POLARIZED LIGHT
OBJECTIVE: To illustrate the use of dichroic polarizers (“polaroids”) to produce and analyze
plane polarized light.
METHOD: Circular polarizers with divided circles are mounted in holders that allow the
polarizing directions of the polarizers to be set at any angle with an error of about plus or minus
2 degrees. The polarizing direction of the polarizers is the same as that of the diameter passing
through the 0 to 180 marks on the divided circle. Quantitative variations in light intensity are
measured with a photocell (a light-sensitive resistor) connected to an ohmmeter. We will use
the DMM as the ohmmeter. The photocell resistance varies inversely with the light intensity.
Thus, as light intensity goes up, the resistance reading goes down. Since these light meters will
not be calibrated, absolute intensities cannot be measured. However, we can write that the
intensity and corresponding resistance reading are related through the following equation,
R
BR '
I B
(1)
where B is a constant characteristic of the photocell circuit. Note that R’ is the resistance with
no light on the detector (I = 0). Solving this equation for the intensity gives
 R' 
I  B  1
R

(2)
While an absolute intensity cannot be determined from one reading, we can use Eq. (2) to find
a relative intensity (the ratio of intensity I at one time to an initial or reference intensity Io)
relative intensity 
I
R' / R  1 Ro (R'R )


Io R' / Ro  1 R(R'Ro )
(3)
where Ro is the resistance corresponding to Io and R is the resistance corresponding to I.
Observe that if R’ >> R and R’ >> Ro, then Eq. (3) reduces to the simpler form of
relative intensity 
I Ro

Io
R
(4)
Part 1: Beam Polarization
THEORY: Inserting a polarizer in an unpolarized beam should reduce the intensity by a factor of
two, and the intensity should be independent of the polarizer orientation.
PROCEDURE:
1) Place the photocell directly in front of the lamp without any polarizers in between and record
the resistance reading. Call this reading Ro .
2) Place a polarizer in the beam in front of the detector and check to see if there is any
Polarized Light 2
variation in intensity as the polarizer is slowly rotated. Note the angles at which intensity
maxima and minima occur, if there are any. Record the average resistance reading with the
polarizer in the beam, and call it R.
3) Now switch off the lamp and read the resistance. Call this meter reading R' .
REPORT:
1. Did you observe a variation of intensity as great as 10% as the polarizer was rotated? If not
you can assume the beam was unpolarized.
2. Are we justified in using Eq. (4) to calculate relative intensities? (Is R’ >> Ro? Is R’ >> R?)
3. Calculate the average relative intensity of the transmitted beam and compare with the
theoretical value of 0.5. What could cause any discrepancy?
Part 2: Law of Malus (I  cos²)
THEORY: Dichroic polarizers polarize light by selectively absorbing energy from the incoming
light. Energy is absorbed from an electromagnetic wave most efficiently if the electric field in
the wave is oscillating perpendicular to the polarization axis. The electric field of an incoming
wave can be broken down into two components: one perpendicular to the polarization axis and
one parallel to the axis. Thus, you can view the original wave as the superposition of two waves
whose electric fields are perpendicular. The polarizer effectively passes only the wave whose
electric field is parallel to the polarization axis. (It absorbs the energy from the wave whose field
component is perpendicular to the axis.) This “vector nature” of the polarization process is
reflected in the Law of Malus and demonstrated nicely with this simple experiment.
The Law of Malus states that the intensity of light transmitted through two polarizers is
proportional to the square of the cosine of the angle between the polarization axes of the
polarizers,
I = Io cos2
(5)
It is easy to derive. If the electric field passing through the first polarizer has an amplitude of Eo,
then the light passing through the second polarizer has an amplitude of Eocos where  is the
angle between the polarization axes of the polarizers. (See Figure 1.) This Eocos is simply
the component of the maximum field that makes it through the second polarizer. Since the
intensity of light is proportional to the square of the electric field amplitude, Malus’ Law follows.
Eo

Eo cos
Fig. 1: Law of Malus with Two Polarizers
PROCEDURE:
1) Switch the lamp back on and insert a second polarizer between the detector and the first
polarizer and set both at the same angle. The meter reading in this case is called Ro .
Polarized Light 3
2) Now rotate the second polarizer 90 in 10 increments and record the meter reading and
angle at each increment; in the calculations these meter readings will be called R.
3) Obtain a meter reading R' by turning off the lamp.
REPORT:
1. Calculate the relative intensities for the various values of .
2. PLOT a graph of I / Io versus cos2. The Law of Malus for two polarizers, Eq. (5), predicts
that the points should lie on a straight line with a slope of unity. Do they? Obtain the equation
of the best-fit line to answer this question. What could be the major sources of error to account
for any differences?
Part 3: Rotating the Polarization
THEORY: If two polarizers are oriented with their polarization axes crossed at 90, then no light
is predicted to pass through the second polarizer. However, if a third polarizer is placed
between the original two polarizers in the right orientation, then the transmitted light will
increase.
This phenomenon also can be understood by using the vector nature of polarization and the
Law of Malus. Examine Figure 2. The first and last polarizers have their polarization axes
crossed. The angle between the polarization axes of the first and middle polarizer is . The
angle between the polarization axes of the middle and last polarizer is . Note that  = 90-.
Io

Io cos2

Io cos2 cos2
Fig. 2: Law of Malus with Three Polarizers
According to the Law of Malus, the intensity of the light passing through the middle polarizer is
Io cos2 . Applying the law again gives the final transmitted intensity as Io cos2 cos2. Since
 = 90- and cos(90-) is equal to sin, we expect this transmitted intensity to be
I = Io cos2 sin2
(6)
In essence, the middle polarizer is rotating the polarization axes of the light that comes through
the first polarizer so that a component of the rotated electric field is now passed by the final
polarizer.
Polarized Light 4
PROCEDURE:
1) Place all three polarizers between the lamp and the detector. Initially set all three
polarization axes parallel to one another. Record the resistance. This value will serve as Ro.
2) Now rotate the last polarizer so that its axis is perpendicular to the axes of the first two
polarizers. Note that you now have the set-up shown in Figure 2 with  = 0. Record the
resistance R. Now increase  in 5 increments to 45. Record the resistance R for each
setting.
REPORT:
1. Calculate the relative intensities for the various values of  using Eq. (4).
2. PLOT a graph of I / Io versus cos2 sin2. The Law of Malus for three polarizers, Eq. (6),
predicts that the points should lie on a straight line with a slope of unity. Do they? Obtain the
equation of the best-fit line to answer this question.
3. At what angle  is most light transmitted? Does this angle make sense? WHY?
Part 4: Polarization by Reflection
THEORY: When unpolarized light reflects off a surface, the reflected light will become partially
linearly polarized. The degree to which it becomes linearly polarized depends on the reflector
material and the angle of incidence. At one angle of incidence, called the polarization angle or
the Brewster angle (B), the reflected light is completely linear polarized. For light traveling in
air that then reflects off glass with refractive index n, the Brewster angle is given by
tan(B) = n
(7)
PROCEDURE:
1) To observe polarization by reflection, first look at the reflection of the room lights by the
desk. Then look at this reflection through a polarizer. Note that as you rotate the polarizer, the
reflected glare is preferentially reduced at one orientation of the polarizer. Notice the orientation
of the polarization axis of the polarizer when the reflected light is diminished the most. What
can you conclude about the polarization direction of the reflected light?
2) Now let's use the laser to look at this more closely. Place a polarizer in the beam to linearly
polarize the laser light. Next, stand a glass plate on its edge in the polarized beam. Rotate the
glass plate about a vertical axis and look for a position at which the reflected beam vanishes (or
almost vanishes). Try various orientations of the polarizer until you obtain no reflection. The
angle of incidence when this condition is obtained is the Brewster angle. Measure the Brewster
angle. With the glass plate at the Brewster angle, but without the initial polarizer in place, insert
a polarizer in the reflected beam and show that it is polarized. Determine the direction of the
polarization of the reflected light. Check to see if the transmitted beam is polarized.
REPORT:
1. Is reflected light polarized? Describe the direction of the polarization.
2. Calculate the value for the refractive index of the glass plate using Eq. (7) and your
measured value of the Brewster angle. Compare this index to the value obtained in the earlier
experiment involving refraction and virtual image formation by the glass plate.
3. Describe how and why ‘polaroid’ sunglasses reduce glare.