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Lecture 5 Polarization
5- 1
5.0 Introduction
We have already established that light may be treated as a transverse electromagnetic wave. Thus far we have
consider only linearly polarized or plane polarized light, that is light for which the orientation of the electric
field is constant although its magnitude and sign varies in time. The electric field or optical disturbance


therefore resides in what is known as the plane of vibration. That fixed plane contains both E and k , the
electric field vector and the propagation vector in the direction of motion. Imagine now that we have two
harmonic, linearly polarized waves of the same frequency, moving through the same direction. If their
electric field vectors are collinear, the superimposing disturbances will simply combine to form a resultant
linearly polarized wave. Its amplitude and phase were examined in detail, under a diversity of conditions in
the previous chapter when we considered the phenomenon of interference. On contrary, if two light waves are
such that their respective electric field directions are mutually perpendicular, the resultant wave may or may
not be linearly polarized. Exactly what form that light will take(i.e its state of polarization), how we can
observe it, produce it, change it and make use of it will be the concern of this chapter.
Lecture 5 Polarization
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5.1 Nature of polarized light
5.1.1 Linear polarization
Fig. 5.1(a) shows an electromagnetic wave with its electric field oscillating parallel to the vertical y axis.

The plane containing the E vector is called the plane of oscillation or vibration. We can represent the
wave’s polarization by showing the extent of the electric field oscillations in a “head-on” view of the
plane of oscillation, as in Fig. 5.1 (b).
Fig. 5.1 (a) The plane of
oscillation of a polarized
electromagnetic wave. (b) To
represent the polarization,
we view the plane of
oscillation “head-on” and
indicate the amplitude of the
oscillating electric field.
Consider two orthogonal optical disturbances

Ex ( z, t )  iˆE0 x cos(kz  t )
and

E y ( z, t )  ˆjE0 y cos( kz  t   )
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(5.1)
(5.2)
where  is the relative phase difference between the waves, both of which are traveling in the z-direction.
The resultant optical disturbance is the vector sum of these two perpendicular waves:



E ( z, t )  Ex ( z, t )  E y ( z, t ).
(5.3)
If   0 or is an integral multiple of  2 , the waves are said to be in phase. In that
particular case Eq. 5.3 becomes

E ( z, t )  (iˆE0 x  ˆjE0 y ) cos(kz  t ).
(5.4)
The resultant wave therefore has a fixed amplitude i.e it is a linearly polarized wave. , as shown in Fig.
5.2.
This process of addition can be carried out equally well in reverse; that is, we resolve any plane-polarized
wave into two orthogonal components.
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Fig. 5.2 Linear light.
This process of addition can be carried out equally well in reverse; that is, we resolve any plane-polarized
wave into two orthogonal components.
Suppose now that  is and odd integer multiple of
  . The two waves are said to be 1800 out of phase
and

E ( z, t )  (iˆE0 x  ˆjE0 y ) cos(kz  t )
This wave is again a linearly polarized but the plane of polarization has been rotated from that of the
previous case.
5.1.2 Circular polarization
Now we consider another particular case. That is, E0 x  E0 y  E0 , in addition,
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    2  2m ,
where m  0,  1,  2,  . Accordingly,

E x ( z, t )  iˆE0 cos(kz  t )

E y ( z, t )  ˆjE0 sin( kz  t ).
and
(5.5)
(5.6)
The resultant wave is given by

E  E0 [iˆ cos(kz  t )  ˆj sin( kz  t )],
Notice now that the scalar amplitude of

E
(5.7)
, which is equal to E0 is a constant. But the direction of

E
is
time varying and it is not restricted as before to a single plane.
Let see what happens if
3
and wt 
.
4
Case 1: kz 

4
kz 

4
for example, we will consider four cases when wt  0, wt 

4
, wt 
and wt  0


E x ( z , t )  iˆE 0 cos( )
4


E y ( z , t )  ˆjE 0 sin( )
4



E  iˆE 0 cos( )  ˆjE 0 sin( )
4
4

2
Case 2: kz 

4
and wt 

 
E x ( z , t )  iˆE 0 cos(  )
4 4

 
E y ( z , t )  ˆjE 0 sin(  )
4 4
Case 3: kz 

4
and wt 

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4
1
11
1

E  iˆE0

1
21
1
2



E  iˆE 0 cos( )  ˆjE 0 sin( )
4
4

3
Case 4: kz 
and wt 
4
2

E  iˆE0
1
41
1
1
31
1

The resultant electric field vector E is rotating clockwise at an angular frequency of  . Such a wave is
said to be right-circularly polarized. Figure (5.3) below
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Fig 5.3 Right-circular light.
In comparison, if    2  2m , where m  0,  1,  2,  , then

E  E0 [iˆ cos(kz  t )  ˆj sin( kz  t )].

The amplitude is unaffected, but E
(5.8)
now rotates counter-clockwise, and the wave is referred to as left-
circularly polarized.
A linearly polarized wave can be synthesized from two oppositely polarized circular waves of equal
amplitude. In particular, if we add Eq. 5.7 to Eq. 5.8, we get a linearly polarized wave,

E  2E0iˆ cos(kz  t ).
(5.9)
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5.1.3 Elliptical polarization
Both linear and circular light may be considered to be special cases of elliptically polarized, or more

simply, elliptical light. By that we mean that, in general, the resultant electric field vector E will both

rotate and change its magnitude as well. In such cases the endpoint of E will trace out an ellipse, in a

fixed plane perpendicular to k as the wave sweeps by.

We can better see this by actually writing an expression for the curves traverses by the tip of E . To that
end recall that
Ex  E0 x cos(kz  t )
and
E y  E0 y cos( kz  t   ).
(5.10)
(5.11)
The equation of the curve we are looking for should neither be a function of position nor time, i.e we
should be able to get rid of the (kz  t   ) dependence.
Expand the expression of E y into
E y E0 y  cos( kz  t ) cos   sin( kz  t ) sin 
and combine it with Ex E0 x  cos(kz  t ) to yield
Ex E0 x  cos(kz  t )
Ey
E0 y

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Ex
cos    sin( kz  t ) sin  .
E0 x
(5.12)
It follows from Eq. 5.10 that
sin( kz  t )  [1  ( E x E0 x ) 2 ]1 2 ,
So Eq. 5.12 leads to
  E 2 
 Ey

E

 x cos    1   x   sin 2  .
E
 
 0 y E0 x
   E0 x  
2
Finally, on rearranging terms, we have
2
 E y   Ex 
 E x  E y 

 

 cos   sin 2  ..




2
 E  E 
E  E 
 0 x  0 y 
 0 y   0x 
2
This is the equation of an ellipse making an angle
system (Fig. 5.4) such that
tan 2 
2 E0 x E0 y cos 
E02x  E02y
.
(5.13)
 with the (Ex, Ey)-coordinate
(5.14)
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Fig. 5.4 Elliptical light.
If
  0 or equivalently     2,  3 2,  5 2 ,  , we have familiar form
2
 E y   Ex 

 
  1.
E  E 
 0 y   0x 
2
(5.15)
Furthermore, if E0 y  E0 x  E0 , this can be reduced to
E y2  Ex2  E02 .
(5.16)
Clearly, it is a circle.
If

is an even multiple of
 , Eq. 5.13 results in
Ey 
E0 y
E0 x
Ex
(5.17)
And similarly for odd multiples of
Ey  
,
E0 y
E0 x
Ex .
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(5.18)
These are both straight lines having slopes of  E0 y E0 x
; in other words, we have linear light. So, both
linear and circular light may be considered to be special cases of elliptically polarized light.
Fig. 5.5 gives various polarization configurations. This very important diagram is labeled across the bottom
“ E x leads by E y : 0,  4,  2, 3 4 ,  , ” where these are the positive values of

to be used in Eq.
5.2.
Fig. 5.5 (a) Various polarization configurations.
(b) .
leads
by E x , or alternatively ,
leads
Ey
by
Ex
3 2
.
Ey
 2
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We are now in a position to refer to a particular light wave in terms of its specific state of polarization.
If the light is linearly, or plane polarized, then we say it is the P-state. Light that is right circularly
polarized is in the R-state. Light that is left circularly polarized is in the L-state. Finally, elliptically
polarized light is referred to as being in the E-state.
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5.1.4 Natural light
The electromagnetic waves emitted from any common source of light are polarized randomly or
unpolarized. That is, the electric field changes directions randomly.
We can use the mess like that in Fig. 5.6 (a) to represent the unpolarized light. In principle, we can
simplify the mess by resolving each electric field in Fig. 5.6 (a) into y and z components and then finding
the net fields along the two directions. In doing
Fig. 5.6 (a) and (b) Two
different drawings to
represent natural light.
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so, we mathematically change unpolarized light into the superposition of two polarized waves
whose planes of oscillation are perpendicular to each other. The result is the double-arrow
representation of Fig. 5.6 (b), which simplifies drawing of unpolarized light. Actually, light is
generally neither completely polarized nor completely unpolarized. More often, the electric
field vector varies in a way that is neither totally regular nor totally irregular, and one refers to
such an optical disturbance as being partially polarized. For this situation, we can draw one of
the arrows of the double-arrow representation longer than the other arrow.
5.2 Polarizers
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Now that we have some idea of what polarized light is, the next global step is to develop and
understanding of the techniques used to generate it, change it, and in general manipulate it to feet
our needs. An optical device whose input is natural light and whose output is some form of
polarized light is a polarizer.
Fig. 5.7 A linear polarizer.
Fig. 5.7 shows a linear polarizer. Depending on the form of the output, we could also have circular or
elliptical polarizers.
Polarizing direction of a linear polarizer: An electric filed component parallel to the polarizing direction
is passed (transmitted) by a polarizer; a component perpendicular to it is absorbed. Polarizers take on
many different configurations as we shall see, but they are all based on one of four fundamental
physical mechanisms: dichroism or selective absorption, reflection, scattering and birefringence or
double refraction
5.2.1 Malu’s Law
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Fig. 5.7 A linear polarizer.
How do we determine experimentally whether or not a device is actually a linear polarizer?
By definition, if natural light is incident on an ideal linear polarizer Fig(5.7), only light in a p-state will
be transmitted. That P-state will have an orientation parallel to a specific direction which we will call
the transmission axis of the polarizer. In order words, only the component of the optical field parallel
to the transmission axis will pass through the device essentially unaffected.
If the polarizer in Fig(5.7) is rotated about the z-axis the reading of the detector (e.g a photocell) will
be unchanged because of the complete symmetry of the unpolarized light. Because of the very high
frequency of light the detector will, for practical reasons, measure only the incident irradiance. Since
the irradiance is proportional to the square of the amplitude of the electric field(
need only concern ourselves with amplitude.
) we
5.2.1 Malu’s Law
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Now suppose that we introduce a second identical ideal polarizer or analyzer (Fig5.8) whose
transmission axis is vertical. If the amplitude of the electric field transmitted by the polarizer
is E0, only its component E0cosθ, parallel to the transmission axis of the polarizer (Fig 5.9)
will be passed on to the detector(assuming no absorption). The irradiance reaching the
detector is then given by
θ is the angle between the polarizer P1 and the
analyzer P2
Fig. 5.9 Polarized light
approaching a linear polarizer.
Fig. 5.8 A linear polarizer and analyzer.
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, occurs when the angle θ between the transmission
The maximum irradiance,
axes of the analyzer and polarizer is zero. The expression
can acccordingly
be rewritten as
This is known as Malus’s Law.
If θ=90, then I(90)=0. This arises from the fact that the electric field that has passed through
the polarizer is perpendicular to the transmission axis of the analyzer(the two devices so
arranged are said to be crossed). The field is therefore parallel. That is called the extinction
axis of the analyzer and hence obviously has no component along the transmission axis.
The set-up of Fig (5.8)
can be used along with Malus’s Law to determine whether or not a
particular device is in fact a linear polarizer.
5- 19
Fig. 5.10 Polarizing sunglasses
consist of sheets whose
polarizing directions are vertical
when the sunglasses are worn. (a)
Overlapping sunglasses transmit
light fairly well when their
polarizing directions have the
same orientation, but (b) they
block most of the light when ther
are crossed.
5.2.2 Polarization by reflection
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There is another way that we can produce hundred percent linearly polarized light, and we can do
that by reflecting unpolarized light off a dielectric. For instance, water or glass. None of this
follows from Snell’s, Snell’s Law was two hundred fifty years before Maxwell, polarization
wasn’t even known in the days of Snell. But Maxwell’s equations allow you to properly deal with
refraction and reflection including polarization. And I will take no attempt to derive this for ou in
detail but I will present you wit some results so that you can a least appreciate the far-reaching
consequences of the reflection in which we can produce a hundred percent polarized light.
Suppose we have light unpolarized light coming in here, medium one , index of refraction n1.
⏊
⏊
refl
inc
θ1
θ1
θ2
⏊
trans
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The light is coming at an incident angle of θ1 unpolarized. Some of it is reflected into medium 1
and the angle of reflection is θ1 .Some the light is refracted into medium 2(index of refraction n2)
and the angle of refraction is called θ2 .
So this light, unpolarized, comes in, reflects and refracts.
If we want to use Maxwell’s equation at the surface of the two media, we should decompose the
electric vector of the incoming light in to two components. One which perpendicular to the plane
of incidence(Same as the board), and the other parallel to the plane of incidence(in the board). We
will have to do the same for the reflected and refracted light.
The incident is unplarized so there is no preferred direction, that means that in the representation
that we have the strength of the two component is equal, because if one were stronger than the
other, then the light wouldn't be unpolarized.
What Maxwell’s equations now can do for us is a lot of work. It can relate the parallel component
in reflection with the parallel component of incidence, the parallel component of refraction with
the parallel component of incidence. It gives two relations, two equations.
5- 22
It can also relate the perpendicular component of the reflection component with the
perpendicular component of incidence and the perpendicular component of refraction with the
perpendicular component of incidence so we get another two sets of equations.
Only one of the four equations will be given today.
(5.19)
If we apply Snell’s Law equation (5.19) can be simplify to
(5.20)
There is something very special hidden in this equation (5.20)
the
the downstairs of equation (5.19)
parallel component in reflection is zero. So if
and that is, when
is infinitely large and so that means the
in reflection goes to zero, there is only
left, which is not zero and means the reflected light is now hundred percent polarized in
this direction, because we have killed
component completely.
But this only works if the condition
is met. If the condition
5- 23
is met
then it follows from high school math that.
If we remember Snell’s Law we can replace only for this case,
by
. Then we will
have
Equation (5.21) is the secret to getting hundred percent polarized light, and the angle θ1 is call
the Brewster angle. If for instance we look at the transition from air to glass, glass has an index
of refraction approximately 1.5 depending of the kind of glass that you have. For that particular
case the Brewster angle is about
5.2.3 Polarization by Dichroism
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Dichroism is a selective absorption of one polarization plane over the other during the
transmission through a material. In the laboratory, sheet polarizers or polaroids are the most
used type of polarizers. They are manufactured from an organic material imbedded into plastic
sheet. The sheet is stretched aligning molecules and causing them to be birefringent. The
molecules selectively attach themselves to aligned polymer molecules, so that absorption is
high in one plane and weak in the other. The transmitted beam is linearly polarized.
5.2.4 Polarization of scattered light
Let us consider light passing through a gas. The oscillating electrons are separated by large
distances and act independently of one another. If unpolarized light beam falls upon a gas and
we are situated at right angles to them, we will see that scattered light is polarized. This
situation is the one we see when we look at sunlight through a polarizer under a cloudless sky.
We notice that it is partially polarized.
5.2.5 Polarization by double refraction
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In gases, liquids, solids like glass or cubic crystals, the velocity of light, i.e the refraction index
does not depend on the direction of propagation, they are said to be optically isotropic.
Anisotropy means that one or many properties of a substance depend on direction, due to the
arrangement of atoms being different in different directions throughout the volume. For
example in many crystals the refractive index depends on direction of propagation. This is
called “double refraction” or “birefringence”.
A birefringent crystal such as calcite will divide an incident beam of monochromatic light into
two separate beams having polarizations perpendicular to each other.
The difference in behavior between the two beams may be used to make birefringent crystal
polarizers.