EM Reflection & Transmission in Layered Media
Reading - Shen and Kong – Ch. 4
Outline
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Review of Reflection and Transmission
Reflection and Transmission in Layered Media
Anti-Reflection Coatings
Optical Resonators
Use of Gain
1
TRUE or FALSE
2
1. The refractive index of glass is approximately n = 1.5 for
visible frequencies. If we shine a 1 mW laser on glass, more
than 0.5 mW of the light will be transmitted.
Laser
?
?
2
Reflection & Transmission of EM Waves at Boundaries
E
H
incident wave
Normal Incidence
transmitted wave
Medium 1
Medium 2
Additional Java simulation at
http://phet.colorado.edu/new/simulations/
Animation © Dr. Dan Russell, Kettering University. All rights reserved. This
content is excluded from our Creative Commons license. For more information,
see http://ocw.mit.edu/fairuse.
3
Incident EM Waves at Boundaries
E
H
Medium 1
Incident Wave
incident wave
Medium 2
Known
i = x̂E i e−jk1 z
E
o
1
1 i −jk1 z
Hi = ẑ × Ei = yˆ Eo e
η1
η1
4
Normal Incidence
Reflected EM Waves at Boundaries
E
H
Normal Incidence
reflected wave
Medium 1
Reflected Wave
Medium 2
Unknown
DEFINE REFLECTION
COEFFICIENT AS
r = x̂E r e+jk1 z
E
o
r
1
E
r = (−z)
r = −yˆ o e+jk1 z
ˆ ×E
H
η1
η1
5
Transmitted EM Waves at Boundaries
E
H
Normal Incidence
transmitted wave
Medium 1
Transmitted Wave
Medium 2
Unknown
DEFINE TRANSMISSION
COEFFICIENT AS
r = x̂E t e−jk2 z
E
o
t
1
E
t = ẑ × E
t = yˆ o e−jk2 z
H
η2
η2
6
Reflection & Transmission of EM Waves at Boundaries
2 = E
t
E
1 = E
i + E
r
E
Medium 1
1 = H
i + H
r
H
Medium 2
2 = H
r
H
7
Reflectivity & Transmissivity of Waves
•  Define the reflection coefficient as
•  Define the transmission coefficient as
2
8
Thin Film Interference
Incident
light
Constructive
interference
air
oil
water
Image by Yoko Nekonomania http://www.
flickr.com/photos/nekonomania/4827035737/
on flickr
9
Reflection & Transmission in Layered Media
Medium 1 ( k1 , η1 )
Incident
Reflected
Medium 2 ( k2 , η2 )
→
←
Forward
Backward
z=0
Incident:
Reflected:
→
←
Transmitted →
z=L
Ei e−jk1 z
Er e+jk1 z
Omit ejωt
H± = ±E± /η
Backward:
Ef e−jk2 z
Eb e+jk2 (z −L)
Transmitted:
Et e−jk3 (z−L)
Forward:
Medium 3 ( k3 , η3 )
10
√
k ≡ ω μ
μ
η≡
Reflection & Transmission in Layered Media
Apply boundary conditions …
•  E at z = 0 → Ei + Er = Ef + Eb
•  H at z = 0 → Ei /η1 − Er /η1 = Ef /η2 − Eb /η2
•  E at z = L → Ef e−jk2 L + Eb e+jk2 L = Et e−jk3 L
•  H at z = L →Ef e−jk2 L /η2 − Eb e+jk2 L /η2 = Et e−jk3 L /η3
•  … and solve for Er , Ef , Eb and Et as functions of Ei .
Could easily be extended to more layers.
11
Reflection by Infinite Series
Medium 1 ( k1 , η1 )
Medium 2 ( k2 , η2 )
2
t21 r21 r23
t12 Γ4 Ei
2
t21 r23
t12 Γ2 Ei
r12 Ei
Ei
Medium 3
Γ ≡ e−jk2 L
∗r21 r23 Γ2
r23 t12 Γ2 Ei
t12 ΓEi
t12 Ei
z=0
z=L
2 2 4
Er = Ei (r21 + t21 r23 t12 Γ2 (1 + r21 r23 Γ2 + r21
r23 Γ ...))
= Ei (r21 + t21 r23 t12 Γ2 /(1 − r21 r23 Γ2 ))
12
Transmission by Infinite Series
Medium 2 ( k2 , η2 )
Medium 1
Medium 3 ( k3 , η3 )
Γ ≡ e−jk2 L
∗r21 r23 Γ2
Ei
t12 Ei
r21 r23 t23 t12 Γ3 Ei
t23 t12 ΓEi
t23 ΓEi
z=0
z=L
2 2 4
Et = Ei (t23 t12 Γ(1 + r21 r23 Γ2 + r21
r23 Γ ...))
= Ei t23 t21 Γ/(1 − r21 r23 Γ2 ))
13
Is Zero Reflection Possible?
One could solve for conditions under which …
•  Er = 0
… no reflected wave
•  |Et |2 /η3 = |Ei |2 /η1
power
… transmitted wave carries incident
and then determine conditions on L and η2 for which there is no
reflection, for example. This would yield the design of an antireflection coating.
Or, one could use generalized impedances …
14
Todays Culture Moment
GPS
The Global Positioning System (GPS) is a
constellation of 24 Earth-orbiting
satellites.The orbits are arranged so that
at any time, anywhere on Earth, there
are at least four satellites "visible" in the
sky. GPS operations depend on a very
accurate time reference; each GPS
satellite has atomic clocks on board.
Galileo – a global system being developed by the European Union
and other partner countries, planned to be operational by 2014
Beidou – People's Republic of China's regional system, covering Asia
and the West Pacific
COMPASS – People's Republic of China's global system, planned to be
operational by 2020
GLONASS – Russia's global navigation system
15
Image by ines saraiva http://www.flickr.com/
photos/inessaraiva/4006000559/ on flickr
Reflection and Transmission by an Infinite Series
Medium 1 ( k1 , η1 )
Medium 2 ( k2 , η2 )
Medium 3
Ei
Er = Ei (r12 + t21 r23 t12 Γ2 /(1 − r21 r23 Γ2 )) Γ ≡ e−jk2 L
Et = Ei (t23 t12 Γ/(1 − r21 r23 Γ2 ))
How do we get zero reflection?
16
Generalized Impedance
Define a spatially-dependent impedance
(
)
η(z) = −
H (z)
E z
1
2
z=0
In region 1 (z < 0) we have
()
η1 z
=
μ1 e− jkz + re jkz
ε1 e− jkz − re jkz
In region 2 (z > 0) we have
()
η2 z
17
=
μ2
ε2
Generalized Impedance
The incident wave in region 1
now sees an impedance of
regions 2 and 3:
d
1
2
3
( )
η −d
=
z=0
μ2 e jk2 d + r23e− jk2 d
ε 2 e jk2 d − r23e− jk2 d
Reflection of incident wave can be
eliminated if we match impedance
( )
η −d
18
=
μ1
ε1
Matching Impedances
We need
μ1
=
ε1
μ2 e jk2 d + r23e− jk2 d
=
jk2 d
− jk2 d
ε 2 e − r23e
μ2 1 + r23e−2 jk2 d
ε 2 1 − r23e−2 jk2 d
For lossless material, ε and μ are real, so only choices are
e
2 jk2 d
= ±1
Choose -1 and obtain … requires d = λ/4n2
μ1
=
ε1
μ2 1 − r23
ε 2 1 + r23
19
Matching Impedances
Consider impedance at z = 0
μ2 1 + r23
=
ε 2 1 − r23
μ3
ε3
1 + r23
⇒
1 − r23
=
μ3 ε 2
ε 3 μ2
So, we can eliminate the reflection as long as
μ1
=
ε1
μ2
ε2
⎛ μ
2
⎜
⎜⎝ ε 2
ε3 ⎞
⎟
μ3 ⎟⎠
⇒
μ2
=
ε2
μ1 μ3
ε1 ε 3
η2 ⋅ η2 = η1 ⋅ η3
(n2 )2 = n1 n3
20
Anti-reflection Coating
Reflectance(%)
Uncoated glass
Coated glass
wavelength
Image is in the public domain
21
Everyday Anti-Reflection Coatings
Incident light path
no
n1
n2
Ghost image
Destructive
interference
Transmission Again
Ei
Transmitted Wave from a few slides ago
− jk2 L
Ei t 21t12 e
Et =
1 − r21r21e− j 2 k2 L
23
Transmission
|t|
2
Fabry-Perot Resonance
1
0.8
0.6
0.4
0.2
0
1.5
1.52
1.54
1.56
1.58
Wavelength [μm]
Fabry-Perot Resonance:
maximum transmission
minimum reflection
24
Resonators with Internal Gain
What if it was possible to make a material with negative absorption
so the field grew in magnitude as it passed through a material?
Field strength [a.u.]
1.0
i egz gz
E(z)
=E
e
0.5
0
-0.5
-1.0
1
2
3
4
5
6
Propagation distance [cm]
Resonance:
25
Laser Using Fabre-Perot Cavity
Resonant modes
Gain profile
Image is in the public domain
26
Key Takeaways
Reflection and Transmission by an Infinite Series
2 2 4
Er = Ei (r21 + t21 r23 t12 Γ2 (1 + r21 r23 Γ2 + r21
r23 Γ ...))
= Ei (r21 + t21 r23 t12 Γ2 /(1 − r21 r23 Γ2 ))
2 2 4
Et = Ei (t23 t12 Γ(1 + r21 r23 Γ2 + r21
r23 Γ ...))
d = λ/4n2
(n2 )2 = n1 n3
Transmission
Anti-reflective coatings by
impedance matching:
|t|
2
= Ei t23 t21 Γ/(1 − r21 r23 Γ2 ))
1
0.8
0.6
0.4
0.2
0
1.5
1.52
1.54
1.56
1.58
Wavelength [μm]
Fabry-Perot Resonance
27
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6.007 Electromagnetic Energy: From Motors to Lasers
Spring 2011
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