Chapter 24

Electromagnetic Waves

24.6 Polarization

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.

POLARIZED ELECTROMAGNETIC WAVES

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

MALUS’ LAW è θ2cosoSS =

intensity before analyzer intensity after

analyzer

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 )

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

θ251 cos= 51cos =θ 4.63=θ

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

S0 S0,polar S

è è

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 = S1 cos2 θ = (S0/2) cos2 θ

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

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 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.

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