Fresnel Equations and EM Power Flow
Reading - Shen and Kong – Ch. 4
Outline
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Review of Oblique Incidence Review of Snells Law
Fresnel Equations
Evanescence and TIR
Brewsters Angle
EM Power Flow
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TRUE / FALSE
1.  This EM wave is TE
(transverse electric) polarized:
Hr
Et
Er
Ht
Ei
Hi
y•
2. Snells Law only works for
TE polarized light.
3. Total internal reflection only occurs when light goes from
a high index material to a low index material.
2
Refraction E&M Waves
Image by NOAA Photo library
http://www.flickr.com/photos/
noaaphotolib/5179052751/ on flickr
Water Waves
Waves refract at the top where
the water is shallower
Refraction involves a change in the direction of wave propagation
due to a change in propagation speed. It involves the oblique
incidence of waves on media boundaries, and hence wave
propagation in at least two dimensions.
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Refraction in Suburbia Think of refraction as a pair of wheels on an axle going from a sidewalk
onto grass. The wheel in the grass moves slower, so the direction of the
wheel pair changes.
Note that the wheels
move faster (longer space)
on the sidewalk, and slower
(shorter space) in the grass.
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Total Internal Reflection in Suburbia Moreover, this wheel analogy is mathematically equivalent
to the refraction phenomenon. One can recover Snells
law from it: n1sinθ1 = n2sinθ2 .
The upper wheel hits the sidewalk and starts to go faster, which turns the axle until the upper wheel re-enters the grass and wheel pair goes straight
again.
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Oblique Incidence (3D view) ki
kiz
θi
x
y
kix
z
Boundary
kix = ki sin(θi) kiz = ki cos(θi) Identical definitions for kr and kt 6
Oblique Incidence at Dielectric Interface Hr
Et
Er
Transverse Electric Field Ht
Ei
Hi
Hr
Et
Er
Transverse
Magnetic Field
Ht
Ei
Hi
y•
Why do we consider only these two polarizations? 7
Partial TE Analysis Hr
E =E
Eo ej(ωt−kk·kr)
E
ωi = ωr = ωt
z=0
Et
Er
- i = ŷE i e−jkix x−jkiz z
E
o
Ht
Ei
Hi
- r = ŷE r e−jkrx x+jkrz z
E
o
y•
- t = ŷE t e−jktx x−jktz z
E
o
Tangential E must be continuous at the boundary z = 0 for all x and for t.
Eoi e−jkix + Eor e−jkrx x = Eot e−jktx x
This is possible if and only if kix = krx = ktx and ωi = ωr =
ωt. The former condition is phase matching.
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Snells Law Diagram
Tangential E field is continuous … kix = ktx
Refraction
Total Internal Reflection kx
kx
Radius = k2
ktx
θr
θt
θc
kz
θt
θc
kix
Radius = k1
k 1 < k2
k 1 > k2
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θt
kz
Snells Law n1
n2
x
y•
z
kix = krx
kix = ktx
n1 sin θi = n1 sin θr
n1 sin θi = n2 sin θt
θ i = θr
Snells Law
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TE Analysis - Set Up - i = ŷEo ej(−kix x−kiz z)
E
z=0
Hr
kr
Er
Ei
Et
kt
θr
θt
θi
- r = ŷrEo ej(−kix x+kiz z)
E
- t = ŷtEo ej(−kix x−kiz z)
E
∂μH
∇×E =−
∂t
- = −jωμH
−jk × E
- = 1 k×E
H
ωμ
Ht
ki
Hi
Medium 1
y•
Medium 2
To get H, use
Faradays
Law
Eo j(−kix x−kiz z)
- i = (ẑkix − xk
ˆ iz )
e
H
ωμ1
kx2 + kz2 = k 2 = ω 2 μ�
- r = (ẑkix + x̂kiz ) rEo ej(−kix x+kiz z)
H
ωμ1
kx = k sin θ
kz = k cos θ
- t = (ẑktx + x̂ktz ) tEo ej(−ktx x−ktz z)
H
ωμ2
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TE Analysis – Boundary Conditions Incident Wavenumber: ki2 = kr2 = ω 2 μ�
Phase Matching:
kix = krx = ktx
θ r = θi
ki sin θi = kt sin θt
Tangential E:
1+r =t
Normal μH:
1+r =t
Eo
tEo
(1 − r)kitz =
kitz
ωμi
ωμt
√
μt εt μi cos θt
ηi cos θt
(1 − r) = t √
=t
μi εi μt cos θi
ηt cos θi
Tangential H:
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Solution Boundary conditions are: z=0
1+r =t
Hr
kr
ηi cos θt
1−r =t
ηt cos θi
Er
Ei
Et
kt
θr
θt
θi
Ht
ki
2ηt cos θi
t=
ηt cos θi + ηi cos θt
Hi
Medium 1
ηt cos θi − ηi cos θt
r=
ηt cos θi + ηi cos θt
y•
Medium 2
Et = tEi
Er = rEi
Fresnel Equations
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Todays Culture Moment
Sir David Brewster
•  Scottish scientist
•  Studied at University of Edinburgh at age 12
•  Independently discovered Fresnel lens
•  Editor of Edinburgh Encyclopedia and contributor
to Encyclopedia Britannica (7th and 8th editions)
•  Inventor of the Kaleidoscope
•  Nominated (1849) to the National Institute of
France.
1781 –1868
Fresnel Lens
Kaleidoscope
All Images are in the public domain
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TM Case is the dual of TE
E Gauss:
k·E = 0
H Gauss:
k·H = 0
Faraday:
k х E = ωμH
Ampere:
k х H = - ωεE
E→H
H→-E
ε→μ
E Gauss:
k·E = 0
H Gauss:
k·H = 0 Faraday:
k х E = ωμH Ampere:
k х H = - ωεE μ→ε
The TM solution can be recovered from the TE solution.
So, consider only the TE solution in detail.
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TE & TM Analysis – Solution TE solution comes directly from the boundary condition analysis
2ηt cos θi
t=
ηt cos θi + ηi cos θt
ηt cos θi − ηi cos θt
r=
ηt cos θi + ηi cos θt
TM solution comes from ε ↔ μ
ηt−1 cos θi − ηi−1 cos θt
r = −1
ηt cos θi + ηi−1 cos θt
2ηt−1 cos θi
t = −1
ηt cos θi + ηi−1 cos θt
Note that the TM solution provides the reflection and transmission
coefficients for H, since TM is the dual of TE.
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Fresnel Equations - Summary
TE-polarization Eor
Ni − N t
rE = i =
Eo
Ni + Nt
2Ni
Eot
tE = i =
Eo
Ni + Nt
1
cos θi
Ni =
η1
1
Nt =
cos θt
η2
TM Polarization rH
Hor
M i − Mt
= i =
Ho
M i + Mt
tH
Mi = η1 cos θi
2Mi
Hot
= i =
Ho
M i + Mt
Mt = η2 cos θt
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Fresnel Equations - Summary From Shen and Kong … just another way of writing the same results TE Polarization
rT E
Eor
μ2 kiz − μ1 ktz
= i =
Eo
μ2 kiz + μ1 ktz
tT E
2μ2 kiz
Eot
= i =
Eo
μ2 kiz + μ1 ktz
TM Polarization 18
rT M
Eor
�2 kiz − �1 ktz
= i =
Eo
�2 kiz + �1 ktz
tT M
2�2 kiz
Eot
= i =
Eo
�2 kiz + �1 ktz
Fresnel Equations
n1 = 1.0
n2 = 1.5
Hr
Et
TE
Ei
Ht
Hi
Hr
Et
Er
Ei
TM
Hi
tTM
1
tTE
Reflection Coefficients
Er
1.5
0.5
0
rTM
θB
rTE
-0.5
Ht
-1
0
y•
19
30
60
Incidence Angle
90
Fresnel Equations
Total Internal Reflection
n1 = 1.5
n2 = 1.0
1
Hr
Et
TE
Ei
0.8
Ht
Hi
Hr
Et
Reflection Coefficients
Er
0.6
0.4
rTE
0.2
0
rTM
θB
-0.2
Er
Ei
TM
Hi
Ht
-0.4
0
y•
20
θC
θC = sin-1(n2 / n1)
30
Incidence Angle θi
60
90
Reflection of Light
(Optics Viewpoint … μ1 = μ2)
Hr
TE
Ht
Reflection Coefficients
Hr
Et
Er
TM
Hi
TM: r� =
n2 cos θi − n1 cos θt
n2 cos θi + n1 cos θt
θc
Hi
Ei
n1 cos θi − n2 cos θt
n1 cos θi + n2 cos θt
Et
Er
Ei
TE: r⊥ =
Ht
n1 = 1.44
n2 = 1.00
|r⊥ |
θB
|r� |
Incidence Angle θi
y•
21
Brewsters Angle
(Optics Viewpoint … μ1 = μ2)
r⊥ = 0
Snells Law
r� = 0
Image in the Public Domain
Snells Law
Requires
n1 = n2
TE Polarized
Unpolarized
Incident Ray Reflected Ray
Can be
Satisfied with
Sir David Brewster
θB = arctan(n2/n1)
θB + θT = 90°
Slightly Polarized
Reflracted Ray
Light incident on the surface is absorbed, and then reradiated by oscillating electric dipoles (Lorentz
oscillators) at the interface between the two media. The dipoles that produce the transmitted (refracted)
light oscillate in the polarization direction of that light. These same oscillating dipoles also generate the
reflected light. However, dipoles do not radiate any energy in the direction along which they oscillate.
Consequently, if the direction of the refracted light is perpendicular to the direction in which the light is
predicted to be specularly reflected, the dipoles will not create any TM-polarized reflected light.
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Energy Transport A cos θi
T⊥
TE
A
R⊥
ni
nt
T�
TM
A cos θr
θi θr
A cos θt
n1 = 1.0
n2 = 1.5
Cross-sectional Areas
Si = Eoi Hoi cos θi
R� θB
θi
-=E
- ×H
S
[Degrees]
St = Eot Hot cos θt
Sr = Eor Hor cos θr
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Transmitted Power Fraction:
2
(Eot )
�
�2
μ2
cos θt
St
Eot Hot cos θt
2 Nt
�
Ts =
= i i
=
=
t
E
2
�1
Si
Eo Ho cos θi
Ni
i
(Eo )
cos
θ
i
μ1
=
t 2
(Eo )
2
(Eoi )
≡ t2E
t 2
(Ho )
2
(Hoi )
2
≡ tH
24
(Hot )
2
(Hoi )
2
�
�
μ2
�2
cos θt
μ1
�1
cos θi
=
t2H
Mt
Mi
Reflected Power Fraction:
Sr
Rs =
=
Si
(Eor )
(Eoi )
2
2
(Hor )
(Hoi )
2
2
2
≡ rE
2
≡ rH
Eor Hor
Eoi Hoi
2
(Eor )
�
�1
μ1
cos θr
2
�
=
=
r
E
2
�1
cos θi
i
(Eo )
μ1
�
2
μ1
(Hor )
�1
2
�
=
= rH
2
μ1
(Hoi )
�1
… and from ENERGY CONSERVATION we know:
T S + RS = 1
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Summary CASE I: E-field is polarized perpendicular to the plane of incidence … then E-field is parallel (tangential) to the surface,
and continuity of tangential fields requires that:
1 + rE = tE
T S + RS = 1 (tE )
2 Nt
Ni
(1 + rE )
Eor
Ni − Nt
rE = i =
Ni + Nt
Eo
+ (rE )2 = 1
2 Nt
Ni
+ (rE )2 = 1
2Ni
Eot
tE = 1 + rE = i =
Ni + Nt
Eo
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Summary CASE II: H-field is polarized perpendicular to the plane of incidence 1 + rH = tH
… then H-field is parallel (tangential) to the surface,
and continuity of tangential fields requires that:
T S + RS = 1 Mt
2
(tH )
+ (rH ) = 1
Mi
2
Mt
2
(1 + rH )
+ (rH ) = 1
Mi
2
rH
Hor
M i − Mt
= i =
M i + Mt
Ho
tH = 1 + rH
27
2Mi
Hot
= i =
Mi + Mt
Ho
MIT OpenCourseWare
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6.007 Electromagnetic Energy: From Motors to Lasers
Spring 2011
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