Polarization
Light is a transverse wave: the electric and magnetic fields both oscillate along axes that are
perpendicular to the direction of wave motion. The simplest kind of radiation is said to be
linearly polarized. For this radiation the electric field is directed along a single axis everywhere
along the light wave.
An example would be an electromagnetic wave whose fields take the form
E(z,t) = E0 e j (kz - t) i , B(z,t) = B0 e j (kz - t) j
This wave is traveling in the +z direction and is polarized along the x-axis.
Ordinary incandescent light consists of a mixture of polarizations, a mixture which is rapidly and
randomly varying. Such light is called unpolarized. Linearly polarized light can be produced
most easily by sending a beam of unpolarized light through a film of dichroic material. This is a
material that selectively
absorbs the light polarized
in one direction and
transmits the light with
complementary
polarization.
In the figure at the left,
incident unpolarized light
falls on a linear polarizer, a
sheet of dichroic material
that transmits light
polarized parallel to the
transmission axis and
absorbs light polarized at
right angles to that axis.
The emerging light is
6.1
linearly polarized with polarization axis parallel to the transmission axis of the polarizer.
When polarized light falls on
a linear polarizing sheet, only
the component of the incident
light polarized parallel to the
transmission axis is
transmitted while the
perpendicular component is
absorbed. This means that if
E0 is the amplitude of the
incident light and the angle
between the polarization axis
of the light and the
transmission axis of the
polarizer, the amplitude of the
transmitted light is E0 cos .
Since the intensity is
proportional to the square of
the amplitude, the transmitted
intensity is related to the incident intensity by
(13.1)
I = I0 cos2 .
This relationship is called the law of Malus and holds only for incident linearly polarized light.
Dichroism is not difficult to
understand Common dichroic
materials are formed from long
chain-like molecules with electrons
that can roam over the length of the
molecule. A linearly polarizing sheet
has these molecules preferentially
oriented in one particular direction in
the plane of the sheet, as in the
sketch at the left. When a light wave
passes through the sheet, the component of the electric field that is parallel to the molecular axis
will cause electrons to oscillate back and forth along the length of the molecule, thereby
transferring some of the light wave’s energy to electron motion. This electronic motion is resisted
by friction with the molecule and its environment and the electrons’ energy will eventually be
lost to heat. In this way, the component of electric field parallel to the molecule’s long axis is
absorbed. Light polarized oppositely to the long axis of the molecule cannot cause significant
electron motion and so is not absorbed. For the case illustrated, the vertical axis would be the
6.2
transmission axis.
The law of Malus tells us about how a linear polarizer affects unpolarized light. The electric field
of an unpolarized light wave moving in the z direction can be written as
(13.2)
E = E0 ( cos (t) i + sin (t) j ) sin(k z - t + (t))
\where and are a rapidly and randomly varying functions of time. The fluctuations occur on a
time scale of 10-13 s or less and it is usually appropriate to deal with the averaged intensity. The
average intensity of this wave is
(13.3) I0 = 0 c E0 2 [sin2 (t) + cos2 (t)] sin2 (kz - t + (t)) = 0 c E0 2 sin2 (kz - t+ (t)) = 0 c E0 2 / 2
If the transmission axis of a polarizer lies along the y-direction, the transmitted electric field is
(13.3)
E = E0 sin (t) sin(k z - t + (t)) j
so that the instantaneous transmitted intensity is I(t) = 0 c E0 2 sin2 (t) sin2 (k z - t + (t)).
The polarization and phase fluctuations are usually statistically independent. The phase
fluctuations arise from atomic collisions while the polarization fluctuations come from different
atoms contributing to the radiation, each with a different polarization. We are therefore justified
in taking the averages separately:
(13.4)
I =
0
c E0 2 sin2 (t) sin2 (k z - t + (t)) =
so that finally
(13.5)
I = I0 / 2
which holds for unpolarized light moving through a linear polarizer.
6.3
0
c E0 2 /4
Experiment 6.1: The Law of Malus
Apparatus: Two polarizers, one of which is adjustable and equipped with a protractor;
table lamp; pc spectrometer.
Procedure: Put the table lamp’s light through one polarizer and look at this light through
the second adjustable polarizer. Try to line up the transmission axis (TA) of this
polarizer so that the transmitted intensity is a maximum. Then place the spectrometer’s
fiber-probe on a fixed support so that it is receiving the light through both polarizers. Set
the integration time and averaging width for optimum data collection. Fine adjust the TA
so that the intensity received by the probe is a maximum.
Choose three wavelengths that collectively span most of the visible spectrum and
record the intensity at the three selected wavelengths. Then rotate the TA by 10( and
repeat the measurements. Continue in 10( increments until the intensity is almost zero.
Make a table of your data (in MAPLE). Subtract the dark reading from all entries. Note:
This may differ for each wavelength. For each wavelength, plot the intenisty and angle
using pointsplot where is the angle between the TA and its position at maximum
intensity. On the same graph plot I0cos2 where I0 is the maximum intensity. You will
have three graphs that collectively contain all your data as well as the angular
distribution predicted by the law of Malus. Comment on how well this law describes the
data.
6.4
Experiment 6.2: Quarter and Half Wave Plates
Apparatus: Polarizers, two quarter wave plates ( /4 plates) , mercury vapor lamp with
green filter, table light.
Procedure:
1.Using the green mercury light, cross the polarizer and analyzer (i.e. adjust the TA’s so
that no light is transmitted by the analyzer). Place a quarter wave plate between the
polarizer and analyzer. Rotate the plate and describe what happens.
2. Remove the polarizer and study the light coming through the /4 plate and the
analyzer. Then do the same using only the /4 plate and the polarizer. Can you explain
what the /4 plate is doing in terms of dichroism?
3. Next set up the crossed polarizer and analyzer and place two /4 plates between
them. Adjust various angles and describe what you see.
Question: Can you think of a model to describe this behavior?
6.5
Circular and Elliptic Polarization
Linear polarization is not the only way a light wave can be polarized. Consider the following
electric field:
(13.6)
E = E0 [-j i + j] e j (k z - t)
whose real part is E0 [sin(k z - t ) i + cos(k z - t) j ]
This describes a wave moving in the +z-direction but the phase of the x-component of the
radiation differs by 90( from that of the y-component. Examine this wave at the location z=0. At
time t=0, the electric field points along the +y axis. When t = T/4 (where T = period = 2 / ), the
electric field points along the -x-direction. At t = T/2, it points along the -y direction and at t =
3T/4 it points along the +x direction. At the fixed location z = 0 the electric field appears to rotate
counterclockwise.
Problem 1: Show that at a fixed time the electric field rotates clockwise with increasing z.
A wave like that of (13.6) is said to be left circular polarized (LCP). At a fixed location in space,
if one looks into the oncoming wave and points the thumb of the left hand into the wave (opposite
to the direction of wave motion), the fingers curl in the sense of rotation of the electric field. In the
opposite case, where the electric field rotates in the sense in which the fingers of the right hand
curl, we speak of right circular polarization (RCP).
Problem 2: Find whether the below electric fields correspond to LCP, RCP, or linear
polarization.
A.
E = E0 [-j i - j ]e j (k z - t)
B.
E = E0 [ i -j j ] e j (k z - t)
C.
E = -j E0 [ i + j ] e j (k z - t)
D.
E = E0 [e j / 4 i + e 3 j / 4 j ] e j (k z - t)
6.6
Problem 3: Find an expression for an electric field of a RCP wave moving in the -x direction.
(There is more than one correct answer).
A more general kind of polarization corresponds to rotating electric fields with different
amplitudes along different axes. As an example, consider the wave whose electric field is
(13.7) E(r, t) = [ E1 i + E2 e j j ] e j (k z -
t)
where E1 and E2 are real and E1 g E2 . At the location z = 0 this has the real part
(13.8) E(0, t) = E1 cos( t) i + E2 cos( t - ) j
The magnitude of this electric field is not constant. To see how it changes with time, note that
(13.10)
Ey = E2 cos( t - ) = E2 [ cos( t) cos( ) + sin( t) sin( ) ]
Take Ey / E2 from the above and subtract from it the quantity (Ex / E1) cos( ) to get
(13.11)
Ey / E2 - (Ex / E1) cos( ) = sin( t) sin( )
Now, sin2 ( t) = 1 - cos2 ( t) = 1 - (Ex / E1)2 so that
(13.12)
[ Ey / E2 - (Ex / E1) cos( ) ]2 = [1 - (Ex / E1)2] sin2 ( )
Rearrange terms to get
(13.13)
(Ey / E2)2 + (Ex / E1)2 - 2 (Ex / E1) (Ey / E2) cos( ) = sin2 ( )
This is the equation of an ellipse making an angle of
(13.14)
tan(2 ) = 2 E1 E2 cos( ) / [ E12 - E22 ]
This is graphed on the next page.
6.7
with the +Ex axis where
The situation described by (13.7) is called elliptical polarization. At a fixed point in space the
electric field vector rotates with its tip sweeping out the curve of the ellipse in one period of
oscillation. The polarization is either right or left elliptical polarization depending on the direction
of rotation of E.
Circular polarization is a special case of elliptical polarization, occurring when E1 = E2 and
/2.
Problem 4: What sort of polarization occurs when E1 = E2 and
=
=0?
Producing Circularly and Elliptically Polarized Light
Suppose that a material had a polarization-dependent refractive index. Light moving in the +z
direction through the material experiences a refractive index n1 if the light is polarized along the
x- axis and an index n2 if it is polarized along the y-axis where n1 g n2 . Such a material is called
birefringent. Monochromatic light originally polarized along the x-axis enters a thickness t of this
material and emerges still polarized along the x-axis having undergone a phase shift of
1 = 2 n1 t / .
Light initially polarized along the y-axis will emerge still polarized along the y-axis with a phase
shift of 2 = 2 n2 t / . What about light linearly polarized at 45( to the coordinate axes?
6.8
If the entering light has an electric field given by
(13.15)
Ein = (E0 /2) [ i + j ]e j (k z -
t)
,
the emerging light will have an electric field given by
(13.16)
Eout =
E0 j φ1
e i + e j φ2 j]e j ( kz −ωt )
[
2
so that travel through the material leads to a phase difference between the components polarized
along the x and y axes of
= 2 - 1 = 2 (n2 - n1) t / . When
= /2, the light polarized along
one axis will have traveled through an optical path length one quarter of a wavelength longer than
light polarized along the perpendicular axis. The emerging light in that case is circularly
polarized. A piece of material with this property is called a quarter-wave plate.
Light polarized along the x-axis moves through the birefringent material at a speed c/n1 while
light polarized along the y-axis moves at speed c/n2. If n1 < n2 , the x-axis is called the fast axis of
the material, the y-axis the slow one. The fast and slow axes refer to directions within the
birefringent material, not to any particular lines in space, so they might better be called fast and
slow directions. They are determined by the crystal structure of the birefringent material.
Problem 5: Linearly polarized light enters two sequential quarter-wave plates. The two plates
have their fast axes parallel (This is the same as a single half wave plate). The incident light is
polarized at 45( to these axes. Describe the polarization state of the transmitted light.
Problem 6: Linearly polarized light enters two sequential half-wave plates. The two plates have
their fast axes parallel (This is the same as a single full wave plate). The incident light is polarized
at 45( to these axes. Describe the polarization state of the transmitted light
Problem 7: In the below arrangement, light polarized at 45( to the fast and slow axes of a quarterwave plate enters beam splitter #1. The two resulting beams are brought together at beam splitter
#2, so that the emerging light is a 50-50 combination of linearly and circularly polarized light.
Assume that there are no other significant phase differences between the beams and that they have
equal amplitudes. Describe in detail (i.e. in words and with as much mathematical precision as
possible) the state of polarization of the light emerging from each of the exit ports of beam splitter
#2 . (Hint: Equation (A.10) may be useful).
6.9
6.10