Today in Physics 218: optical properties of
conductors
‰ Good and bad conductors
‰ Relative phase of E and B
of waves in conductors
‰ Reflection from
conducting surfaces
‰ The characteristic matrix
of a conducting layer
False-color infrared image of the
Rosebud Nebula, NGC 7129, by Tom
Megeath (CfA), Rob Gutermuth and
Judy Pipher (UR), using the IRAC
instrument on the new Spitzer Space
Telescope.
16 February 2004
Physics 218, Spring 2004
1
Good and bad conductors
The imaginary part of the wavenumber in a conductor is, as
we showed last time,
12
1
2
⎛⎛
⎞
2⎞
µε ω ⎜
⎛ 4πσ ⎞
⎟
⎜1+ ⎜
⎟
−
κ=
1
.
⎟ ⎟
⎜
⎟
⎜
2 c ⎜ ⎝ ⎝ εω ⎠ ⎠
⎟
⎝
⎠
This is small compared to the real part if the term under the
square root is small, which in turn will be true if
εω
σ
4π
( εω = ε r ε 0ω in MKS )
.
We would call this the case of a bad conductor. Note the
frequency dependence: even matter with large conductivity
can be a bad conductor for fields that vary rapidly enough in
time.
16 February 2004
Physics 218, Spring 2004
2
Good and bad conductors (continued)
The opposite, of course, is what we would call a good
conductor:
εω
.
σ
4π
Comparing this to the time constant for attenuation of
electromagnetic waves as they propagate in conducting
media,
ε
τ=
,
2πσ
we see that bad (good) conductors have
τ ()
16 February 2004
2
ω
=
1 ⎛ 2π ⎞
⎜
⎟ .
π⎝ ω ⎠
Physics 218, Spring 2004
3
Good and bad conductors (continued)
For bad conductors, then,
12
1
2
⎛
⎞
µε ω ⎜ ⎛ ⎛ 4πσ ⎞2 ⎞
⎟
⎜1+⎜
⎟
−
κ=
1
⎟
⎟⎟
2 c ⎜⎜ ⎜⎝ ⎝ εω ⎠ ⎟⎠
⎝
⎠
≅
µε ω ⎛
⎞
1 ⎛ 4πσ ⎞
⎜1+ ⎜
− 1⎟
⎟
⎟
2 c ⎜⎝
2 ⎝ εω ⎠
⎠
2πσ
=
c
Similarly,
k≅
16 February 2004
µ σ
= Z
ε
2
µε ω ⎛
2
12
⎛ σ
⎜⎜ =
⎝ 2c
1+ x ≅ 1+ x 2 :
µε ω 1 4πσ
=
2 c 2 εω
⎞
µr µ0 σ
Z in MKS ⎟ .
=
⎟
εrε0
2c
⎠
⎞
1 ⎛ 4πσ ⎞
⎜1+ ⎜
+ 1⎟
⎟
⎟
2 c ⎜⎝
2 ⎝ εω ⎠
⎠
2
For x 1,
12
Physics 218, Spring 2004
≅ µε
ω
c
κ
.
4
Good and bad conductors (continued)
2
And for good conductors, since ( 4πσ εω ) 1,
k⎫
⎬=
κ⎭
=
k =
⎛⎛
µε ω ⎜
⎛ 4πσ
⎜1+ ⎜
2 c ⎜⎜ ⎝⎜ ⎝ εω
⎝
⎛ 1
2πωµσ
⎜⎜ =
c
⎝ c
2 ⎞1 2
⎞
⎟ ⎟⎟
⎠ ⎠
⎞
± 1⎟
⎟⎟
⎠
ωµr µ0σ
2
12
≅
µε ω ⎛ ⎛ 4πσ
⎜⎜
2 c ⎝ ⎝ εω
⎞ ⎞
⎟ ± 1⎟
⎠ ⎠
12
⎞
in MKS ⎟⎟ ;
⎠
2πωµσ
(1 + i) .
c
The real and imaginary parts of the complex wavenumber are
the same size, in good conductors.
16 February 2004
Physics 218, Spring 2004
5
Phases of E and B
For polarization in the x direction, plane waves in conductors
look like this:
i kz −ωt )
E = xˆ E 0 e −κ z e (
,
i kz −ωt )
B = yˆ B e −κ z e (
.
0
However, it turns out that E and B are no longer in phase
because the wavenumber is complex, as we can show from
Faraday’s law:
∂
E
i kz −ωt )
— ¥ E = yˆ x = ( ik − κ ) yˆ E 0 e −κ z e (
∂z
1 ∂B iω −κ z i( kz −ωt )
=−
=
yˆ B0 e e
;
c ∂t
c
16 February 2004
Physics 218, Spring 2004
6
Phases of E and B (continued)
that is,
B = ( k + iκ ) cE =
0
0
ω
B = yˆ
c k
ω
c k eiφ
ω
φ = arctan (κ k )
E 0
i kz −ωt +φ )
E 0 e −κ z e (
.
B is delayed, relative to E. The time-averaged Poynting vector
for these out-of-phase waves (Problem 9.20b) is
c
k
2 −2κ z − iφ
c ∗
c ∗
c
ˆ
E0 e
e
S = Re
E ¥ H = Re
E ¥ B = Re z
8π
8πµ
8πµ ω
⎧ ˆ 1 k 2 −2 κ z
2 k
2
,
c
c k 2 −2κ z ⎪z 2 µ ω E0 e
2
2
−
κ
z
E0 e
E0 e
cos φ = zˆ
.⎨
= zˆ
8πµ ω
8πµ ω
⎪in MKS.
⎩
16 February 2004
Physics 218, Spring 2004
7
Reflection from conducting surfaces
The boundary conditions for light incident on a conductor
aren’t different from the ones for nonconductors.
‰ If J = σ E , then an infinite field would be needed to
produce a surface current:
J f = K f δ ( z) = σ E .
So we get
ε 1E1⊥ − ε 2 E2 ⊥ = 4πσ f = 0
B1⊥ − B2 ⊥ = 0
E1& − E2& = 0
1
µ1
16 February 2004
B1& −
1
µ2
B2&
4π
=
K f × nˆ = 0 (still).
c
Physics 218, Spring 2004
8
Reflection from conducting surfaces (continued)
Let us therefore consider light incident from a non-conductor
onto a conductor, and for simplicity let’s restrict ourselves to
normal incidence. The fields are
i k z −ωt )
E I = xˆ E 0 I e ( 1
i − k z −ωt )
E R = xˆ E 0 R e ( 1
(
ˆ
ET = xE0T e
i k2 z −ωt
)
i k z −ωt )
B I = yˆ µ1ε 1 E 0 I e ( 1
i − k z −ωt )
B R = yˆ µ1ε 1 E 0 R e ( 1
i ( k2 z −ωt )
ck
2
E 0T e
B T = yˆ
ω
As usual we put the surface at z = 0, and the boundary
conditions there give:
16 February 2004
Physics 218, Spring 2004
9
Reflection from conducting surfaces (continued)
E 0 I + E 0 R = E 0T
(
µ1
1
µ1ε 1 E 0 I − µ1ε 1 E 0 R
)
,
1 ck2 E0T
=
µ2 ω
.
Note that we’ve written this in terms of the complex k2 ;
we can still divide through by ε 1 µ1 and rearrange to get
E + E = E
,
0I
0R
0T
E 0 I − E 0 R = β E 0T
where
µ
ck
1
2
β =
ε 1 µ2ω
,
.
is complex, this time.
16 February 2004
Physics 218, Spring 2004
10
Reflection from conducting surfaces (continued)
These equations have the usual solutions,
2 1 − β E0T =
E0 I , E0 R =
E0 I
1+ β
1+ β
.
It is interesting to specialize this for the most common case,
that of a good conductor (= an ordinary metal mirror). Here,
2πωµσ
µ1 c
µ1ε 2 2πσ
β=
(1 + i) =
(1 + i) ≡ γ (1 + i)
c
ε 1 µ2ω
ε 1 µ 2 ε 2ω
= β non-conductor
µ1
γ=
ε 1 µ2
16 February 2004
2πσ
ω
2πσ
ε 2ω
(1 + i)
⎛
µ1
⎜⎜ =
ε 1 µ2
⎝
, where
⎞
σ
in MKS ⎟ .
⎟
2ω
⎠
Physics 218, Spring 2004
11
Reflection from conducting surfaces (continued)
‰ If σ were infinite, β would be too, so
E 0T = 0 , E 0 R = −E 0 I .
No transmission, and reflected light 180° out of phase
(upside-down) with respect to the incident light.
‰ If σ were merely large (as in Problem 9.21),
E 0 R
IR
=
II
E 0 I
=
2
2
1 − β 1 − β ∗ 1 −
=
=
∗
1 + β 1 + β
1+
1 − ( 2γ ) + 2γ 2
1 + ( 2γ ) + 2γ
16 February 2004
2
(
)
+ β ∗ + ββ
∗
β
(
)
∗
β + β ∗ + ββ
= 0.923 for aluminum at λ =0.55 µ m .
Physics 218, Spring 2004
12
Reflection and transmission by conducting layers
Consider a plane-parallel metal layer, with conductivity σ
and thickness d, bounded on either side by nonconducting
media (see next slide). What are the transmittance and
reflectance of this structure?
‰ For simplicity, let’s stick to normal incidence. The
boundary conditions at the first surface are the ones we
already used for a single surface, plus an extra reflected
wave:
E& ,1 = E 0 I + E 0 R 1 = E 0T 1 + E 0 R 2
H & ,1 =
16 February 2004
1
µ0
(
µ0ε 0 E 0 I − E 0 R1
)
,
1 ck1 =
E0T 1 − E 0 R 2
µ1 ω
Physics 218, Spring 2004
(
)
.
13
Reflection and transmission by conducting layers
(continued)
ε 0 , µ0
E I
1
kI 1
B I
ε 1 , µ1 , σ
E T 1
kT 1
B T 1
E R 1
E T 2
kT 2
B T 2
B R 1
kR 1
ε 2 , µ2
2
B R 2
k R 2
E R 2
d
16 February 2004
Physics 218, Spring 2004
14
Reflection and transmission by conducting layers
(continued)
‰ As a wave crosses the layer it travels a distance d.
Compared to the undisplaced wave that would have
resulted if the slab were not there, it undergoes a phase
change of
δ 1 = k1 d .
‰ Thus the E& and H& boundary conditions at surface 2 are
E& ,2 = E 0T 1 eiδ 1 + E 0 R 2 e −iδ 1 = E 0T 2 eiδ 1
H & ,2
16 February 2004
1 ck1 E0T 1 eiδ 1 − E 0 R 2 e −iδ 1 =
=
µ1 ω
(
)
Physics 218, Spring 2004
,
ε2 E0T 2 eiδ 1
µ2
15
Reflection and transmission by conducting layers
(continued)
‰ By now, this probably looks familiar: the boundary
conditions, and the expressions for E& and H& at the
surfaces, are just the same as in the case of nonconductors,
with
ε1
1 ck1
,
→
Y1 =
µ1
2π n1 d
δ1 =
λ
µ1 ω
→ k1 d .
Thus the conducting layer has a characteristic matrix that
looks very similar to that of a nonconducting layer. The
major differences are that the admittance Y and the phase
shift δ are complex, not real.
16 February 2004
Physics 218, Spring 2004
16
Characteristic matrix of a conducting layer at
normal incidence
With the definitions
1 ck1
Y1 =
µ1 ω
where
and δ 1 = k1 d ,
⎧
2πωµ1σ 1
(1 + i)
⎪
c
⎪
k1 = ⎨
⎪ µ ε ω + i 2πσ 1 µ1
⎪ 1 1 c
ε1
c
⎩
good conductor
,
bad conductor
the characteristic matrix is as usual
⎡ cos δ 1
M1 = ⎢
⎣ −iY1 sin δ 1
16 February 2004
−i sin δ 1 /Y1 ⎤
.
⎥
cos δ 1
⎦
Physics 218, Spring 2004
17
Characteristic matrix of a conducting layer at
normal incidence (continued)
This can be used in transmittance and reflectance calculations
just as before, in combination with other layers, conducting
or non conducting. In particular, we still can use
2Y0
t=
,
m11Y0 + m12Y0Yp + 1 + m21 + m22Yp + 1
r=
m11Y0 + m12Y0Yp + 1 − m21 − m22Yp + 1
,
m11Y0 + m12Y0Yp + 1 + m21 + m22Yp + 1
ρ=
SR 1,⊥
= r
2
and τ =
ST , p + 1,⊥
=
Yp + 1
t
2
Y0
SI , ⊥
SI , ⊥
Note, however, that ρ + τ no longer makes 1; there is
absorption in the conducting layer, due to its resistance.
16 February 2004
Physics 218, Spring 2004
.
18