Wave Propagation in Media
Typically, the speed of light, the wavelength, and the amplitude decrease.
Vacuum (or air)
Medium
n=1
n=2

Absorption depth = 1/
Absorptive
nk
k
n
E(x,t) = E0 exp[i(kx – t)]
Wavelength decreases
Dispersive
E(x,t) = E0exp[(–/2)x]exp[i(nkx– t)]
where  is the "absorption coefficient" and n is the "refractive index."
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I(z) = I(0) exp(- x)
r  E0 r / E0i   ni cos(q i )  nt cos(q t ) /  ni cos(q i )  nt cos(q t )
t  E0t / E0i  2ni cos(q i ) /  ni cos(q i )  nt cos(q t ) 
ki
Er
Ei
Bi
Interface
ni
Br
qi qr
qt
Et
Bt
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kr
nt
kt
r||  E0 r / E0i   ni cos(qt )  nt cos(qi ) /  ni cos(qt )  nt cos(qi )
t||  E0t / E0i  2ni cos(qi ) /  ni cos(qt )  nt cos(qi )
ki
kr
Ei
Bi
Br
qi qr
ni
×
Er
Interface
Beam geometry
for light with its
electric field
parallel to the
plane of incidence
(i.e., in the page)
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qt
Et
Bt
kt
nt
Reflection & Transmission Coefficients
for an Air-to-Glass Interface
nair  1 < nglass  1.5
Total reflection at q = 90°
for both polarizations
Zero reflection for parallel
polarization at 56.3°
“Brewster's angle”
(For different refractive
indices, Brewster’s angle
will be different.)
Reflection coefficient, r
Note:
1.0
Brewster’s angle
.5
r||=0!
0
r
┴
-.5
-1.0
0°
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r||
30°
60°
90°
Incidence angle, qi
Reflection Coefficients for a Glassto-Air Interface
1.0
Note:
Total internal reflection
above the "critical angle"
qcrit  arcsin(nt /ni)
(The sine in Snell's Law
can't be > 1!)
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Reflection coefficient, r
nglass  1.5 > nair  1
Critical
angle
r
┴
.5
Total internal
reflection
0
Brewster’s
angle
-.5
Critical
angle
r||
-1.0
0°
30°
60°
90°
Incidence angle, qi
Reflectance (R)
2
  0 c0 
I  n
E
 0
 2 
R  Reflected Power / Incident Power

wi
ni
nt
qi qr
I r Ar
I i Ai
A = Area
wi
Because the angle of incidence = the angle of reflection,
the beam area doesn’t change on reflection.
Also, n is the same for both incident and reflected beams.
R  r2
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2
 c 
I   n 0 0  E0
 2 
Transmittance (T)
T  Transmitted Power / Incident Power 
If the beam
has width wi:
wi
I t At
I i Ai
qi
ni
nt
qt
wt
A = Area
At wt cos(qt )


Ai wi cos(qi )
The beam expands in one dimension on refraction.
2
  0 c0 
nt
E0t
2


n
E
wt nt wt 2
 wt 
I t At 
2 
t
0t
T



t since


2
2
I i Ai   0 c0 
 wi  ni E0i wi ni wi
n
E
 i
 0i
2



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  nt cos qt    2
T 
t
  ni cos qi   
E0t
E0i
2
2
 t2
Reflectance and Transmittance for
an Air-to-Glass Interface
Perpendicular polarization
1.0
Parallel polarization
1.0
T
T
.5
.5
R
R
0
0
0°
30°
60°
90°
Incidence angle, qi
Note that
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R+T =1
0°
30°
60°
Incidence angle, qi
90°
Practical Applications of Fresnel’s
Equations
Windows look like mirrors at night (when you’re in the brightly lit room)
Lasers use Brewster’s angle components to avoid reflective losses:
R = 100%
0% reflection!
Laser medium
R = 90%
0% reflection!
Optical fibers use total internal reflection. Hollow fibers use highincidence-angle near-unity reflections.
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Optical fiber types
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Wave Propagation in Media
Typically, the speed of light, the wavelength, and the amplitude decrease.
Vacuum (or air)
Medium
n=1
n=2

Absorption depth = 1/
Absorptive
nk
k
n
E(x,t) = E0 exp[i(kx – t)]
Wavelength decreases
Dispersive
E(x,t) = E0exp[(–/2)x]exp[i(nkx– t)]
where  is the "absorption coefficient" and n is the "refractive index."
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I(z) = I(0) exp(- x)
EM Waves in Dielectrics
2E
2


E  0
2
z
We'll assume that the wave won't be dramatically affected by the
induced polarization, so the wave won’t change much.
The Slowly Varying Envelope Approximation
E ( z, t )  E0 ( z ) exp i  kz  t  and P( z, t )  P0 ( z ) exp i  kz  t 
Specifically, the envelopes, E0 and P0, are assumed to vary slowly, and
the fast variations will all be in the complex exponentials,exp i  kz  t  


E ( z, t )  E0


 ik E0 ( z )  exp i  kz  t  
z
 z


E0
 2 E ( z , t )   2 E0
2
  2  2ik
 k E0  exp i  kz  t  
z 2

z

z


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EM Waves in Dielectrics
2E
2


E  0
2
z
Substituting the derivatives into the inhomogeneous wave equation:
E0
2ik
  2 0  0 E0   2E0
z
Now, because k c and
P   0 ( r  1) E
E0
2ik
  0 2 P0
z
If P0 is constant, the integration is trivial. Usually, however, P0 = P0 (E0).
How to describe polarization in solids?
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A Charged Simple Harmonic Oscillator
Polarization of a single charge oscillator
p (t )  er (t )
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Dipole Oscillator Model of A Solid
The microscopic treatment of polarization
(one oscillator)
(solid)
 (r )  unk (r )eikr + many-body interactions
Polarization of a medium:
sum of dipole oscillators (charged harmonic oscillators)
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P(t )  Ner (t )
The Damped Forced Oscillator
As we talked in Chapter 3, a damped forced oscillator is a harmonic
oscillator experiencing a sinusoidal force and viscous drag.
d 2 xe
dxe
me 2  me
 me 02 xe  eE0 exp(i t )
dt
dt
The solution is now:


(e / me )
xe (t )   2
 E (t )
2
 (0    i ) 
The electron still oscillates at the light frequency, but with an amplitude
and a phase that depend on the relative frequencies.
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Recall P(t )  Ner (t )
Ne 2 / me
P( z, t ) 
E ( z, t )
2 (0    i / 2)
Ne 2 / me

E0 ( z ) exp i  kz  t  
2 (0    i / 2)
Slowly Varying Envelope
Ne 2 / me
P0 ( z ) 
E0 ( z )
2 (0    i / 2)
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Solving For the Wave Envelope
E0
Ne2 / me
2
2ik
  
E0 ( z )
z
2 (0    i / 2)
or:
E0
Ne2 / me
 2
i
E0 ( z )
z
2k 2 (0    i / 2)
This differential equation is of the form:
dy / dx  iay
which has the solution: y ( x)  y (0) exp(iax)
Here, the constant is complex, so write:
So:
a  (n  1)k  i / 2
E0 ( z )  E0 (0) exp i[i / 2  (n  1)k ]z
where a is the "absorption coefficient" and n is the "refractive index."
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Why do we define a  (n  1)k  i / 2 ?
The electromagnetic wave in the medium becomes (combining the
slowly varying envelope with the complex exponential):
E ( z, t )  E0 (0) exp i[i / 2  (n  1)k ]z exp[i(kz  t )]
Simplifying:
E ( z , t )  E0 (0) exp[( / 2) z ]exp[i (nkz  t )]
Absorption causes
attenuation of the field
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Refractive index
changes the k-vector
Refractive index and Absorption coefficient
n comes from the real part
  2

Ne 2 / me
(n  1)k  Re 

2
k
2

(




i

/
2
0


and  comes from the imaginary part of the constant:
  2

Ne 2 / me
 / 2  Im 

2
k
2

(




i

/
2
0


Simplifying:
Ne 2 / me
(0   )
n 1 
4 0 (0   ) 2  ( / 2) 2

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Ne 2 / me
 /2

2 0 c0 (0   ) 2  ( / 2) 2
Complex Lorentzian
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Refractive index and Absorption coefficient
Ne2 / me
 /2

2 0 c0 (0   ) 2  ( / 2) 2
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Ne 2 / me
0  
n 1 
4 0 (0   ) 2  ( / 2) 2
Refractive Index vs. Wavelength
Since resonance frequencies exist in many spectral ranges,
the refractive index varies in a complex manner.
Electronic resonances usually occur in the UV; vibrational and
rotational resonances occur in the IR, and "inner-shell" electronic
resonances occur in the x-ray region.
n increases with frequency, except in "anomalous dispersion" regions.
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Quiz: Reflection at Normal Incidence
When qi = 0,
 nt  ni 
R  

n

n
i 
 t
and
T

2
4 nt ni
 nt  ni 
2
For an air-glass interface (ni = 1 and nt = 1.5),
R = 4% and T = 96%
The values are the same, whichever direction the light travels, from
air to glass or from glass to air.
The 4% has big implications for photography lenses.
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