Fresnel Equations and Light Guiding
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
1
TRUE / FALSE
1.  The Fresnel equations describe reflection and
transmission coefficients as a function of intensity.
2.  This is the power reflection
and transmission plot for an
EM wave that is TE
(transverse electric) polarized:
T
n1 = 1.0
n2 = 1.5
R
θi
θB
[Degrees]
3. The phase matching condition for refraction is a direct
result of the boundary conditions.
2
Oblique Incidence at Dielectric Interface
Hr
Et
Er
Transverse
Electric Field
Ht
Ei
Hi
Hr
Er
Et
Transverse
Magnetic Field
Ht
Ei
Hi
y•
3
Partial TE Analysis
i = ŷE i e−jkix x−jkiz z
E
o
z=0
Hr
Et
Er
r = ŷE r e−jkrx x+jkrz z
E
o
Ht
Ei
Hi
t = ŷE t e−jktx x−jktz z
E
o
ωi = ωr = ωt
y•
Tangential E must be continuous at the boundary z = 0 for all x and for t.
Eoi e−jkix x + 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 kix = krx = ktx
4
Snells Law
Er
n1
n2
Et
Ei
y•
kix = krx
kix = ktx
n1 sin θi = n1 sin θr
n1 sin θi = n2 sin θt
Snells Law
θ i = θr
5
TE Analysis - Set Up
Hr
kr
Er
Ei
i = ŷEo ej(−kix x−kiz z)
E
z=0
Et
r = ŷrEo ej (−kix x+kiz z)
E
kt
t = ŷtEo ej(−kix x−kiz z)
E
θr
θt
θi
∂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
i = (ẑkix − x̂kiz ) Eo ej(−kix x−kiz z)
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
6
TE & TM Analysis – Solution
TE solution comes directly from the boundary condition analysis
ηt cos θi − ηi cos θt
r=
ηt cos θi + ηi cos θt
2ηt cos θi
t=
ηt cos θi + ηi cos θt
TM solution comes from ε μ
2ηt−1 cos θi
t = −1
ηt cos θi + ηi−1 cos θt
ηt−1 cos θi − ηi−1 cos θt
r = −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.
7
Fresnel Equations - Summary
From Shen and Kong … just another way of writing the same results
TE Polarization
rTE
Eor
μ2 kiz − μ1 ktz
= i =
μ2 kiz + μ1 ktz
Eo
tTE
Eot
2μ2 kiz
= i =
μ2 kiz + μ1 ktz
Eo
TM Polarization
8
rTM
Eor
2 kiz − 1 ktz
= i =
2 kiz + 1 ktz
Eo
tTM
Eot
22 kiz
= i =
2 kiz + 1 ktz
Eo
Reflection of Light
(Optics Viewpoint … μ1 = μ2)
Hr
Hi
y•
Et
Er
Ht
Ei
TM
Hi
TM: r =
n2 cos θi − n1 cos θt
n2 cos θi + n1 cos θt
E-field
parallel to
the plane of
incidence
9
θc
Reflection Coefficients
E-field
perpendicular
to the plane of
Ht incidence
TE
Hr
n1 cos θi − n2 cos θt
n1 cos θi + n2 cos θt
Et
Er
Ei
TE: r⊥ =
n1 = 1.44
n2 = 1.00
|r⊥ |
θB
|r |
Incidence Angle θi
Brewsters Angle
Incident ray
(unpolarised)
Reflected ray
(TE polarised)
θB + θt = 90°
n1
n2
Image in the Public Domain
Sir David Brewster (1781 –1868)
was a Scottish scientist, inventor
and writer. Rediscovered and
popularized kaleidoscope in 1815.
n1 sin (θB ) = n2 sin (90 − θB )
Refracted ray
(slightly polarised)
= n2 cos (θB )
θB = arctan
@ θi = θB
TE: r⊥ =
n1 cos θB − n2 cos θt
= 0
n1 cos θB + n2 cos θt
n2 cos θB = n1 cos θt
n1 sin θB = n2 sin θt
n2 cos θB − n1 cos θt
TM: r =
=0
n2 cos θB + n1 cos θt
10
n2
n1
Total Internal Reflection
Beyond the critical angle,
refraction no longer occurs
–  thereafter, you get total internal reflection
TOTAL INTERNAL REFLECTION
n2sinθ2 = n1sinθ1 θcrit = sin-1(n1/n2)
Image in the Public Domain
–  for glass (n2 = 1.5), the critical internal angle is 42°
–  for water, its 49°
–  a ray within the higher index medium cannot escape at
shallower angles (look at sky from underwater…)
incoming ray hugs surface
n1 = 1.0
n2 = 1.5
42°
11
Snells Law Diagram
Tangential field is continuous …
kix = kit
Refraction
Total Internal Reflection
kx
kx
Radius = k2
ktx
θr
θt
θc
kz
θt
θc
kix
Radius = k1
k 1 < k2
k 1 > k2
12
θt
kz
Total Internal Reflection & Evanscence
Snells Law dictates n1 sin(θi) = n2 sin(θt) , or
equivalently, kix = ktx . For n1 > n2 , θt = 90° at
θi = sin-1(n2/n1) θC. What happens for θi > θC ?
n1 > n2
kx
kr
θc
kt
ktz2 = kt2 – ktx2 < 0 ktz = ± j αtz , with αtz real.
The refracted, or transmitted, wave takes
the complex exponential form
θt
kz
θc
ki
1 2
exp(- j ktx x - αtz z) .
This is a non-uniform plane wave that travels in
the x direction and decays in the z direction. It
carries no time average power into Medium 2.
This phenomenon is referred to as total internal
reflection. This is the similar to reflection of
radio waves by the ionosphere.
13
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.
14
Frustrated Total Internal Reflection In Suburbia
D
k1
d
An evanescent field can propagate once the
field is again in a high-index material.
15
Applications of Evanescent Waves
fingertip
Image in the Public Domain
light
source
camera
The camera observes TIR from a
fingerprint valley and blurred TIR
from a fingerprint ridge.
Image in the Public Domain
16
Light Propagating
Through a Multimode Optical Fibre
Jacket
400 μm
Buffer
250 μm
Cladding
125 μm
Core
8 μm
Single Mode Fibre Structure
Image in the Public Domain
The optic fiber used in undersea cables is chosen for its exceptional clarity, permitting runs of more than
100 kilometers between repeaters to minimize the number of amplifiers and the distortion they cause.
Image in the Public Domain
A cross-section of a
submarine communications cable:
Submarine communication cables
crossing the Scottish shore
1. Polyethylene
2. "Mylar" tape
3. Stranded steel wires
4. Aluminum water barrier
5. Polycarbonate
6. Copper or aluminum tube
7. Petroleum jelly
8. Optical fibers
Typically 69 mm in diameter and weigh around 10 kg per meter
17
Image by Jmb at http://en.wikipedia.org/
wiki/File:Submarine_Telephone_Cables_
PICT8182_1.JPG on Wikipedia.
Optical Waveguides Examples
Image by Apreche
http://www.flickr.com/photos/
apreche/69061912/ on flickr
Image by Rberteig
http://www.flickr.com/photos/
rberteig/89584968/ on flickr
LCD screen lit by two backlights coupled
into a flat waveguide
Optical fiber
18
Image by Mike Licht
http://www.flickr.com/photos/notionscapital/
2424165659/ on flickr
Todays Culture Moment
Global Fiber Optic Network
Image in the Public Domain
Image by MIT OpenCourseWare
Image by Paul Keleher http://commons.wiki
media.org/wiki/File:Trench_USA-fiber.jpg
on Wikimedia Commons.
19
Todays Culture Moment
Laying Transcontinental Cables
Image in the Public Domain
20
METAL REFLECTION
© Kyle Hounsell. All rights reserved.
This content is excluded from our
Creative Commons license. For more
information, see
http://ocw.mit.edu/fairuse
Three Ways to Make a Mirror
MULTILAYER REFLECTION
Image is in the public domain
Image is in the public domain:
http://en.wikipedia.org/wiki/
File:Dielectric_laser_mirror_from_a_dye_las
er.JPG
TOTAL INTERNAL REFLECTION
21
Transporting Light
We can transport light along the z-direction by bouncing it between two mirrors
y
Mirror
x
Mirror
d
z
θ
..the ray moves along both y- and z-axes..
Ex (y, z) = Ae±jky y e−jβz
…where
ky = nko sin θ
kz = nko cos θ
22
nko
θ
kz
ky
Transverse Electric (TE) Modes
d
−jβz
+jky y
−jky y
Ex (y, z) = A1 e
+ A2 e
e
= am um (y)e−jβm z
y
x
Mirror
Mirror
STANDING WAVE
IN y-DIRECTION
d/2
0
−d/2
z
m=1
2
3
6
23
Perfect Conductor Waveguide
y
Mirror
x
Mirror
d/2
0
−d/2
z
m=1
2
3
6
2/d cos ky y if m = odd
um (y) = 2/d sin ky y if m = even
kym
Boundary
Conditions
π
=m
d
mπy 2/d cos
d um (y) = mπ
y
2/d sin
d
24
if m = odd
if m = even
Transporting Light
y
Mirror
x
Mirror
d/2
0
−d/2
z
m=1
2
3
6
2/d cos ky y if m = odd
um (y) = 2/d sin ky y if m = even
kym = nko sin θm
nko
ky
θ
kz,m
π
=m
d
kz,m = nko cos θm
25
2
kz,
m
2
π
= k 2 − m2 2
d
Transporting Light
y
Mirror
x
Mirror
d/2
0
−d/2
z
m=1
2
3
6
nko
ky
θ
kz,m
The solutions can be plotted along a circle of radius k=nko…
ky = ko sin θ
nko
M
kym = nko sin θm
m
π
d
kz,m = nko cos θm
3
2
1
kz,m kz,m
nko
= nko cos θm
π
=m
d
26
Waveguide Mode Propagation Velocity
Mirror
Mirror
Velocity along the direction of the guide…
…steeper angles take longer to travel through the guide
27
Lowest Frequency Guided Mode
Number of Modes
Cut-off Frequency
Frequency of light (color)
Mirror
Mirror
28
Solutions for a Dielectric Slab Waveguide
y
d
2
m=1
2
3
4
8
0
d
−
2
z
What does it mean to be a mode of a waveguide?
29
Slab Dielectric Waveguides
ky
n2 k o
n1 k o
M
n1 ko sin θc
steepest incidence angle
m
θc
1
θm
0
0
0
n2 k o
n1 k o
30
shallowest incidence angle
kz = β
Comparison of Mirror Guide and Dielectric Waveguide
Metal Waveguide
Dielectric Waveguide
θ
n2
n1
n2
31
Key Takeaways
n1 > n2
kx
Total Internal Reflection. What happens for θi > θC ?
ktz2 = kt2 – ktx2 < 0 ktz = ± j αtz , with αtz real.
kr
kt
θc
Evanescent field
θt
kz exp(- j ktx x - αtz z)
θc
ki
Waveguide Modes
mπy 2/d cos
d um (y) = mπy
2/d sin
Mirror
d
1 2
x
y
Mirror
d/2
0
−d/2
m=1 2
3
z
6
32
if m = odd
if m = even
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6.007 Electromagnetic Energy: From Motors to Lasers
Spring 2011
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