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In the absence of sources, the Maxwell equations in an
infinite medium are
Assuming solutions with harmonic time dependence,
the equations for the amplitude E(ω,x), etc. read
For uniform isotropic linear media we have
where and may in general be complex
functions of ω. Assuming that they are real, positive
and spatially constant. The equations for E and H are
By combining the two equations we get the Helmholtz
wave equation, real, positive and spatially constant
Consider as a possible solution a plane wave traveling
in x direction. From (7.3) we find the requirement that
the wave number k and the frequency ω are related by
The phase velocity of the wave is
The quantity n is called the index of refraction and is
usually a function of frequencyμ=μ(ω)
The primordial solution in one dimension is
Using
from (7.5), this can be written
If the medium is non[dispersive μ(ω)], the Fourier
superposition theorem (2.44) and (2.45) can be used to
construct a general solution of the form
A(k) μ(k)
Where f(z) and g(z) are arbitrary functions.
If the medium is dispersive, equation (7.7) no longer
holds. The wave changes shape as it propagates.
We now consider an electromagnetic plane wave of
frequency ω and wave vector
and require that it
satisfy not only the Helmholtz wave equation (7.3) but
also all the Maxwell equations.
With the convention that the physical electric and
magnetic fields are obtained by taking the real parts of
complex quantities, we write the plane wave fields as
where
and n are constant vectors.
Each component of E and B satisfies (7.3) provided
To recover (7.4) it is necessary that n be a unit vector.
With the wave equation satisfied, there only remains
the fixing of the vectorial properties so that the
Maxwell equations (7.1) are valid. The divergence
equations in (7.1) demand that
This means that E and B are both perpendicular to the
direction of propagation n.
The curl equations provide a further restriction,
namely
The factor
can be written
the index of refraction defined in (7.5).
where n is
In engineering literature the magnetic field H is often
displayed in parallel to E instead of B. The analog of
(7.11) for H is
where
is an impedance.
In vacuum,
impedance of free space.
ohms, the
If n is real, (7.11) implies that and have the same
phase. It is then useful to introduce a set of real
mutually orthogonal unit vectors
, as shown
in Fig. 7.1.
In terms of these unit vectors the field strengths
are
and
or
The real part of the complex Poynting vector is:
The energy flow (energy per unit area per unit time) is
The time-averaged energy density u is correspondingly
This gives, |S|=uv
The ratio of the magnitude of (7.13) to (7.14) shows
that the speed of energy flow is
as expected
from (7.5).
In the discussion that follows (7.11) we assumed that n
\in R. This does not yield the most general possible
solution for a plane wave.
Let n \in C, and write
Then the exponential in (7.8) becomes, grows/decays
in certain directions
The relations (7.10) and (7.11)
still hold. The requirement
imaginary parts, and implies that
has real and
The second of these conditions shows that and
are orthogonal. The coordinate axes can be oriented so
that is in the x direction and in the y direction.
The first equation in (7.15) can be satisfied generally
by writing, ch2x- sh2x=1,
The most general vector
satisfying
is then
It is easily verified that for θ= 0, the solutions (7.12)
and (7.12’) are recovered with Ch 0=1, sh 0=0.
The plane wave (7.8) and (7.12) is a wave with its
electric field vector always in the direction .
Evidently the wave described in (7.12’) is linearly
polarized with polarization vector and is linearly
independent of the first.
Thus the two waves,
with
can be combined to give the most general homogeneous plane wave propagating in the direction
Note that the amplitudes E1 and E2 are complex
numbers. If E1 and E2 have the same (different)
phases, (7.19) represents a linearly (elliptically)
polarized wave.
If E1 and E2 have the same (different) phases,
(7.19) represents a linearly (elliptically) polarized
wave.
As shown in Fig. 7.2, its polarization vector making an
angle
with and a magnitude
, if E1 and E2 are real.
Consider the case of circularly polarized wave where
E1 and E2 have the same magnitude, but differ in phase
by 90°. (E1, E2) = E0 (1, i ). The wave (7.19)
becomes:
with E0 the common real amplitude.
The components of the actual electric field, obtained
by taking the real part of (7.20), are (t upx down, y up)
Ey=sin (ωt-kz) for + helicity
Ey=sin (ωt-kz) for + helicity
At a fixed point space, the fields (7.21) are such that E
is constant in magnitude, but sweeps around in a circle
at frequency omega as shown in Fig. 7.3.
circularly polarized, +(-) left(right)polarized, counter-clockwise, positive(negative)
helicity, for such a wave
has a positive projection
of L on the z-axis.
The two circularly polarized wave (7.20) form an
equally acceptable set of basic fields for description of
a general state of polarization. We introduce the
complex orthogonal unit vectors:
with properties, ε1, ε2 \in R, <a,b>=a*‧b
Then a general representation, equivalent to (7.19), is
where E+ and E- are complex amplitudes. If E+ and Ehave different magnitudes, but the same phase, (7.24)
represents an elliptically polarized wave with principal
axes of the ellipse in the directions of and .
If the amplitudes have a phase difference
between them, then it is easy to show that the
ellipse traced out by the E vector has its axes
rotated by an angle (α/2), as shown in Fig.7.4.
HomeWork:
 Ex 
 E' x 

  R /2  ,
- E 
 E' 
y


 y
 E' x  | E  |  (1  r) cos (    /2) 
 


 E' 
2  (1 - r) sin (    /2) 
 y
 E    | E  | exp[ i ] 
   
,
    k  x - ωt 
 cos/2 , s/2
R /2  
 - s/2 , c/2



 Ex 
 E' x 

  R /2  ,
- E 
 E' 
y


 y
 E' x  | E  |  (1  r) cos (    /2) 
 


 E' 
2  (1 - r) sin (    /2) 
 y
elliptical orbit
The Stokes parameters can be motivated by observing
that for a wave propagating in the z direction, the
scalar products,
are the amplitudes of radiation, respectively, with
linear polarization in the x, y directions, +, - helicity.
For the latter purpose we define each of the scalar
coefficients in (7.19) and (7.24) as a magnitude times
a phase factor:
E1=?
E2=?
In terms of the linear polarization basis
Stokes parameters are
, the
If the circular polarization basis
is used
instead, the definitions read, same si as (7.27),
The parameter s0 measures relative intensity of the
wave in either case.
The parameter s1 gives the
preponderance of x-linear polarization over y-linear
polarization,
s2 and s3 in the linear basis give phase information.
RMK: The four Stokes parameters are not
independent, since they depend on only three
quantities. They satisfy the relation:
Monochromatic radiation, in practice, contains a range
of frequencies and are not completely monochromatic.
One way of viewing this is to say that the magnitudes
and phases in (7.26) vary slowly in time, slowly, that
is, when compared to the frequency ω. The observable
Stokes parameters become averages over a relatively
long time interval, and are written as
T
One consequence of the averaging process is that the
Stokes parameters for a quasi-monochromatic beam
satisfy an inequality, by Schwartz inequality,
rather than the equality. Just Mention This Part！
The reflection and refraction of light at a plane surface
between two media of different dielectric properties
are familiar phenomena. The various aspects of this
phenomena divide themselves into two classes.
1.Kinematic properties: continuity of phase: k‧x|z=0
(a) Angle of reflection equals angle of incidence.
(b) Snell’s law: (sin i)/(sin r) = n’/n
2. Dynamic properties: [from Maxwell equations]
(a) Intensities of reflected and refracted radiation.
(b) Phase changes and polarization.
The coordinate system and symbols appropriate to the
problem are shown in Fig. 7.5. The media below and
above the plane z = 0 have permeabilities and
permittivities
and
, respectively.
C
C
According to (7.18), the three waves are:
Incident
Refracted
Reflected
The wave number of the magnitudes
The existence of boundary conditions at z = 0, which
must be satisfied at all points on the plane at all times.
We must have the phase factors all equal at z = 0,
Independent of the nature of the boundary conditions.
In the notation of Fig. 7.5, (7.34) reads,
Since
, we find
; the angle of incidence
equals the angle of reflection. Snell’s law is
The dynamic properties are contained in boundary
conditions. In terms of fields (7.30)-(7.32) these
boundary conditions at z = 0 are:
First we consider the electric field perpendicular to the
plane of incidence, as shown in Fig. 7.6a. The third
and fourth equations in (7.37) give, E⊥ ,
E⊥
E//
Using Snell’s law, the relative amplitudes of the
refracted and reflected waves can be found from
(7.38).
For E perpendicular to plane of incidence,
the Fresnel’s formula: E⊥
For optical frequencies it is usually permitted to put
.
If the electric field is parallel to the plane of incidence,
as shown in Fig. 7.6b., E//
Using Snell’s law.
For E parallel to plane of incidence:
For normal incidence (i = 0), both (7.39) and (7.41)
reduce to
a -- sign for E⊥
Where the results on the right hold for
.
Two aspects of the dynamical relations on reflection
and refraction are worthy of mention. The first is
Brewster’s angle, and the second is total internal
reflection.
E//
For polarization parallel to the plane of incidence
there is an angle of incidence, called Brewster’s angle,
for which there is no reflected wave. In (7.41), the
amplitude of reflected wave vanishes when
For typical ratio
.
Total internal reflection:
Snell’s law (7.36) shows that, if n > n’, then r > i.
Consequently,
when
where
What happens if
? For
, r is a complex
angle with a purely imaginary cosine.
Consider the propagation factor for the refracted wave.
This shows that the wave is attenuated exponentially
beyond the interface.
There is no energy flow through the surface. The lack
of energy flow can be verified by calculating the timeaveraged normal component of the Poynting vector
just inside the surface:
with
, we find
But n · k’=k’ cos r is purely imaginary, so that S · n
=0.
E⊥
E//
Total reflection:
n cos i - i n' (sin i /sin i 0 ) 2 - 1
n cos i - n' cos r
R ~

n cos i  n' cos r n cos i  i n' (sin i /sin i 0 ) 2 - 1
a-ib

 exp[ i   ], tan   b/a
a  ib
The evanescent wave penetrating into the region z > 0
has an exponential decay in the perpendicular
direction,
where
Goos-Hächen effect: If a beam of radiation having a
finite transverse extent undergoes total internal
reflection, the reflected beam emerges displaced
laterally with respect to the prediction of a geometrical
ray reflected at the boundary. As shown in Fig. 7.7, the
beam should emerge with a transverse displacement of
Goos-Hächen effect:
The first-order expressions for D for the two states of
linear polarization are, kλ=2π,
Where λ is the wavelength in the medium of higher
index of refraction.