cos 0
A
Ey
D cos 0 cos  sin 0 sin 
B
6.68
6.69
In order to find an expression that is invariant of time and location, 0 is eliminated
by putting Eq. (6.68) into (6.69):
Ey
D
B
Ex
A
cos  1
Ex
A
2
sin 
6.70
Rearranged, Eq. (6.70) is
Ex
A
2
C
Ey
B
2
2
Ex Ey
cos  D sin2 
A B
6.71
In order to facilitate the manipulation, let’s rewrite Eq. (6.71) as
gX, Y D sin2 
6.72
2
2
gX, Y D a11 X C a22 Y C 2a12 XY
6.73
where
X D Ex ,
a11 D
Y D Ey
1
,
A2
a22 D
1
,
B2
a12 D cos 
AB
6.74
Let us express Eq. (6.73) in matrix form so that various rules [16] associated with
the matrix operation can be utilized. As shown in Fig. 6.47, let the position vector v
of a point X, Y on the ellipse be represented by
X
vD
Y
vt D [X Y]
6.75
432
POLARIZATION OF LIGHT
Y
Mv
y
(X,Y )
v2
v
x
v1
^
v2
^
v1
0
X
Figure 6.47 The normal to the ellipse is Mv, and the directions of the major and minor axes are those
v that satisfy Mv D v. X D Ex and Y D Ey .
and define a symmetric matrix M as
a
M D 11
a12
a12
a22
6.76
Realize that an equality exists as
vt Mv D a11 X2 C a22 Y2 C 2a12 XY
6.77
The normal vN from the circumference of the ellipse is obtained by taking the
gradient of Eq. (6.73) (see Eq. (4.89)) and is expressed in vector form as
vN D 2
a11 X C a12 Y
a12 X C a22 Y
6.78
Hence,
vN D 2Mv
6.79
As shown in Fig. 6.47, if the vector v were to represent the direction of the major
or minor axis of the ellipse, vN should be parallel to v or vN D v, and hence, the
STATES OF POLARIZATION AND THEIR COMPONENT WAVES
433
condition for v to be along the major or minor axis is, from Eq. (6.79),
Mv D v
6.80
where the factor 2 was absorbed in . Thus, the eigenvectors of matrix M provides the
directions of the major and minor axes, and the eigenvectors are given by Eqs. (6.60)
and (6.61).
Next, the actual lengths a and b of the major and minor axes of the ellipse will be
found. The position vector v in Fig. 6.47 of a point (X,Y) on the ellipse in the X–Y
coordinates is expressed in the new x –y coordinates taken along the major and minor
axes as
v D x vO 1 C y vO 2
6.81
where vO 1 and vO 2 are the unit vectors of v1 and v2 .
Inserting Eq. (6.82) into (6.77) gives
vt Mv D x vO t1 C y vO t2 Mx vO 1 C y vO 2 D 1 x 2 C 2 y 2
6.82
where use was made of
MOv1 D 1 vO 1
6.83
MOv2 D 1 vO 2
6.84
vO t1 ·vO 2 D 0
Combining Eqs. (6.72), (6.73), (6.77) and (6.83) gives
y2
x2
C
D1
sin 2 /1
sin 2 /2
6.85
Thus, in the new x –y coordinates along v1 and v2 , the major and minor axes of the
ellipse a and b are
j sin j
aD p
1
and
j sin j
bD p
2
6.86
Since a is conventionally taken as the length of the major axis, the smaller eigenvalue
is taken for 1 . That is, the negative sign of Eq. (6.57) will be taken for 1 , and the
positive sign for 2 .
In summary, the eigenvalues and eigenvectors of M have given the lengths as well
as the directions of the major and minor axes.
Before going any further, we will verify that Eq. (6.71) is indeed the expression
of an ellipse, and not that of a hyperbola, as both formulas are quite alike. Note
that if 1 2 > 0, then Eq. (6.85) is an ellipse, but if 1 2 < 0, it is hyperbola. From
Eq. (6.57), the product 1 2 is
2
1 2 D a11 a22 a12
6.87
434
POLARIZATION OF LIGHT
Note that Eq. (6.87) is exactly the determinant of M. Thus, the conclusions are
det M > 0,
ellipse
det M < 0,
hyperbola
6.88
The value of the determinant is, from Eqs. (6.74) and (6.76),


1 2 D det 

1
A2
cos 
AB
cos  
2
AB 
 D sin 

A2 B 2
1
6.89
B2
Thus,
1 2 > 0
6.90
and Eq. (6.71) is indeed the expression of an ellipse.
6.9.2
Azimuth of the Principal Axes of an Elliptically Polarized Wave
Figure 6.48 shows the general geometry of an ellipse. Capital letters will be used for the
quantities associated with the X and Y components of the E field, and lowercase letters
Y
y
2a
x
v2
b
v1
a
q
X
0
2B
2b
2A
Figure 6.48 Parameter associated with an ellipse.
STATES OF POLARIZATION AND THEIR COMPONENT WAVES
435
for those quantities expressed in x –y coordinates. The x –y coordinates correspond to
the directions of the major and minor axes of the ellipse. EX does not exceed A and
EY does not exceed B, and the ellipse is always bordered by a rectangle 2A ð 2B. The
ratio B/A is often expressed in terms of the angle ˛ as
tan ˛ D
B
A
6.91
Since the right-hand side of Eq. (6.91) is a positive quantity, ˛ must lie in the range
0 ˛ /2
6.92
The vector v1 points in the direction of the x axis. The azimuth angle of the major
axis with respect to the X axis will be found.
From Eq. (6.60), tan is expressed as
tan D a11 1 a12
6.93
Inserting Eqs. (6.57) and (6.74) into Eq. (6.93) gives
tan D t C
t2 C cos2 
1
cos 
6.94
where
tD
A2 B 2
2AB
6.95
Equations (6.94) and (6.95) will be simplified further. Using the double-angle
relationship of the tangent function given by
tan 2 D
2 tan 1 tan2 6.96
cos 
t
6.97
Eq. (6.94) is greatly simplified as
tan 2 D
Applying the double-angle relationship of Eq. (6.96) to the angle ˛, and making use
of Eq. (6.91), Eq. (6.95) becomes
tD
1
tan 2˛
6.98
The final result is obtained from Eqs. (6.97) and (6.98):
tan 2 D tan 2˛ cos 
6.99
436
POLARIZATION OF LIGHT
As seen from Fig. 6.48, all configurations of the principal axes can be expressed by
0 180° .
The following conclusions are immediately drawn from Eq. (6.99):
1. If the amplitudes A and B are identical and ˛ D 45° , then the azimuth can only
be 45° or 135° , regardless of the value of :
D 45°
D 135°
for cos  > 0
for cos  < 0
This agrees with previous discussions involving Fig. 6.4.
2. For any value of A and B, if  D 90° , the azimuth is either 0° or 90° .
6.9.3
Ellipticity of an Elliptically Polarized Wave
Ellipticity is another quantity that describes the shape of an ellipse. The ellipticity is
defined as
b
6.100
D
a
where a is the length of the major axis, and b is the length of the minor axis of the
ellipse.
From Eqs. (6.57), (6.86), and (6.100) the ellipticity is
1Y
D
6.101
1CY
where
YD
a11 a22
a11 C a22
2
C
2a12
a11 C a22
2
6.102
From Eq. (6.74), the quantities under the square root of Eq. (6.102) are simplified as
a11 a22
D cos 2˛
a11 C a22
2a12
D sin 2˛ cos 
a11 C a22
6.103
where the trigonometric relationships
1 tan2 ˛
1 C tan2 ˛
2 tan ˛
sin 2˛ D
1 C tan2 ˛
cos 2˛ D
were used.
6.104
6.105
STATES OF POLARIZATION AND THEIR COMPONENT WAVES
437
Insertion of Eq. (6.103) into (6.102) gives
YD
1 sin2 2˛ sin2 
6.106
Thus, insertion of Eq. (6.106) into (6.101) gives the ellipticity. A few more manipulations will be made on the expression for , but first, observe the following behavior
of for given values of B/A and :
1. With zero retardance , the value is always zero and the wave is linearly
polarized.
2. Only when B/A D 1 and  D 90° , can the wave be circularly polarized.
Returning to the manipulations on the ellipticity expression, will be rewritten
further in terms of trigonometric functions. Referring to Fig. 6.48, can be represented
by the angle ˇ:
tan ˇ D
Since
6.107
is a quantity between 0 and 1
0 ˇ /4
6.108
The trigonometric relationship
2 tan ˇ
1 C tan2 ˇ
sin 2ˇ D
6.109
is applied to Eqs. (6.101) and (6.107) to obtain
1 Y2
6.110
sin2 2˛ sin2 
6.111
sin 2ˇ D
Insertion of Eq. (6.106) into (6.110) gives
sin 2ˇ D
sin 2ˇ D sin 2˛j sin j
6.112
Because of the restrictions imposed on ˛ and ˇ in Eqs. (6.92) and (6.108), both
sin 2ˇ and sin 2˛ are positive and the absolute value of sin  has to be taken.
6.9.4
Conservation of Energy
When the state of polarization is converted, the light power neither increases nor
decreases, aside from the loss due to nonideal optical components. Conservation of
energy dictates that
a 2 C b 2 D A2 C B 2
6.113
438
POLARIZATION OF LIGHT
Equation (6.113) will now be verified. From Eq. (6.86), a2 C b2 is expressed in terms
of the eigenvalues 1 and 2 and  as
a2 C b2 D
1
1
C
1
2
sin2 
6.114
From Eqs. (6.57) and (6.74), the sum of the eigenvalues is
1 C 2 D
1
1
C 2
2
A
B
6.115
Insertion of Eqs. (6.89) and (6.115) into Eq. (6.114) finally proves the equality of
Eq. (6.113).
Next, area relationships will be derived from Eq. (6.86). The product ab is
sin2 
ab D p
1 2
With Eq. (6.89), a substitution for
p
6.116
1 2 is found and
ab D ABj sin j
6.117
where the absolute value sign was used because all other quantities are positive. Note
area ab of the ellipse becomes zero when  D 0, and a maximum when  D š/2,
for given values of A and B.
Furthermore, the difference a2 b2 will be calculated. From Eq. (6.86), a2 b2 is
a2 b2 D
1
1
1
2
sin2 
6.118
With Eq. (6.89), Eq. (6.118) becomes
a2 b2 D 2 1 A2 B2
6.119
In the following, 2 1 will be calculated. From Eq. (6.57), the difference 2 1 is
2 1 D a22 a11 1 C
2a12
a22 a11
2
6.120
Manipulation of Eqs. (6.74), (6.95), (6.98), and (6.99) gives
2a12
D tan 2
a22 a11
6.121
From Eq. (6.120) and (6.121), the difference becomes
2 1 D
A2 B 2 1
A2 B2 cos 2
6.122
STATES OF POLARIZATION AND THEIR COMPONENT WAVES
439
Inserting Eq. (6.122) back into (6.119) gives the final result of
a2 b2 cos 2 D A2 B2
6.123
This relationship is used later on in converting between X–Y and x –y components.
6.9.5 Relating the Parameters of an Elliptically Polarized Wave to Those of
Component Waves
So far, parameters such as D tan ˇ and have been derived from B/AD tan ˛ and
. In this section, ˛ and  will conversely be obtained from ˇ and .
Using the trigonometric identity of Eq. (6.104) for ˇ instead of ˛ and using
Eqs. (6.107), (6.113), and (6.123), cos 2ˇ is expressed as
cos 2ˇ D
1
A2 B 2
Ð
A2 C B2 cos 2
6.124
Dividing both numerator and denominator by A2 , and using the trigonometric
relationship Eq. (6.104) for tan ˛, Eq. (6.124) becomes
cos 2˛ D cos 2 cos 2ˇ
6.125
Next, the expression for the retardance  will be derived. The derivation makes use
of Eq. (6.112), which contains j sin j, and the absolute value cannot be ignored.
j sin j D sin 
for sin  > 0; left-handed
j sin j D sin 
for sin  < 0; right-handed
6.126
where the handedness information is given by Eq. (6.15). The ratio between Eqs. (6.99)
and (6.112), and the use of Eqs. (6.125) and (6.126) leads to
tan  D šcosec 2 tan 2ˇ
6.127
The plus and minus signs are for left-handed and right-handed elliptical polarization,
respectively.
6.9.6
Summary of Essential Formulas
The formulas derived in the last few sections are often used for calculating the state
of polarization and will be summarized here.
 D y x
 > 0, y component is lagging
 < 0, y component is leading
B tan ˛ D
0˛
A
2
6.7 and 6.8
6.91
440
POLARIZATION OF LIGHT
b
tan ˇ D D
0ˇ
6.100 and 6.107
a
4
6.94
tan D t C t2 C cos2  / cos  0 < A2 B2
2AB
t D 1/ tan 2˛
tD
6.95
6.98
tan 2 D tan 2˛ cos  0 2 < 2
6.99
sin 2ˇ D sin 2˛j sin j
6.112
a 2 C b 2 D A2 C B 2
6.113
ab D ABj sin j
6.117
a2 b2 cos 2 D A2 B2
6.123
2
2
a b sin 2 D 2AB cos 
Prob. 6.12a
a2 b2 cos 2 D A2 C B2 cos 2˛
Prob. 6.12b
cos 2˛ D cos 2 cos 2ˇ
sin  > 0
and j sin j D sin ; left-handed
sin  < 0
and j sin j D sin ; right-handed
tan  D šcosec 2 tan 2ˇ
6.125
6.126
C is for left-handed and for right-handed 6.127
Example 6.9 A linearly polarized wave is incident onto a retarder whose fast axis
is along the x axis. The retardance  is 38° and the amplitudes of the Ex and Ey
components are 2.0 V/m and 3.1 V/m, respectively. Calculate the azimuth and the
ellipticity of the emergent elliptically polarized wave. Also, determine the lengths a
and b of the major and minor axes. Find the solution graphically as well as analytically.
Solution
For the given parameters,
A D 2.0 V/m
B D 3.1 V/m
 D 38°
and
will be found.
B
D 1.55
A
˛ D 57.2°
tan ˛ D
From Eq. (6.99), the angle is obtained:
tan 2 D tan 2˛ cos 
D 2.20.788
STATES OF POLARIZATION AND THEIR COMPONENT WAVES
441
D 1.733
D 30° or 60°
Since 0 , D 60° is the answer.
From Eq. (6.112)
sin 2ˇ D sin 2˛j sin j
D 0.9100.616
D 0.560
ˇ D 17.0°
D tan ˇ D 0.31
Next, a and b are calculated from Eqs. (6.113) and (6.123):
a2 C b2 D A2 C B2 D 13.61
1
5.61
a2 b2 D
A2 B2 D
cos 2
0.5
D 11.22
a D 3.52
b D 1.09
The circle diagram is shown in Fig. 6.49a and the calculated results are summarized
in Fig. 6.49b.
Example 6.10
The parameters of an elliptically polarized wave are A D 10 V/m, B D 8 V/m,
a D 12.40 V/m, and b D 3.22 V/m.
(a) Find the azimuth and the retardance .
(b) For the given values of A and B, what is the maximum ellipticity that can be
obtained by manipulating the retardance?
(c) For the given values of A and B, is it possible to obtain an ellipse with D 0.26
and azimuth D 50° by manipulating the retardance ?
Solution
From A and B, tan ˛ is obtained:
A D 10 V/m
B D 8 V/m
tan ˛ D 0.8
From a and b, tan ˇ is obtained:
a D 12.4 V/m
b D 3.20 V/m
tan ˇ D 0.26
442
POLARIZATION OF LIGHT
y
1
2
1
4
4
x
0
2
3.1
C2
3
3
4
2.
0
3
38°
1
C1
2
(a)
y
3.1
q = 60°
a = 57°
1.0
0
2.0
x
3.5
2
9
(b)
Figure 6.49 Solutions of Example 6.9 obtained graphically as well as analytically. (a) Graphical
solution. (b) Summary of calculated results.
STATES OF POLARIZATION AND THEIR COMPONENT WAVES
443
(a) and  are computed as follows. From Eq. (6.123), is calculated:
a2 b2 cos 2 D A2 B2
36
D 0.25
143.52
D 37.8°
cos 2 D
From Eq. (6.117),  can be found:
ab D ABj sin j
ab
D 0.496
AB
 D š29.7°
j sin j D
(b) From Eq. (6.112), the value of  that maximizes 2ˇ for a given value of ˛ is
 D 90° . The maximum value of is
D tan ˇ D tan ˛ D 0.8
(c) Let us see if Eq. (6.125) is satisfied:
cos 2˛ D cos 2 cos 2ˇ
with tan ˛ D 0.8,
˛ D 38.7°
and with tan ˇ D 0.26
ˇ D 14.6°
D 50°
0.218 D 1.74 ð 0.873
0.218 6D 1.52
For given A and B, the value of and are mutually related, and one cannot arbitrarily
pick the two values.
p
Example 6.11 A right-handed elliptically polarized wave with a D 3 V/m and
b D 1 V/m and D 22.5° is incident onto a /4 plate with its fast axis oriented at
 D 45° with respect to the X axis. Find the state of polarization of the wave emergent
from the /4 plate.
The circle diagram was used to solve the same question in Section 6.2.5, Fig. 6.10.
Solution
The following steps will be taken:
1. Calculate A, B, and  of the incident wave.
444
POLARIZATION OF LIGHT
2. Find B0 /A0 , and 0 of the emergent wave from the /4 plate by means of the
Jones matrix.
3. Convert B0 /A0 , and 0 into a0 , b0 , and 0 .
Step 1. The given parameters are
p
aD
3 V/m
b D 1 V/m
D 22.5°
1
tan ˇ D b/a D p
3
ˇ D 30°
From Eq. (6.125), ˛ is calculated:
cos 2˛ D cos 2 cos 2ˇ
D 0.354
˛ D 34.6°
Next, the retardance  will be found using Eq. (6.127):
tan  D š cosec2 tan 2ˇ
D š2.45
where the C sign is for left-handed and the sign is for right-handed. Since the problem
specifies right-handed, tan  D C2.45 is eliminated.
The two possibilities for  are
 D 67.8° or C 112.2°
From Eq. (6.126), the correct choice of  is
 D 67.8°
From Eq. (6.113), A and B are found:
a 2 C b 2 D A2 C B 2
D A2 1 C tan2 ˛
1C3
1 C tan2 34.6°
A D 1.65
A2 D
B D 1.14
STATES OF POLARIZATION AND THEIR COMPONENT WAVES
445
Step 2. The Jones matrix of the /4 plate whose fast axis azimuth angle  is 45°
is used to calculate emergent wave.
Ex
Ey
1
1 j
1.65
Dp
1.14ej67.8°
2 j 1
1
1
ej90°
1.65
Dp
j90°
1.14ej67.8°
1
2 e
1.65 C 1.14ej157.8°
1
Dp
2 1.65ej90° C 1.14ej67.8°
[1.65 C 1.14 cos157.8° ]2 C [1.14 sin157.8° ]2 ejx
1
Dp
2
[1.14 cos67.8° ]2 C [1.65 C 1.14 sin67.8° ]2 ejy
0.5952 C 0.432 ejx
1
Dp
2
0.432 C 2.712 ejy
0.733
1 0.73 ej35.5°
ej35.5°
p
Dp
D
2.74e45.5°
2 2.74 ej81.0
2
0 D 45.5°
2.74
D 3.74
tan ˛ D
0.733
˛0 D 75.0°
Step 3. From Eq. (6.99), is found:
tan 2 0 D tan 2˛0 cos 0
D 0.5660.72 D 0.407
2 D 22.2° or157.8°
0
Since 0 0 180° , and 0 2 0 360° , the negative value is rejected and 0 D 79° .
From Eq. (6.112), ˇ0 is calculated:
sin 2ˇ0 D sin 2˛0 j sin 0 j
D 0.5 ð j0.707j
D 0.354
ˇ D 10.36°
0
D tan ˇ0 D 0.18
sin 0 D 0.707 < 0
The emergent wave is right-handed.
a02 C b02 D a02 1 C
2
446
POLARIZATION OF LIGHT
From Eq. (6.113), the above equation is rewritten
A2 C B2 D a02 1 C
2
and conservation of energy gives
A2 C B2 D a2 C b2 D a02 C b02
p 2
3 C1
02
a D
D 3.87
1 C 0.182
a0 D 1.96
b0 D 0.35
Compare with the answer shown in Fig. 6.10 using a circle diagram.
PROBLEMS
p
6.1 A linearly polarized wave with A D 3 V/m, and B D 1 V/m is incident
onto a /4 plate with its fast axis along the x axis. A is the amplitude of
the Ex component wave, and B is the amplitude of the Ey component wave.
Obtain the emergent elliptically polarized wave by using the two different
conventions of E D ejωtjˇz and EC D ejωtCjˇz representing a forward wave,
and demonstrate that both results are the same.
6.2 A linearly polarized wave with azimuth 1 D 63.4° is incident onto a retarder
with  D 315° whose fast axis is oriented along the x axis. Graphically
determine the azimuth angle 2 , which is the angle between the major axis
of the emergent ellipse and the x axis, and the ellipticity of the ellipse.
6.3 In Example 6.1, the direction of polarization of the incident light was fixed
and the fast axis of the /4 plate (retarder with  D 90° ) was rotated. The
results were drawn in Fig. 6.8. Draw the results (analogous to Fig. 6.8) for
q
0
Figure P6.3 A linearly polarized wave is incident onto a quarter-waveplate with its fast axis oriented
horizontally.
PROBLEMS
447
the case when the direction of the fast axis is fixed horizontally, and the
direction of the incident linear polarization is rotated as shown in Fig. P6.3
at D 0° , 22.5° , 45° , 67.5° , 90° , 112.5° , 135° , 157.5° , and 180° :
6.4 Decompose graphically an elliptically polarized wave into component waves
that are parallel and perpendicular to the major or minor axis, and verify that
the phase difference between the component waves is 90° .
6.5 Obtain graphically the state of polarization with the same configuration as that
shown in Fig. 6.10, but with the opposite handedness of rotation of the incident
wave, that is, left-handed.
6.6 A linearly polarized light wave is incident normal to a pair of polarizers P1
and P2 whose transmission axes are oriented at 1 and 2 (Fig. P6.6). Assume
k1 D 1 and k2 D 0 for both polarizers.
(a) The light is incident from P1 to P2 . What is the orientation of the linearly
polarized eigenvector?
(b) What is the orientation of the linearly polarized eigenvector when the light
is incident from P2 to P1 ?
6.7 The horseshoe crab’s eyes are known to be polarization sensitive. It is believed
that this sensitivity is used as a means of navigating in sunlight, and the principle
involved is that of the polarization of sunlight by scattering from particles in
the water. Referring to the configuration in Fig. P6.7, how can the horseshoe
crab orient itself along a north–south line in the early morning? What is the
direction of polarization that the horseshoe crab sees when facing south in the
early morning?
6.8 Linearly polarized laser light ( D 0.63 µm) is transmitted through a quartz
crystal along its optical axis. Due to Rayleigh scattering from minute irregularities in the crystal, one can observe a trace of the laser beam from the side of
the crystal. One may even notice a spatial modulation of the intensity along the
P2
P1
q2
E
q
q1
0
Figure P6.6 Direction of the eigenvector (looking from the source).
448
POLARIZATION OF LIGHT
West
East
Figure P6.7 The horseshoe crab is known to be sensitive to polarized light.
L
Laser
Quartz
Figure P6.8 Modulation of the intensity of Rayleigh scattering inside a quartz crystal due to the
optical activity of the crystal.
trace, as illustrated in Fig. P6.8. If one assumes that the spatial modulation is
due to the rotary power, what is the period of the modulation? The rotary power
of quartz is 19.5 deg/mm at D 0.63 µm and at a temperature of 20° C.
6.9 Figure 6.11 shows a diagram of a /4 plate. If one assumes d1 > d2 , is the
birefringence of the crystal in the figure positive or negative?
6.10 Devise a scheme to determine the directions of the fast and slow axes of a
retarder.
6.11 The ellipse shown in Fig. P6.11 was made with B/A D 1 and  D 0, just like
the ellipses shown in Fig. 6.4. Prove that the retardance  is identical to the
angle 6 ABC D 2ˇ on the ellipse.
6.12 Prove the following equalities:
(a) a2 b2 sin 2 D 2AB cos .
(b) a2 b2 cos 2 D A2 C B2 cos 2˛.
REFERENCES
449
y
B
2b
A
45°
x
0
C
Figure P6.11
Prove that 2ˇ is identical to the retardance  when B/A D 1 and  D 0.
REFERENCES
1. W. A. Shurcliff, Polarized Light, Harvard University Press, Cambridge, MA, 1962. A
condensed version was published with the same title, with S. S. Ballard as coauthor,
Book #7, Van Nostrand Momentum, Princeton, 1964.
2. H. G. Gerrard, “Transmission of light through birefringent and optically active media; the
Poincaré sphere,” J. Opt. Soc. Am. 44(8), 634–640 (1954).
3. P. C. Robinson and S. Bradbury, Quantitative Polarized-Light Microscopy, Oxford University Press, Oxford, 1992.
4. R. Ulrich, S. C. Rashleigh, and W. Eickhoff, “Bending-induced birefringence in singlemode fibers,” Opt. Let. 5(6), 273–275 (1980).
5. Y. Imai, M. A. Rodorigues, and K. Iizuka, “Temperature insensitive fiber coil sensor for
altimeters,” Appl. Opt. 29(7), 975–978 (1990).
6. H. C. Lefevre, “Single mode fiber fractional wave devices and polarization controllers,”
Electron. Let. 16(20), 778–780 (1980).
7. K. Shiraishi, H. Hatakeyama, H. Matsumoto, and K. Matsumura, “Laminated polarizers
exhibiting high performance over a wide range of wavelength,” J. Lightwave Technol. 15(6),
1042–1050 (1997).
8. K. Baba, Y. Sato, T. Yoshitake and M. Miyagi, “Fabrication and patterning of
submicrometer-thick optical polarizing films for 800 nm using streched silver island
multilayers,” Opt. Rev., 7(3), 209–215 (2000).
9. C. Viney, Transmitted Polarized Light Microscopy, McCrone Research Institute, Chicago,
1990.
10. K. Okamoto, T. Hosaka, and J. Noda, “High birefringence fiber with flat cladding,” J.
Lightwave Technol. LT-3(4), 758–762 (1985).
11. M. Takagi, Y. Kubo, E. Sasaoka, and H. Suganuma, “Ultra-wide bandwidth fiber polarizer,”
Symposium Digest of the 8th Fiber Sensor Conference, Monterey, CA, pp. 284–287, 1992.
12. H. Fujikake, K. Takizawa, T. Aida, H. Kikuchi, T. Fujii, and M. Kawakita, “Electricallycontrollable liquid crystal polarizing filter for eliminating reflected light,” Opt. Rev. 5(2),
93–98 (1998).
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POLARIZATION OF LIGHT
13. A. Yariv and P. Yeh, Optical Waves in Crystals, Wiley, New York, 1984.
14. D. S. Kliger, J. W. Lewis, and C. E. Randall, Polarized Light in Optics and Spectroscopy,
Academic Press, Boston, 1990.
15. T. J. Kane and R. L. Byer, “Monolithic unidirectional single-mode Nd:YAG ring laser,”
Opt. Let. 10(2), 65–67 (1985).
16. G. James, Advanced Modern Engineering Mathematics, 2nd ed. Addison-Wesley, Reading,
MA, 1999.
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