Download Electromagnetic Waves Plane Waves Polarization

Electromagnetic Waves
We begin with waves in a non-conducting uniform linear
medium, so we are discussing solutions of Maxwell’s
equations without sources. As we are assuming no
time-dependence of the properties of the medium, we will
fourier transform in time and consider the “harmonic”
fields, so
~ ·B
~ =0
~ ×E
~ − iω B
~ =0
~ = µH
~
∇
∇
B
~ ·D
~ =0
∇
~ ×H
~ + iω D
~ =0
∇
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~ = E
~
D
where the permittivity and permeability µ are constant
in space.
So
~ = −∇
~ × ∇×
~ E
~ = −∇
~ × (iωB) = −iωµ∇
~ ×H
~
∇2 E
~ = −ω 2 µE,
~
= −ω 2 µD
! 2
~ = 0, and by taking the curl
which tells us ∇ + µω 2 E
~
of this, the same equation holds for B.
Thus a general solution of Maxwell’s sourceless equations
will be a linear superposition of (the real parts of)
)
~ x, t) = Ee
~ i~k·~x−iωt
E(~
with k 2 = µω 2 ,
~ x, t) = Be
~ i~k·~x−iωt
B(~
~ coming from the rest of
but with constraints on E~ and B
Maxwell’s equations.
~=0
The divergence equations require ~k · E~ = 0 and ~k · B
~ or
while one curl equation gives i~k × E~ = iω B
√
~ The magnetic field H
~ = B/µ
~
µk̂ × E~ = B.
so
p
~ = k̂ × E/Z
~
H
where Z = µ/ is
an
impediance.
p
The impediance of free space is µ0 /0 = 376.7 Ω.
We have not specified that k and ω are real, which one or
the other might not be, as the permittivity and
permeability are in general complex. Still, in many
contexts they are close to real and if we take ~k to be real,
~ and E
~ equal in
E and B will be in phase, with v B
magnitude.
But if E2 /E1 = Aeiφ is not real, the electric field
components in the two directions are out of phase, and at
a given ~x the field sweeps out an ellipse in time.
If |A| = 1 and φ = π/2, this is a circle, we have the
~ x, t) = E1 (1 + i2 )ei~k·~x−iωt . The real field
complex E(~
∝ 1 cos ~k · ~x − 2 sin ~k · ~x then spirals clockwise as ~x moves
along ~k, if we are looking into the wave. It spirals
counterclockwise as time progresses. This is called a left
circularly polarized wave, or a wave of positive helicity. Of
course the opposite phase, with E1 /E2 = −i, is a right
circularly polarized wave of negative helicity.
Define ~± = √12 (~1 ± i~2 ). Then the electric field can be
decomposed into ~± components rather than ~j (j = 1, 2)
components, and each of these is complex. In either case
there are four real parameters giving the amplitude of the
wave. I think we will not need to discuss the Stokes
parameters that give these in terms of measurable
quantities.
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Plane Waves
i(~k·~
x−ωt)
~ x, t) ∝ e
If we have a plane wave, E(~
, this will
satisfy
provided the wave number
pthe equation
√
k := ~k 2 = µ ω. A fixed phase of this wave moves at
√
2
~
~v = kω/k so v = 1/ µ, which is called the phase
r
µ
velocity. The index of refraction is defined as n =
,
µ0 0
and so v = c/n.
If we consider plane waves in the x direction, uniform in y
and z, we have uk (x, t) = aeik(x−vt) + be−ik(x+vt) ,
corresponding to right and left moving sinusoidal waves
respectively.
If the medium is nondispersive, so n is constant, we may
superimpose these waves with different k to have vaves of
arbitrary shape, u(x, t) = f (x − vt) + g(x + vt), but if
there is dispersion, having created such a wave packet at
t = 0 will not produce pulses of unchanged shape at later
times, because the vt terms in the phase will vary with k.
Polarization
From these constraints we see that E~ is a vector
perpendicular to the wavenumber ~k, and if we set up
orthonormal basis vectors ~1 and ~2 for that plane, with ~1
and ~2 for that plane, with ~1 × ~2 = k̂, we have
~ = √µ(E1~2 − E2~1 ).
E~ = E1~1 + E2~2 , and then B
Physics 504,
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Reflection
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Physics 504,
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Electromagneti
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Reflection
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The amplitudes E1 and E2 may be complex.
If onlyone of
~ x, t) = Re E1 ei~k·~x−iωt ~1 =
them is nonzero, say E1 , E(~
|E1 | cos(~k · ~x − ωt + arg E1 )~1 so the argument of E1 is
just a phase shift, pretty much irrelevant. In this case, the
electric field is linearly polarized, oscillating but always in
the direction ~1 . The same is true if E1 and E2 are not
~
zero but have the same phase, with the resulting E
oscillating in the direction |E1 |~1 + |E2 |~2 .
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Reflection and Refraction
Consider a planar interface between two uniform linear
media. In each medium,
z
the fields must be a comr
k’
n
bination of plane waves.
µ’, ε ’
Suppose a wave
i~k·~
x−iωt
~ = E
~0 e
E
,
√
~
~
B =
µ k̂ × E
µ,ε
x
k
i
r’
k’’
(k̂ is a unit vector) is incident from below, inducing a
refracted wave in the upper medium:
p
~ 0=E
~ 00 ei~k 0 ·~x−iωt ,
~ 0 = µ0 0 k̂ 0 × E
~0
E
B
and a reflected wave in the lower medium
~ 00 = E
~ 00 ei~k 00 ·~x−iωt ,
~ 00 = √µ k̂ 00 × E
~ 00 .
E
B
0
As all the equations are linear with time-independent
parameters, only the one fourier component is involved.
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Kinematics
That is, all the waves have the same frequency. This can
also be viewed as saying the boundary values must
oscillate together in time.
The x and y dependence of the fields at the boundary will
also need to match whatever the boundary conditions.
This tells us kx = kx0 = kx00 and ky = ky0 = ky00 . The
magnitudes of the three k’s are determined by ω and the
√
material parameters, k = |~k| = |~k 00 | = ω µ,
√
k 0 = |~k 0 | = ω µ0 0 . The x, y matching means
k sin i = k 0 sin r = k 00 sin r0 , so k = k 00 implies i = r0 , or
the angle of reflection is equal to the angle of
incidence.
s
sin i
µ0 0
k0
n0
But we also have
=
=
= , where n and
sin r
k
µ
n
n0 are the indices of refraction above and below the
interface. This is Snell’s Law.
Plane of Incidence
The transverse conditions on the k’s means that they all
lie in the plane of incidence, defined by the incident
direction and the interface normal ~n (assuming i 6= 0).
The solution of the interface equations in general is
messy, but we can consider separately the two linear
polarizations of the incident electric field, in that plane
and perpendicular to that plane. Take that plane to
be the xz plane.
z
First consider the incident
B’
~
~
r
k’
E into the plane, so B is
n
E’
µ’, ε ’
in the direction shown, and
µ,ε
x
E0 x = B0 y = E0 z = 0.
B
B’’
Consider a reflection in the
i i
k E
E’’ k’’
plane of incidence, where
all Ey , Bx and Bz should change sign, but not the other
~ is a vector and B
~ a pseudovector.
components, because E
But this simply reverses the incident fields, and therefore
all components must change sign.
~ in the plane of incidence
E
The argument for the opposite linear polarization,
~ in the plane of inwith E
µ’, ε ’
~ perpendicucidence and B
µ,ε
lar to it proceeds similarly.
~
The E × ~n equation gives
(E − E 00 ) cos i = E 0 cos r
~ × ~n equation gives us
and the H
z
n
B’
k
B
i
E’’
i
B’’ k’’
1
k0
n0
1
1
(E0 + E000 ) = 0 E00 = 0 E00
µ
µ
k
µ
n
Again the results are not simple. We get
E00
E0
E000
E0
=
=
k’
2nµ0 cos i
n0 µ cos i + nµ0 cos r
n0 µ cos i − nµ0 cos r
n0 µ cos i + nµ0 cos r
Reflection
and
Refraction
The Boundary Conditions
Assume we are considering non-conducting materials with
no free charges, so within each material the fields must be
differentiable, and at the interface, Gauss’ law tells us the
~ and D
~ are continuous, while the
normal components of B
other two, integrated on a path just below and just above
~ and vecH
the interface, tells us the components of E
perpendicular to ~n are continuous. Of course below the
interface we need to add the incident and reflected fields,
so we have
h
i
~ · ~n continuous:
~0 + E
~ 00 · ~n = 0
~ 000 ) − 0 E
D
(E
~k×E
~ · ~n continuous:
~ 0 + ~k 00 ×E
~ 00 − ~k 0 ×E
~ 0 · ~n = 0
ωB
0
0
~ × ~n continuous:
~0 + E
~ 000 − E
~ 00 × ~n = 0
E
E
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~ × ~n continuous:
H
1 ~ ~
~ 000 − 1 ~k 0 × E
~ 00 × ~n = 0
k × E0 + ~k 00 × E
0
µ
µ
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~ perpendicular (continued)
E
Thus Ex0 , Ez0 , By0 , Ex00 , Ez00 and By00 must all vanish, and the
reflected and transmitted waves are linearly polarized as
~ ⊥ the plane of incidence.
shown, with all E’s
~ × ~n, and
Thus E0 + E000 − E00 = 0 p
from the continuity of E
p
/µ cos i (E0 − E000 ) = 0 /µ0 cos r E00 from the
~ × ~n. The two equations enable solving for
continuity of H
the ratios E00 /E0 and E000 /E0 whose squares give the
transmission and reflection coefficients.
E00
2nµ0 cos i
=
,
E0
nµ0 cos i + n0 µ cos r
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nµ0 cos i − n0 µ cos r
E000
=
E0
nµ0 cos i + n0 µ cos r
We see they depend on the ratios n0 /n and µ0 /µ of the
indices of refraction and the permeabilities, but for
optical frequencies we may usually take µ0 /µ = 1. The
expression is still somewhat complicated. Jackson
eliminatesp
the r dependence using
p
n0 cos r = n0 2 − n0 2 sin2 r = n0 2 − n2 sin2 i.
Normal Incidence
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x
E
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E’
r
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If the angle of incidence goes to zero, the two results must
converge, as the plane of incidence is not well defined. As
~ ⊥ results give
i = r = 0 in that limit, the E
E00
=
E0
2
,
n0 µ
1+
n µ0
E000
=1−2
E0
1
,
n µ0
1+ 0
n µ
~ in the plane, except that the sign of
and so does the E
E000 /E0 is reversed, but that is because the directions
defining E000 are reversed in the two cases.
Electromagneti
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Simplification if µ = µ0
We are often interested in visible light in dielectric
materials which have little magnetic susceptibility, so let
~ ⊥,
us assume µ = µ0 . Then for E
E00
E0
E000
E0
=
=
2n cos i
2 cos i sin r
=
n cos i + n0 cos r
sin(i + r)
sin(r − i)
n cos i − n0 cos r
=
.
n cos i + n0 cos r
sin(r + i)
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=
=
For the electric field in the plane of incidence, there is an
angle for which E000 vanishes, called Brewster’s angle.
Assuming µ0 = µ for simplicity (and it is nearly true for
optics in dielectric materials), this is when
n0 cos i = n cos r, and as n sin i = n0 sin r, multiplying
them, we see sin 2i = sin 2r, or r = π/2 − i, and the
reflected and refracted waves are perpendicular to each
other. Then n0 = n sin i/ sin r = n tan i, so
0
n
Brewster’s angle:
iB = tan−1
.
n
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~ · ~n =
S
=
=
i
h 1
~ 0∗
~ 0×H
Re ~n · E
2
i
h
1
~ 0 × ~k 0 × E
~ 0∗
Re ~n · E
2ωµ0
2
1
~ 0 Re ~n · ~k 0 E
0 = 0,
2ωµ0
as ~n · ~k 0 is pure imaginary.
The naive geometrical picture of well defined beams that
are reflected exactly at the boundary should be doubted.
Perhaps it is effectively off a plane somewhat inside the
boundary, and the reflected ray would intersect the
incident ray inside the second medium. This is called the
Goos-Hänchen effect. To really describe how much the
reflected beam is moved from the naive path, we would
need wave packets in x rather than pure wavenumber. We
are not going to pursue this.
=
2n
,
n + n0
E000
E0
=
n0
n−
.
n + n0
Total Internal Reflection
Earlier we derived Snell’s law, n sin i = n0 sin r, which
shows that, if n > n0 , for an angle of incidence i greater
than sin−1 (n0 /n) we need a refraction angle with a sine
greater than 1. How do we interpret that? Really we got
0
0
0
this by
p requiring kx = kx with k = n k/n, from which
kz0 = k 0 2 − kx2 which is imaginary (say iκ under these
circumstances. Then the fields
~ 0=E
~ 00 eikx x−κz−iωt ,
E
~ 0 = 1 ~k 0 × E
~ 0.
H
µ0 ω
Note the fields fall off exponentially with depth z into the
less dense medium for angles beyond that of total internal
reflection, so there is no continuing beam of refracted
light, though there is field close to the interface.
This is a way to get complete linear polarization of light.
It also indicates that even at other angles, the
transmission of one polarization will be greater than the
other.
There is no flow of energy into the n0 medium, because
the Poynting vector’s z component is
E00
Electromagneti
Waves
Reflection
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Refraction
Note that if the reflection is off a more dense medium
~ is reversed on reflection, which is
(n0 > n), the sign of E
the origin of the rule we use in elementary discussions of
interference, that there is an extra phase shift by π under
reflection off a more-dense medium.
2n cos i
4 sin r cos i
=
n0 cos i + n cos r
sin(2i) + sin(2r)
sin(2i) − sin(2r)
tan(i − r)
=
.
sin(2i) + sin(2r)
tan(i + r)
Brewster angle
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For normal incidence, we need to not use the sines, which
~ ⊥ definition where positive values are
vanish. Using the E
all in the same direction, we see
E0
~ k,
and for E
E00
E0
E000
E0
Normal incidence, again, with µ = µ0
Physics 504,
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Electromagneti
Waves
Reflection
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Physics 504,
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Electricity
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Shapiro
Electromagneti
Waves
Reflection
and
Refraction