Download - Lecture15 Polarization
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Chapter 24
ElectromagneticWaves
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24.6 Polarization
A transverse wave is linearly polarized
when its vibrations always occur along
one direction.
A linearly polarized wave on a rope
canpass through a slit that is parallel
to the direction of the rope vibrations.
The rope wave cannotpass througha slit that is perpendicular tothe
vibrations.
POLARIZED ELECTROMAGNETIC WAVES
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24.6 Polarization
In polarized light, the electric field
of the electromagnetic wave fluctuates
along a single direction.
Unpolarized lightconsists of short
bursts of electromagnetic waves emitted
by many different atoms. The electricfield directions of these bursts are
perpendicular to the direction of wave
travel, as in polarized light, but are
distributed randomly about it.
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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.
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24.6 Polarization
MALUS
LAW !2
coso
SS =
intensity beforeanalyzer
intensity after
analyzer
Consider two sheets of polarizing material, one to polarize unpolarized light
(Polarizer) and one at an angle !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)
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24.6 Polarization
Example 7 Using Polarizers and Analyzers
What value of !should be used so the average intensity of the polarized
light reaching the photocell is one-tenththe average intensity of the
unpolarized light?
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24.6 Polarization
!2
5
1cos=
5
1cos =! !
4.63=!
Applying MalusLaw at the Analyzer: S= S0,polarcos2!
But S0,polar= (1/2)S0 and we want S= (1/10)S0
So (1/10)S0= (1/2)S0cos2
!
S0 S0,polar S
! !
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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 S2 S3
S1= S0/2
S2= S1cos2!
= (S0/2) cos2!
S3= S2cos2(90 - !)
= (S0/2) cos2!sin2!
Clearly, S3> 0 as long as
!!0 or 90o
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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.
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Polaroid Sun Glasses
Besides sunlight re-radiated from atmospheric molecules, there are
other sources of horizontally polarizedlight that occur in nature, such
as sunlight reflected from horizontal surfaces such as lakes.
Polaroid sun glassestake advantage of this fact by using polarizers
with their axes oriented vertically:
!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!!.).
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24.6 Polarization
IMAX movie projector Movie viewer using polarized glasses
Another application using polarized glasses: watching 3-D movies.
In a 3-D movie, two separate rolls of film are projected using a projectorwith 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.
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Example. Partially polarized and partially unpolarized light.
A light beam passes through a polarizer whose transmission axis makes
angle !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/m2when != 20.0oand has a maximum
value of 8.0 W/m2when the angle is != !max.
!Find a) S0,unpolar, and b) S0,polar.
Incident light
S0= S0,polar+ S0,unpolar
Transmitted light
S= Spolar+ Sunpolar
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Incident lightS0= S0,polar+ S0,unpolarTransmitted lightS= Spolar+ Sunpolar
a) Minimum transmitted intensity S= Smin= 2.0 W/m2at != 20.0o.
Sis minimum when Spolar= 0 since Sunpolaris not effected by !.
Sunpolar= Smin- Spolar= 2.0 - 0 = 2.0 W/m2
Sunpolar= (1/2)S0,unpolar ! S0,unpolar= 2Sunpolar= 2(2.0) = 4.0 W/m2
b) Maximum transmitted intensity S= Smax= 8.0 W/m2occurs at !max.
Sis maximum when Spolar= S0,polar(when != 20o+ 90o= 110o= !max)
Smax= S0,polar+ Sunpolor!S0,polar= Smax- Sunpolar= 8.0 - 2.0 = 6.0 W/m2