W9_parallel_resonance.eps

Week 11: Light 383

examples of linearly polarized light propagating in thez-direction with frequencyω:

Light linearly polarized in thex-direction:

E~(z, t) =E 0 xxˆsin(kz−ωt) (935)

(The associated magnetic fieldmustbe:

B~(z, t) =B 0 yyˆsin(kz−ωt) =E^0 x

c

yˆsin(kz−ωt) (936)

according to the rules derived in the previous chapter, because

|B~|=|

E~|

c

(937)

and because

xˆ×ˆy=zˆ (938)

in the Poynting vector.)

Light linearly polarized in they-direction:

E~(z, t) =E 0 yyˆsin(kz−ωt) (939)

(The associated magnetic fieldmustbe:

B~(z, t) =−B 0 xxˆsin(kz−ωt) =−E^0 y

c

ˆxsin(kz−ωt) (940)

according to the rules derived in the previous chapter, because

yˆ×−xˆ=zˆ (941)

in the Poynting vector.)

Finally, light linearly polarized along the line atπ/4 above thex-axis is::

E~(z, t) =

√

2

2

E 0 xˆsin(kz−ωt) +

√

2

2

E 0 yˆsin(kz−ωt) (942)

The amplitude of the electric field isE 0 (why?). What must the direction and magnitude

of the associated magnetic field?

Circularly Polarized Light

There is no reason that the magnitudes of the electric polarization components in the two

independent directions have to bethe sameor to bein phase. We start by considering

the case where they have the same magnitude but areπ/2 out of phase:

E~(z, t) =

√

2

2

E 0 xˆsin(kz−ωt±π/2) +

√

2

2

E 0 yˆsin(kz−ωt)

E~(z, t) =

√

2

2

E 0 (±ˆxcos(kz−ωt) +yˆsin(kz−ωt)) (943)

These two components describe a vector of constant length thatsweeps around in a

circle, either counterclockwise (-) or clockwise (+). We call thiscircularly polarized

light. Note that the two components must have equal amplitudes and must beπ/ 2

out of phase to be circularly polarized. There are two independenthelicitiesof circularly

polarized light: right (clockwise/+) and left (anticlockwise/-) when facinginthe direction

of propagation).