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Transcript
Transmission and reflection in a very thin foil
An EM plane wave of frequency ω = 2πc/λ impinges at normal
incidence on a thin metal foil of thickness d ≪ λ. Taking the limit
of an infinitely thin foil, the electron density may be written as
ne (x) = n0 dδ(x) where n0 d is the surface density and, analogously,
the current density may be written as J(x, t) = j(t)dδ(x). The aim
of the problem is to evaluate the transmission and reflection of the
wave through the foil within the above outlined approximation.
a) Show that the boundary conditions at the foil position for the
EM field components parallel to the foil surface are given by
Ek (0+ ) − Ek (0− ) = 0,
Bk (0+ ) − Bk (0− ) = −
Ei
kt
ki
kr
0
1
jd .
ǫ0 c 2
x
(1)
b) Assuming a linear dependence of the current density J = J̃e−iωt on the electric field, i.e. J = σE
(where σ is in general a complex scalar quantity), find the expression of the EM field in the whole
space as a function of σ.
b) Now use the classical equation of motion for electrons in the metal
me
dv
= −eE − me νv ,
dt
(2)
to work out an expression for σ and evaluate the cycle-averaged absorbed power in the limits ν ≫ ω
and ν ≪ ω, respectively.
c) Verify the conservation of energy for the system by showing that the flux of EM energy inside the
foil equals the absorbed power.
1
Solution
a) Since the problem of finding the transmission and reflection coefficient is linear and the medium
is isotropic, the choice of polarization is arbitrary. For definiteness let us assume the electric (E) and
magnetic (B) fields of the impinging wave to be along the y and z directions, respectively, and drop
the vector sign in the following.
We apply Stokes’s theorem for both E and B to a closed rectangular path C (delimiting the
surface A), extended along x from x = −ǫ/2 to x = +ǫ/2 and of side length ℓ in the direction
parallel to the field:
I
Z
E · dl = [E(+ǫ/2) − E(−ǫ/2)] ℓ = +iω BdA = iω B̄ℓǫ,
(3)
C
A
I
Z
B · dl = [B(+ǫ/2) − B(−ǫ/2)] ℓ = µ0
(J − iωǫ0 E) dA = −µ0 Jdℓ + iω Ēℓǫ,
(4)
C
A
where we used the average value theorem. In the limit ǫ → 0, being |Ē| and |B̄| limited, the terms
proportional to ǫ vanish and we obtain the relations (1).
b) The most general expression for the field is the sum of the incident and the reflected wave for
x < 0 and the transmitted wave for x > 0:
E(x, t) = Ei eikx−iωt + Er e−ikx−iωt (x < 0) ,
=
Et eikx−iωt
(x > 0) .
(5)
(6)
The amplitudes Er and Et must be determined as a function of Ei and other parameters by imposing
the bondary conditions (1). Noticing that j = σE(0) = σEt and that ∂x E(x, t) = −∂t B(x, t), we
have
Et − Ei − Er = 0,
Et − Ei + Er = −
σd
Et ,
ǫ0 c
(7)
so that, posing σd/2ǫ0 c = η for brevity
Er = −
η
Ei ,
1+η
Et =
1
Ei .
1+η
(8)
c) In the limit ν ≫ ω the conductivity is given by σ = n0 e2 /me ν and is a real number (Ohmic
conductor). The mechanical power P is the cycle average of J · E integrated over the volume of the
foil, thus we obtain (per unit surface)
P = hσE(0)2 id =
η
1 σd
2
E
=
ǫ
c
E2 .
0
i
2 (1 + η)2
(1 + η)2 i
(9)
d) In the limit ν ≪ ω the conductivity is σ = in0 e2 /me ω = iǫ0 ωp2 /ω and is thus imaginary (that is
equivalent to a real permittivity ǫ = 1 − ωp2 /ω 2 ). This implies that J and E have opposite phase,
thus hJ · Ei = 0 as it may be verified directly.
2
e) The energy flux through the foil is given by the difference between the values of the Poynting flux
at the two surfaces,
(10)
S(0+ ) − S(0− ) = ǫ0 c2 E(0+ )B(0+ ) − E(0− )B(0− ) .
Using the boundary conditions we may write
1
E(0+ )B(0+ ) − E(0− )B(0− ) = E(0) B(0+ ) − B(0− ) = −E(0) 2 Jd ,
ǫ0 c
(11)
S(0+ ) − S(0− ) = −JE(0)d ,
(12)
so that
i.e. the energy flux through the foil is equal at any time to the mechanical power depostied in the
foil (all quantities have been defined per unit surface).
Alternatively, we may compute directly the energy flux and compare it with the mechanical
power. For the cycle-averaged Poynting vector at the two surfaces we have
1
1
1
E2 ,
S(0+ ) = ǫ0 chE 2 (0+ )i = ǫ0 c|Et |2 = ǫ0 c
2
2 |1 + η|2 i
1
1
1
S(0− ) = ǫ0 c2 hE(0− )B(0− )i = ǫ0 cRe [(Ei + Er )(Ei∗ − Er∗ )] = ǫ0 c
Re(2η ∗ + 1)Ei2 .(13)
2
2
2 |1 + η|
If ν ≪ ω then η is purely imaginary and S(0− ) = S(0+ ): there is no net flux of energy inside the
foil, which is consistent with the vanishing of the mechanical power.
If ν ≫ ω then η is real and the net flux of energy is given by
1 1 − (2η + 1) 2
η
S(0+ ) − S(0− ) = ǫ0 c
Ei = −ǫ0 c
E 2,
2
2
(1 + η)
(1 + η)2 i
which is equal to the power, Eq.(9).
3
(14)