Chapter 24
Electromagnetic
Waves
24.6 Polarization
POLARIZED ELECTROMAGNETIC WAVES
A transverse wave is linearly polarized
when its vibrations always occur along
one direction.
A linearly polarized wave on a rope
can pass through a slit that is parallel
to the direction of the rope vibrations.
The rope wave cannot pass through
a slit that is perpendicular to the
vibrations.
24.6 Polarization
In polarized light, the electric field
of the electromagnetic wave fluctuates
along a single direction.
Unpolarized light consists of short
bursts of electromagnetic waves emitted
by many different atoms. The electric
field directions of these bursts are
perpendicular to the direction of wave
travel, as in polarized light, but are
distributed randomly about it.
24.6 Polarization
Polarized light may be produced from unpolarized light with the aid of
polarizing material.
The intensity of the transmitted polarized light has 1/2 the intensity of
the incident unpolarized light (perpendicular component blocked).
The transmission axis of the material is the direction of polarization
of the light that passes through the material.
24.6 Polarization
Consider two sheets of polarizing material, one to polarize unpolarized light
(Polarizer) and one at an angle θ to the polarized light (Analyzer). The
intensity of the light after the analyzer is related to the intensity before the
analyzer by Malus’ Law:
( Intensity ~ E2 )
MALUS’ LAW è
intensity after
analyzer
S = S o cos 2 θ
intensity before
analyzer
24.6 Polarization
Example 7 Using Polarizers and Analyzers
What value of θ should be used so the average intensity of the polarized
light reaching the photocell is one-tenth the average intensity of the
unpolarized light?
24.6 Polarization
S0
S0,polar
S
Applying Malus’ Law at the Analyzer:
S = S0,polar cos2 θ
But
S0,polar = (1/2)S0 and we want S = (1/10)S0
So
(1/10)S0 = (1/2)S0 cos2 θ
1
5
= cos 2 θ
è
cos θ =
1
5
è
θ = 63.4
24.6 Polarization
Conceptual Example 8 How Can a Crossed Polarizer and
Analyzer Transmit Light?
Suppose that a third piece of polarizing material is inserted between the
crossed polarizer and analyzer. Does light now reach the photocell?
S0
S1 = S0/2
S2 = S1 cos2 θ
= (S0/2) cos2 θ
S1
S2
S3
S3 = S2 cos2 (90 - θ)
= (S0/2) cos2 θ sin2 θ
Clearly, S3 > 0 as long as
θ ≠ 0 or 90o
24.6 Polarization
Polarized light is produced by the scattering of unpolarized
sunlight by molecules in the atmosphere.
Molecules re-radiate sunlight.
As you look at larger and larger
angles with respect to the
incident sunlight, the re-radiated
light becomes more and more
horizontally polarized.
Polaroid Sun Glasses
Besides sunlight re-radiated from atmospheric molecules, there are
other sources of horizontally polarized light that occur in nature, such
as sunlight reflected from horizontal surfaces such as lakes.
Polaroid sun glasses take advantage of this fact by using polarizers
with their axes oriented vertically:
è in addition to 1/2 of the unpolarized light which is already blocked by
the polarizers, all of the horizontally polarized light is completely blocked,
thus blocking out some of the reflected light which can confuse you
during some outdoor activities (e.g. driving, piloting, fishing…….).
24.6 Polarization
Another application using polarized glasses: watching 3-D movies.
In a 3-D movie, two separate rolls of film are projected using a projector
with two lenses, each with its own polarizer. The two polarizers are
crossed. Viewers watch the action on-screen through glasses that have
correspondingly crossed polarizers for each eye.
IMAX movie projector
Movie viewer using polarized glasses
Example. Partially polarized and partially unpolarized light.
A light beam passes through a polarizer whose transmission axis makes
angle θ with the vertical. The beam is partially polarized and
partially unpolarized, and the average intensity of the incident light,
S0, is the sum of the average intensity of the polarized light, S0,polar,
and the average intensity of the unpolarized light, S0,unpolar.
As the polarizer is rotated clockwise, the intensity of the transmitted light
has a minimum value of 2.0 W/m2 when θ = 20.0o and has a maximum
value of 8.0 W/m2 when the angle is θ = θmax.
è Find a) S0,unpolar, and b) S0,polar.
Incident light
S0 = S0,polar + S0,unpolar
Transmitted light
S = Spolar + Sunpolar
Incident light
S0 = S0,polar + S0,unpolar
Transmitted light
S = Spolar + Sunpolar
a) Minimum transmitted intensity S = Smin = 2.0 W/m2 at θ = 20.0o.
S is minimum when Spolar = 0 since Sunpolar is not effected by θ.
Sunpolar = Smin - Spolar = 2.0 - 0 = 2.0 W/m2
Sunpolar = (1/2)S0,unpolar è S0,unpolar = 2Sunpolar = 2(2.0) = 4.0 W/m2
b) Maximum transmitted intensity S = Smax = 8.0 W/m2 occurs at θmax.
S is maximum when Spolar = S0,polar (when θ = 20o + 90o = 110o = θmax)
Smax = S0,polar + Sunpolor è S0,polar = Smax - Sunpolar = 8.0 - 2.0 = 6.0 W/m2