( ) [ ]

Consider the case of Total Internal Reflection (TIR):
nt
1
nti = =
ni 1.5
ni sin θ i = nt sin θ t
When θ t = 90°,
⇒ θ c = 41.8°
[sin θi ]θ
(
Since cos θ t = 1 − sin θ t
and
ni = 1.5
2
)
1/ 2
(
θi
= 1 − nit sin θ i
2
2
)
1/ 2
ni nt
nit ⋅ nti = ⋅ = 1, with a little algebra, we can write:
nt ni
 Eor
r⊥ ≡ 
 Eoi
and
nt = 1
nt
1
= =
ni 1.5
c
θt
(
(

ni cos θ i − nt cos θ t cos θ i − nti2 − sin 2 θ i
 =
=
2
2
 ⊥ ni cos θ i + nt cos θ t cos θ i + nti − sin θ i
(
(
nt cos θ i − ni cos θ t n cos θ i − n − sin θ i
r|| =
=
ni cos θ t + nt cos θ i nti2 cos θ i + nti2 − sin 2 θ i
2
ti
2
ti
2
)
)
)
)
1/ 2
1/ 2
1/ 2
1/ 2
[ni sin θ i ]θ =θ
i
Since sin θc = nti , when θi > θc,
c
= [nt sin θ t ]θ t =90° ⇒ sin θ c =
Consider nti =
nt
= nti
ni
nt
1
=
ni 1.5
sin θi > nti and r⊥ and r|| become
complex and so r⊥ r⊥∗ = r|| r||∗ = 1
and R = 1 which gives Ir= Ii and It = 0
since R + T = 1.
When It = 0, this means that on
average the transmitted wave cannot
carry energy across the interface
boundary in the case of TIR.
tan θ p' =
nt
1
=
ni 1.5
⇒ θ p' = 33.7°
θ c = sin −1 (1 / 1.5) = 41.8°
Examples of Total Internal Reflection (TIR)
Consider nti =
nt
1
=
ni 1.5
θ c = sin −1 (1 / 1.5) = 41.8°
Consider now the transmitted E-field:
(
y
)
Et = Eot exp i kt ⋅ r − ωt
and kt ⋅ r = ktx x + kty y
with ktx = kt sin θ t
θt
kt cos θ t
kt sin θ t
x
and
kty = kt cos θ t = ± kt 1 − sin 2 θ t
= ± kt 1 −
nt = 1
θi
ni = 1.5
1
2
sin
θi
2
nti
However, when sin θi > nti we have
kty = ±ikt
1
2
θ i − 1 = ± iβ
sin
2
nti
with β = kt
1
2
sin
θi − 1
2
nti
and
and
ktx =
thus
kt
sin θ i
nti
with i 2 = −1
Et = Eot exp i (ktx x + kty y − ωt )
Upon substituting ktx and kty:
Et = Eot exp i (ktx x + kty y − ωt ) = Eot exp i (kt x sin θ i / nti ± iβy − ωt )
= Eot exp(∓ βy ) exp i (kt x sin θ i / nti − ωt )
Et
ni=1.5
Eot exp(− βy )
Only the decaying solution is
physically reasonable.
nt=1 (less dense medium)
y
y = 0 (interface) The E-M wave propagates in the x-direction a Surface
or Evanescent Wave.
The amplitude decays rapidly in the y-direction, becoming negligible at a
distance of only a few wavelengths from the interface.
Frustrated Total Internal Reflection (FTIR)
Form a sandwich having a glass-air-glass interface:
Precision Spacer
n = 1.5
kt
n ≈ 1 ke
n = 1.5
ki
kr
∆y
y
x
3D representation with plane waves
near the interface boundary.
The amount of energy in the transmitted wave can be controlled by adjusting
the spacer thickness ∆y. The phenomenon is therefore analogous to barrier
penetration or tunneling of electron waves in quantum mechanics.
(a) Beamsplitters using FTIR
(c) Commercial Beamsplitter
cubes commonly found in
optics laboratories.
(b) Conventional
beamsplitter
arrangement used to
take photographs
through a microscope
and thus a current
application of FTIR.
Optical Properties of Metals
The presence of an electric field
causes a current, owing to the
E(r,t)
conductivity σ. Note that j = σE which is Ohm’s law at the microscopic
level. We can write Maxwell’s equations for an homogeneous, isotropic
medium containing charge carriers, such as electrons in a non-magnetic metal:
∂B
∇⋅ B = 0 ∇× E = −
∂t
∂E
∇⋅E = ρ /ε
∇ × B = µσE + µε
∂t
rd
th
We can easily manipulate the 3 and 4 equations:
2
∂ ∂E
∂ E
∇ × ∇ × E = − (∇ × B) = − µσ
− µε 2
∂t
∂t
∂t
(
)
(
) (
)
2
Also, we will make use of the vector identity: ∇ × ∇ × A = ∇ ∇ ⋅ A − ∇ A
2
∂E
∂ E
2
∇ ∇ ⋅ E − ∇ E = − µσ
− µε 2 and ∇ ⋅ E = ρ = 0
∂t
∂t
2
∂ E
∂E which is the wave equation in an uncharged
2
⇒ ∇ E = µε 2 + µσ
∂t
∂t conducting medium (i.e. a metal).
(
)
(
)
~2 2
Since E = E0 exp i k ⋅ r − ωt , − k E = − µεω E − iωµσE
~2
~ ω
c ~
2
~
⇒ k = µω (ε + iσ / ω ); n = k = nR − inI ; k = (nR − inI )
ω
c
Thus, the wave equation for a metal gives a complex dispersion relation k(ω)
in which k (~ tilda) here is complex; which leads to a complex index of
refraction containing real (nR) and imaginary (nI) parts. Note that the last
term in the wave equation is like a damping force and is responsible for
absorption. Let’s write the propagating wave as
 k  E = Eo cos(ωt − ky ) = Eo cos ω  t − y  = Eo cos ω (t − n~y / c )
 ω 
Putting this in a more general form involving a complex exponential:
~
E = Eo exp iω (t − n y / c ) = Eo exp iω (t − (nR − inI ) y / c )
= Eo exp(− ωnI y / c ) exp iω (t − nR y / c )
Re E = Eo exp(− ωnI y / c ) cos ω (t − nR y / c )
2
Since I ∝ E
and
I ( y ) = I o exp(− αy )
with I o = I (0),
comparison gives α ≡ 2ωnI / c
This is the absorption or attenuation coefficient α and d = 1/α is known as the
skin or penetration depth. Note that α depends on the frequency or
wavelength. For materials that are transparent, 1/α >> thickness.
For metals, take Cu for example: 1/α ≈ 6Å at λ = 100 nm and 60 Å at λ =
10,000 nm (IR). This explains the opacity of metals in which the penetration
depth is very small and most of the energy is reflected.
Imagine the metal as a collection of driven and damped oscillators. Some of
the electrons in the metal (valence electrons) are free to move and thus possess
no restoring force. Other electrons are bound to the atoms such as in a
dielectric material like glass. Remember for an oscillating dipole we derived
the classical expression from Newton’s second law:
d 2x
FE − k s x = qe Eo cos ωt − meωo x = me 2
dt
2
⇒ qe Eo cos ωt − meωo xo cos ωt = −meω 2 xo cos ωt
2
qe E o
qe E o
qe E (t )
; x=
cos ωt ; x =
⇒ xo =
2
2
2
2
2
me ωo − ω
me ωo − ω
me ωo − ω 2
(
)
(
)
(
)
For the free electrons in a metal, there is no restoring force and ωo=(ks/me)1/2 ≈ 0.
It is as though electrons are attached by very weak springs. Therefore we have
qe E (t )
FE
x(t ) = −
=−
2
meω
meω 2
The minus sign indicates that the free electrons oscillate 180° out-of-phase relative
to the E-field of the incident light. This oscillation therefore creates light waves that
will cancel (interfere destructively with the incident wave) when radiating in the
same direction as the incident wave. The result is a rapidly decaying refracted
wave, as we have just seen with the attenuation. We can extend the dispersion
relation to include both types of electrons (bound and free valence e-’s:
P (t )
ε
2
P = (ε − ε o ) E ⇒ ε = ε o +
, n =
E (t )
εo
Understanding the relative phases for oscillating electrons
qe E (t )
x=
2
me ωo − ω 2 + iγω
(
)
exp(i∆ϕ ) = exp(i 0) = 1
exp(i∆ϕ ) = exp(iπ / 2) = i
exp(i∆ϕ ) = exp(iπ ) = −1
exp(i∆ϕ ) = exp(iπ ) = −1
free e-’s
2
Therefore
qe N
2
n (ω ) = 1 +
ε o me
bound e-’s



fj
fe

+ ∑ 2

2
2


j  ωoj − ω + iγ j ω  
 − ω + iγ eω

where N is the number of atoms per unit volume; fe is the number of free
valence electrons per atom. The second term involving the sum again refers
to the bound electrons, as we have seen previously for dielectric materials.
If a metal has a particular color (Gold and copper are reddish yellow) it
indicates that the atoms of the metal are involved with the selective
absorption via bound electrons. We can approximate the above equation
further by assuming a negligible contribution from bound e-’s and γe ≈ 0 for
large ω, i.e. a small dissipation at high frequencies:
 ωp 
qe N 1
2
n (ω ) = 1 −
= 1 −  
2
ε o me ω
ω 
2
2
2
with ω p =
qe N
ε o me
Consider the behavior of n(ω) for limiting cases in metals:
(i) ω < ωp ⇒ n2(ω) < 0 and n is complex. Since α = 2ωnI/c, the
absorption will be large.
(ii) ω > ωp ⇒ n2(ω) > 0 and n is real, absorption is small and we
observe a transparency condition. Notably, x-rays (100 Å > λ > 1Å)
will penetrate metals.
(iii) Some metals, such as Alkali metals, are transparent even in the
ultraviolet. The table below illustrates the calculated plasma
frequencies and wavelengths for some Alkali metals.
In general, n is complex and so there is absorption in metals when there is a
finite nI.
Consider reflection from a metal surface at normal incidence (θi = 0).
We derived
 nt − ni 
Ir
2

R = = r = 
Ii
 nt + ni 
2
We derived this for a
dielectric.
For an air-metal interface, take
ni = 1 and
nt = n~ = nR − inI
*
~
~
n −1
 n − 1  n − 1 
Since r =
⇒ R = r = r *r =  ~  ~ 
n +1
 n + 1  n + 1 
2
 nR − 1 − inI
= 
 nR + 1 − inI
 nR − 1 + inI

 nR + 1 + inI
 (nR − 1)2 + nI 2
 =
2
2
(
)
n
+
1
+
n

R
I
Typical values of nR, nI, and R at λ=589 nm
Notice that the reflectance of Na has a
larger R despite its smaller nI.
Na
Sn
Ga
nR 0.04
1.5
3.7
nI
2.4
5.3
5.4
R
0.9
0.8
0.7
1) If the conductivity σ→0 which gives a dielectric with nI ≈ 0,
nI << nR, and α→0, nR ≈ nt
2) If σ is large, nI is large and nI >> nR and R→1.
nI(λ),
nR(λ)
Ag is transparent at
λp ≈ 320 nm, at
which point nI < nR
and R << 1.
Ag
Note that Au and Cu exhibit an increasing R in the range 500 nm < λ < 700 nm,
and this provides the reddish-yellow color when the metals are illuminated with
white light.
Note that the dip in R|| is similar to the case of
Brewster’s angle (θp) for dielectric interfaces
except that R|| > 0 at its minimum, as shown.
Phase shifts occur during reflection from a
metal surface. Both E|| and E⊥ experience
phase shifts which are in general between
0 and 180°. When θi = 90°, both E|| and E⊥
experience ∆ϕ = 90°.
Bulk Plasmons and the dispersion relation ω(k) for a metal
We can write the E-M wave equation as
∂ D
2
∇ E = µ o 2 and D = ε (ω , K ) E
∂t
and E = Eo exp i K ⋅ r − ωt
⇒ ε (ω , K ) µ o ω 2 = K 2
2
(
)
Note that ε(ω, K) can be complex in general and lead to K which is
complex, as we have seen in FTIR. We saw already solutions in
which we assumed that the contribution from bound electrons and
damping were assumed negligible:
ω2
ε (ω )
n2 =
εo
= 1−
ω
p
2
Inserting this into the first equation above, we get
 ω p2 
⇒ 1 − 2 ε o µ o ω 2 = k 2
 ω 


and ω 2 − ω p2 = c 2 k 2
We assume propagating solutions in which K=k is real.
Bulk Plasmon Dispersion Relation
ω
ω = ω p2 + c 2 k 2
ωp
Solutions lie above the light line.
Typical values for the plasmon
energy are
e
in
l
t
h
g ck
i
L =
ω p ≈ 10eV …Metals
ω p < 0.5eV...Semiconductors
(depends on dopant concentration).
ω
No allowed propagating
modes
k
Surface Plasmon Polaritons
Consider an E-M wave that can propagate along the interface
between metal and vacuum (i.e. the surface of a metal) or
between the metal and a dielectric material. Again, we assume
solutions to the E-M wave equation.
The solutions E(r,t) are of the form:
E x = A exp(ik x x ) exp(− k zm z ), E y = 0, E z = B exp(ik x x ) exp(− k zm z ), z > 0
E x = C exp(ik x x ) exp(+ k zd z ), E y = 0, E z = D exp(ik x x ) exp(+ k zd z ), z < 0
K m = k x + ik zm and K d = k x − ik zd
z
We need to invoke continuity conditions
of
E|| and (εE)⊥, along with Gauss’s law: ∇ ⋅ E = 0
Metal εm < 0
0
Dielectric εd > 0
∇ ⋅ E = 0 ⇒ ik x E x = k zm E z ⇒ ik x A = k zm B ( z > 0)
∇ ⋅ E = 0 ⇒ ik x E x = −k zd E z ⇒ ik x C = −k zd D ( z < 0)
k zm
AD
⇒
=−
k zd
BC
x
We can further remove the time dependence in the E-M wave
equation by inserting E = E (r ) exp(−iωt )
2
∂ E
2
2
2
2
2
∇ E = µ oε (ω , K ) 2 = −ω µ oε (ω , K ) E ⇒ − k x + k zm = − ω µ oε m (ω , K )
∂t
(z > 0) (1)
2
2
2
and − k x + k zd = −ω µ oε d (ω , K ) (z < 0) (2)
Continuity of Ex requires A = C and for Ez requires εmB =εdD. Therefore
k zm
ε
AD
=−
=− m
k zd
BC
εd
(3)
Subtracting Eqs. (1) and (2) and eliminating kzd using Eq. (3) we get
k
2
zm
= −ω µ o
2
ε m2
εm + εd
ε mε d
and k x = ω µ o
εm + εd
εreal
εd
Examine the dielectric constants for both the dielectric and metal
layers and determine ω(kx) for surface plasmons.
2
ω
(
)
ε
ω
p
n2 =
= 1− 2
εo
ω
ωsp
0
µ
d
ε
/(
ωp
-εd
ω
k x = ω µo
ε mε d
εm + εd
1) For small ω, εm→ -∞ and
kx ≈ ω(εdµo)1/2 = ω/vd
2) At ω = ωsp , εm→ - εd
(from below) and kx → +∞
Li
ω
e
lin
t
gh
=
2
1/
)
o
kx
ωsp
Surface plasmon
dispersion is below
the light line.
kx
Surface Plasmon polaritons (SPPs) in metal nanostructures: the optoelectronic
route to nanotechnology, M. Salerno et. al., Opto-electronics Rev. 10(3), 217 (2002).
(a)
klight
θ
kSPP
nkG
a
ω
(b)
ωSPP
Light dispersion
nkG
klight⋅sinθ
k
kSPP
Light is injected from the left, creating
surface plasmons (SPs) which propagate
to the right. When they reach the end of
the gold strip (point S), a transfer of
momentum to the SPs causes energy to
be transferred back to light, which is
subsequently detected with a photon
scanning tunneling microscope (PSTM).