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PHY4324: EMII: SOLUTIONS
TEST II
12 November, 2009
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly.
Every (algebraic) final result must be supplemented by a check of units. Without such a check, no
more than 75% of the credit will be given even for an otherwise correct solution. On the other hand,
an answer obtained just by the dimensional analysis may give you up to 25% of the credit.
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FIG. 1: Problem 1.
θi
n
N
1. Consider a layer of glass with refraction coefficient n > 1 on the surface of an infinitely thick slab of another
glass with refraction coefficient N > 1. To the left of the layer there is air (refraction coefficient = 1). Light
falls from air at angle θi to the normal. Find if total internal reflection is possible at any of the boundaries of
this system.
Solution: Light impinging on the air/glass n boundary from the left is transmitted at an angle θtn to the
normal, where
sin(θtn ) = sin(θi )/n ≤ 1.
The same angle is also the angle of incidence for light impinging on the glass n/glass N boundary. The angle
transmitted light makes with the normal is θtN , where
sin(θtN ) = (n/N ) sin(θtn ) = sin(θi )/N ≤ 1.
Finally, light reflected from the n/N boundary falls on the n/air boundary at angle θtn and comes out back to
air at θi .
Total internal reflection is possible if the sine of the angle of a transmitted wave is formally larger than one,
which is not the case here. Therefore, total internal reflection is not possible at any of the boundaries for light
coming from the left.
2. A plane monochromatic electromagnetic wave falls normally on a flat metallic surface from vacuum. The
frequency of the wave ω is within the interval
1/τ ≪ ω ≪ ωp ,
p
where τ is the mean free time of electrons and ωp = ne2 /mǫ0 is the plasma frequency of the metal (n is the
number density of electrons and m is the electron mass). The (energy) reflection coefficient R is almost but not
exactly equal to unity. Find R to leading order in 1/τ , assuming that the magnetic permeability and dielectric
permittivity of the metal are µ = µ0 and ǫ = ǫ0 .
Hint: At finite frequency, the conductivity of a metal is given by
σ=
where σ0 = ne2 τ /m is the conductivity at ω = 0.
σ0
,
1 − iωτ
(1)
3
Solution: A complex wavenumber for the wave inside a metal with ǫ = ǫ0 and µ = µ0 is given by
k̃ 2 =
ω2
+ iσωµ0 ,
c2
(2)
where σ is given by Eq. (??). Because we were told that ωτ ≫ 1, we can expand the denominator in Eq. (??)
to leading order in 1/τ as
σ0
σ0
σ =
=
i
1 − iωτ
−iωτ 1 + ωτ
σ0
≈i
(1 − i/ωτ ) .
(3)
ωτ
On substituting the equation for σ0 and using the definition of ωp , the last result reduces to
σ ≈ iǫ0
ωp2
(1 − i/ωτ ) .
ω
(4)
Substituting Eq. (??) into Eq. (??), we obtain
k̃ 2 =
ωp2
ω2
−
(1 − i/ωτ ) .
c2
c2
Now recall that the reflection coefficient for a metal is given by
1 − β̃ 2
1 + |β̃|2 − 2Reβ̃
,
R=
=
1 + β̃
1 + |β̃|2 + 2Reβ̃
(5)
(6)
where β̃ = k̃c/ω. Let’s analyze Eq. (??). If we neglect the 1/ωτ term completely, k̃ 2 = (ω 2 −ωp2 )/c2 ≈ −ωp2 /c2 <
0, which means that k̃ is purely imaginary, and so is β̃. Then Reβ̃ = 0 and R = 1. Therefore, a deviation of R
from 1 is due to the 1/ωτ term. Also, we notice that |β̃|2 ≈ ωp2 /ω 2 ≫ 1. Therefore, we can expand R as
R≈1−4
Reβ̃
.
|β̃|2
(7)
To find β̃, we re-write k̃ as
!1/2
1/2
ωp2
ωp2
i
ωp
1−
=i
k̃ = − 2 + i 2
c
c ωτ
c
ωτ
ωp
i
ωp
ωp 1
≈i
1−
=i
+
.
c
2ωτ
c
c 2ωτ
(8)
This gives
β̃ = k̃
ωp
ωp
c
=i
+
.
ω
ω
2ω 2 τ
(9)
The first term is the dominant one and it determines the magnitude of |β̃|2
whereas
2
2 2 ωp2
|β̃|2 = Reβ̃ + Imβ̃ ≈ Imβ̃ = 2 ,
ω
Reβ̃ =
ωp
.
2ω 2 τ
(10)
(11)
On substituting these results into Eq. (??), we obtain
R=1−4
ωp /2ω 2 τ
2
=1−
.
ωp2 /ω 2
ωp τ
(12)
In a typical metal, ωp ∼ 1015 − 1016 rad/s, whereas τ ∼ 10−12 s, so the correction to unity in R is about
10−3 − 10−4 . Notice that this correction does not depend on the frequency but is determined entirely by the
properties of the metal.
4
3. A plane monochromatic electromagnetic wave falls normally on a metal from vacuum (again!). Assume that the
metal occupies half-space z > 0. Also, assume that the frequency of the wave ω is so low that the conductivity
can be approximated by its zero frequency value σ0 = e2 nτ /m, which is very large. Let the amplitude of the
electric field just below the surface of the metal, i.e., for z → 0+, be E0 . Assume that the magnetic permeability
and dielectric permittivity of the metal are µ = µ0 and ǫ = ǫ0 . Find the Poynting vector, averaged over a period
of oscillations, at distance z inside the metal. If you find that the Poynting vector decreases with the distance,
explain what happens with the energy of the electromagnetic wave.
Solution: In a metal,
k̃ 2 =
ω2
+ iσ0 ωµ0 ,
c2
where now σ0 is independent of ω. Because we were told that σ0 is large, neglect the first term. Then,
√
√√
2√
σ0 ωµ0 ≡= (1 + i)κ,
k̃ = i σ0 ωµ0 = (1 + i)
2
p
where κ = σ0 ωµ0 /2. Therefore,
Ẽ = E0 eik̃z=ωt = E0 e−κz eiκz−ωt .
(13)
(14)
(15)
As a vector, E is perpendicular to the z axis. The magnetic field is perpendicular to E and its magnitude is
given by
√ κ
k̃
κ
Ẽ = (1 + i) E0 e−κz eiκz−ωt = 2 E0 e−κz eiκz−ωt+π/4 ,
ω
ω
ω
√
where we used that (1 + i)/ 2 = eiπ/4 .
B̃ =
(16)
Now, we can either take the real and imaginary parts of Ẽ and B̃, multiply them and average over the period
or to use a compact result of Problem (9.11) which says that hSi = Re(Ẽ B̃ ∗ )/2µ0 . Choosing a longer but more
straightforward route,
E = ReẼ = E0 cos(κz − ωt) exp(−κz)
√ κ
B = ReB̃ = E0 2 cos(κz − ωt + π/4) exp(−κz)
ω
(17)
When multiplying E by B and averaging over the period, notice that
"
√
√ #
2
2
hcos(κz − ωt) cos(κz − ωt + π/4)i = hcos(κz − ωt) cos(κz − ωt)
i
− sin(κz − ωt)
2
2
√
√
2
2
2
=
hcos (κz − ωt)i =
,
2
4
(18)
because hsin(φ) cos(φ)i = 0. Therefore,
hSi =
κ
E 2 exp(−2κz).
2µ0 ω 0
(19)
As a vector, S is along the z axis. The energy lost by the electromagnetic wave is dissipated in the metal, i.e., it is
eventually transformed into heat.