Page 418 - Engineering Electromagnetics, 8th Edition

400 ENGINEERING ELECTROMAGNETICS

can be passed through a polarizer of any orientation, thus yielding linearly polarized
light in any direction (although one loses half the original power this way). Other
uses involve treating linearly polarized light as a superposition of circularly polarized
waves, to be described next.
Circularly polarized light can be generated using an anisotropic medium—a
material whose permittivity is a function of electric field direction. Many crystals
have this property. A crystal orientation can be found such that along one direction
(say, the x axis), the permittivity is lowest, while along the orthogonal direction
(y axis), the permittivity is highest. The strategy is to input a linearly polarized wave
with its field vector at 45 degrees to the x and y axes of the crystal. It will thus have
equal-amplitude x and y components in the crystal, and these will now propagate in
the z direction at different speeds. A phase difference (or retardation) accumulates
between the components as they propagate, which can reach π/2if the crystal is long
enough. The wave at the output thus becomes circularly polarized. Such a crystal, cut
to the right length and used in this manner, is called a quarter-wave plate, since it
introduces a relative phase shift of π/2 between E x and E y , which is equivalent to λ/4.
It is useful to express circularly polarized waves in phasor form. To do this, we
note that (96) can be expressed as
e
E(z, t) = Re E 0 e jωt − jβz a x + e ± jπ/2 a y

Using the fact that e ± jπ/2 =± j,we identify the phasor form as:

E s = E 0 (a x ± ja y )e − jβz (100)
where the plus sign is used for left circular polarization and the minus sign for right
circular polarization. If the wave propagates in the negative z direction, we have

E s = E 0 (a x ± ja y )e + jβz (101)

where in this case the positive sign applies to right circular polarization and the minus
sign to left circular polarization. The student is encouraged to verify this.

EXAMPLE 11.7
Let us consider the result of superimposing left and right circularly polarized fields
of the same amplitude, frequency, and propagation direction, but where a phase shift
of δ radians exists between the two.
Solution. Taking the waves to propagate in the +z direction, and introducing a
relative phase, δ, the total phasor field is found, using (100):
E sT = E sR + E sL = E 0 [a x − ja y ]e − jβz + E 0 [a x + ja y ]e − jβz jδ
e
Grouping components together, this becomes
jδ
E sT = E 0 [(1 + e )a x − j(1 − e )a y ]e − jβz
jδ
Factoring out an overall phase term, e jδ/2 ,we obtain
E sT = E 0 e jδ/2 (e − jδ/2 + e jδ/2 )a x − j(e − jδ/2 − e jδ/2 )a y e − jβz