W9_parallel_resonance.eps

384 Week 11: Light

Elliptically Polarized Light

If the amplitudes of the two waves are (potentially) differentandthe two waves are

(potentially) out of phase, the most general polarization state is that ofellipticalpolar-

ization:

E~(z, t) =E 0 xxˆsin(kz−ωt+δx) +E 0 yyˆsin(kz−ωt+δy) (944)

In this expression,E 0 xandE 0 ymay or may not be equal, and the phasesδxandδymay

or may not be zeroorequal. The amplitudes of thexandylimits define a rectangular

box. The electric field vector rotates within that box wit the box tipped at an angle

relative determined by the relative phase differenceδ=δx−δy(where ifδ= 0 orδ=π

one has linear polarization).

To see a lovely animation of the electric field vector for various flavors of polarization,

visit:

http://www.nsm.buffalo.edu/∼jochena/research/opticalactivity.html

Polarization by Absorption (Malus’s Law)

A polaroid filter is made by putting oriented conducting threads into atransparent

medium in such a way that long currents in those threads created bythe polarization

component of light parallel to the thread heats the threads, absorbing and attenuat-

ingonlythat component of the incident polarized or unpolarized light and passing the

component perpendicular to the threads (thetransmission axisof the filter).

The rules for transmission are simple. If the incident light is unpolarized, on average

half its energy is polarized in either polarization direction. Therefore(assuming that the

filter is “ideal” and otherwise fully transparent):

Itransmitted=

Iincident

2

(945)

The transmitted light is fully linearly polarized in the direction of the transmission axis

of the filter.

If the light that is incident on the filter is already polarized, then only thecomponentof

the electric field vector that isparallelto the transmission axis is transmitted. That is:

Etransmitted=E~·tˆ=Eincidentcos(θ) (946)

whereθis the angle between the direction of linear polarization of the incidentlight and

a unit vector along the transmission axis.

To find the transmittedintensity, we need just remember the relation between the electric

field strength and the intensity that follows from the intensity beingthe time-average

magnitude of the Poynting vector:

I=

∣∣

∣∣^1

2 μ 0

E~×B~

∣∣

∣∣=^1

2 μ 0 c

E^2 (947)

The intensity is directly proportional to the electric field amplitude, squared, so that:

Itransmitted=Iincidentcos^2 (θ) (948)

This result is known asMalus’s law.