Upload
others
View
0
Download
0
Embed Size (px)
Citation preview
Electromagnetic Wave PropagationLecture 12: Oblique incidence I
Daniel Sjoberg
Department of Electrical and Information Technology
October 11, 2012
Outline
1 Introduction
2 Snel’s law
3 Transverse impedance and propagation
4 Critical angle, Brewster angle
5 Evanescent and complex waves
6 Zenneck surface wave
7 Conclusions
2 / 46
Outline
1 Introduction
2 Snel’s law
3 Transverse impedance and propagation
4 Critical angle, Brewster angle
5 Evanescent and complex waves
6 Zenneck surface wave
7 Conclusions
3 / 46
Key questions
I How to analyze the oblique incidence of waves on aninterface?
I What are typical results?
I What are special characteristics of lossy media?
4 / 46
Outline
1 Introduction
2 Snel’s law
3 Transverse impedance and propagation
4 Critical angle, Brewster angle
5 Evanescent and complex waves
6 Zenneck surface wave
7 Conclusions
5 / 46
The spelling
Willebrord Snel van Royen (1580–1626)
6 / 46
Oblique incidence
(Fig. 7.1.1 in Orfanidis)
The waves can be written
E+e−jk+·r, E−e
−jk−·r, E′+e−jk′+·r, E′−e
−jk′−·r
7 / 46
Matching
Matching tangential fields on the boundary z = 0 implies
ET+e−jk+·r +ET−e
−jk−·r = E′T+e−jk′+·r +E′T−e
−jk′−·r
ET+e−jkx+x +ET−e
−jkx−x = E′T+e−jk′x+x +E′T−e
−jk′x−x
Since this applies for all x on the boundary z = 0, we must have
kx+ = kx− = k′x+ = k′x−
and similarly for any y components. Since kx = k sin θ = k0n sin θ,this implies
θ+ = θ− = θ
θ′+ = θ′− = θ′n sin θ = n′ sin θ′
where we used k = nk0 and k′ = n′k0. Since k · k = k2 = ω2εµwe have
kz =√k2 − k2x − k2y
8 / 46
A graphical argument
The condition k2 = ω2εµ describes a sphere (or circle) in k-space.
kx
kz
k2 = ω2ǫµ
k20 = ω2ǫ0µ0
k0
k
Can also be used for photonic crystals.
9 / 46
Sometimes no solutions!
When the wave is incident from a denser medium, it may not bepossible to satisfy the phase matching with real wave vectors.
kx
kz
k20 = ω2ǫ0µ0
k2 = ω2ǫµ
k
k0 =?
Corresponds to total internal reflection.10 / 46
Outline
1 Introduction
2 Snel’s law
3 Transverse impedance and propagation
4 Critical angle, Brewster angle
5 Evanescent and complex waves
6 Zenneck surface wave
7 Conclusions
11 / 46
Transverse impedance
The complete components of the forward field is
E+(r) = [(x cos θ − z sin θ)A++yB+]e−jk+·r
H+(r) =1
η[yA+−(x cos θ − z sin θ)B+]e
−jk+·r
The transverse components can then be written
ET+(r) = [xC++yB+]e−jk+·r
HT+(r) = [yC+
ηTM−xB+
ηTE]e−jk+·r
where C+ = cos θA+ and
ηTM = η cos θ TM, parallel, p-polarization
ηTE =η
cos θTE, perpendicular, s-polarization
12 / 46
Transverse refractive index
For dielectric media, that is, µ = µ0, we have
η =
√µ0ε
=η0√εr
=η0n
This motivates the introduction of the transverse refractive indexvia ηT = η0/nT, or
nTM =n
cos θTM, parallel, p-polarization
nTE = n cos θ TE, perpendicular, s-polarization
13 / 46
Similarity with normal incidence
Making the substitutions (where T can be either TM or TE)
η → ηT, e±jkz → e±jkzz = e±jkz cos θ
everything we derived on propagation in layered structures fornormal incidence remain valid. For instance,(
ET1+
ET1−
)=
(ejkz` 00 e−jkz`
)(ET2+
ET2−
)and (
ET1
HT1
)=
(cos(kz`) jηT sin(kz`)
jη−1T sin(kz`) cos(kz`)
)(ET2
HT2
)
14 / 46
Reflection at an interface
In particular, at an interface we can define the matching matrix(ET+
ET−
)=
1
τT
(1 ρTρT 1
)(E′T+
E′T−
)where ρT and τT are the Fresnel coefficients
ρT =η′T − ηTη′T + ηT
=nT − n′TnT + n′T
τT =2η′T
η′T + ηT=
2nTnT + n′T
which take different values depending on polarization.
15 / 46
Fresnel coefficients, explicit form
Writing out nTM = n/ cos θ and nTE = n cos θ, the explicit formof the Fresnel reflection coefficient is (after some algebra)
ρTM =n cos θ′ − n′ cos θn cos θ′ + n′ cos θ
=
√(n′
n )2 − sin2 θ − (n
′
n )2 cos θ√
(n′
n )2 − sin2 θ + (n
′
n )2 cos θ
ρTE =n cos θ − n′ cos θ′
n cos θ + n′ cos θ′=
cos θ −√
(n′
n )2 − sin2 θ
cos θ +√
(n′
n )2 − sin2 θ
From the rightmost expressions, we find
ρTM → 1, ρTE → −1, as θ → 90◦
regardless of n and n′.
16 / 46
Demo
17 / 46
Outline
1 Introduction
2 Snel’s law
3 Transverse impedance and propagation
4 Critical angle, Brewster angle
5 Evanescent and complex waves
6 Zenneck surface wave
7 Conclusions
18 / 46
Critical angle
Refraction Reflection
(Fig. 7.5.1 in Orfanidis)
sin θ′c =n
n′sin θc =
n′
n
19 / 46
Examples
Prism
Optical manhole
Optical fiber
(Figs. 7.5.2, 7.5.3, 7.5.5 in Orfanidis)
20 / 46
Optical manhole — Snel’s window
http://www.uwphotographyguide.com/snells-window-underwater,photo taken using fisheye lens to cover the angle.
21 / 46
Total internal reflection: numbers
Using the critical angle, the reflection coefficients can be written
ρTM =
√sin2 θc − sin2 θ − sin2 θc cos θ√sin2 θc − sin2 θ + sin2 θc cos θ
ρTE =cos θ −
√sin2 θc − sin2 θ
cos θ +√sin2 θc − sin2 θ
With θ > θc this is (using the branch√−1 = −j)
ρTM =−j√
sin2 θ − sin2 θc − sin2 θc cos θ
−j√sin2 θ − sin2 θc + sin2 θc cos θ
= −1 + jxn2
1− jxn2
ρTE =cos θ + j
√sin2 θc − sin2 θ
cos θ − j√sin2 θc − sin2 θ
=1 + jx
1− jx
where x =
√sin2 θ−sin2 θc
cos θ , and sin θc = 1/n. Thus |ρTE,TM| = 1.
22 / 46
Phase shift at total reflection
The TM and TE polarization are reflected with different phase
ρTM = −1 + jxn2
1− jxn2= ejπ+2jψTM
ρTE =1 + jx
1− jx= e2jψTE
wheretanψTM = xn2, tanψTE = x
The relative phase change between the polarizations is
ρTM
ρTE= ejπ+2jψTM−2jψTE
Thus, if θ is chosen so that ψTM − ψTE = π/8, we have
ρTM
ρTE= ejπ+jπ/4 ⇒
(ρTM
ρTE
)2
= e2jπ+jπ/2 = ejπ/2
23 / 46
The Fresnel rhomb
Thus, after two reflections the TM and TE polarizations differ inphase by π/2.
(Fig. 7.5.6 in Orfanidis)
Using a glass with n = 1.51, we have θc = 41.47◦. The angle54.6◦ results in ψTM − ψTE = π/8. The angle 48.6◦ would alsowork, see Example 7.5.6.
Ideally there is no frequency dependence, that is, the Fresnelrhomb can convert linear to circular polarization in a much widerband than a quarter wavelength plate.
24 / 46
Goos-Hanchen shift
The phase shift in the reflection coefficient also gives rise to theGoos-Hanchen shift (see Example 7.5.7).
(Fig. 7.5.7 in Orfanidis)
DTE =2 sin θ0
k0n√sin2 θ0 − sin2 θc
, DTM =(n′)2DTE
(n2 + 1) sin2 θ0 − (n′)2
25 / 46
Goos-Hanchen shift
The phase shift in the reflection coefficient also gives rise to theGoos-Hanchen shift (see Example 7.5.7).
(Fig. 7.5.7 in Orfanidis)
DTE =2 sin θ0
k0n√sin2 θ0 − sin2 θc
, DTM =(n′)2DTE
(n2 + 1) sin2 θ0 − (n′)2
26 / 46
Goos-Hanchen shift
40 50 60 70 80 90Angle of incidence
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5D/λ
Goos-Hänchen shift (n=1.50)
TETM
27 / 46
The Brewster angle (TM polarization)
ρTM =
√(n′
n )2 − sin2 θ − (n
′
n )2 cos θ√
(n′
n )2 − sin2 θ + (n
′
n )2 cos θ
(Fig. 7.6.1 in Orfanidis)
tan θB =n′
nθB + θ′B =
π
2tan θ′B =
n
n′
28 / 46
Brewster angle, reflection
For glass with n = 1.5, we have θB = 56.3◦ and θ′B = 33.7◦, andθc = 41.8◦.
(Fig. 7.6.2 in Orfanidis)
Can be used to obtain linear polarization, but loses power throughpartial transmission of TE component.
29 / 46
Measuring Brewster’s angle between classes
Hastings A. Smith, Jr, The Physics Teacher, Feb 1979, p. 109.
30 / 46
Outline
1 Introduction
2 Snel’s law
3 Transverse impedance and propagation
4 Critical angle, Brewster angle
5 Evanescent and complex waves
6 Zenneck surface wave
7 Conclusions
31 / 46
What happens on the other side at total reflection?
Even though we have total reflection, there are fields on the farside of the interface. The transmission coefficients are
τTM = 1 + ρTM, τTE = 1 + ρTE
which are nonzero unless ρTM = ρTE = −1. The z wavenumbersare
kz =√ω2µ0ε− k2x
k′z =√ω2µ0ε′ − k2x
Since kx = k sin θ, k = ω√µ0ε, and ε′ = ε sin2 θc, we have
k′z = k√sin2 θc − sin2 θ = −jk
√sin2 θ − sin2 θc = −jα′
Note the branch√−1 = −j must be taken in order to have
exponential decay e−jk′zz = e−α
′z.
32 / 46
Exponential decay
The transmitted wave has spatial dependence
e−jk′zze−jkxx = e−α
′ze−jβ′x
Exponential attenuation in the z-direction, same transverse phaseas in incident wave (β′ = kx).
(Fig. 7.8.1 in Orfanidis)
33 / 46
Evanescent waves
An evanescent wave oscillates so quickly in x (kx > k′) that it isexponentially attenuated in z (kz = −jα) due to k2x + k2z = (k′)2.
(Fig. 7.8.1 in Orfanidis)
Thus, there is a region close to the surface containing reactivefields (non-propagating). The size of the region is on the order
δ =1
α=
1
k√sin2 θ − sin2 θc
=λ
2π√sin2 θ − sin2 θc
34 / 46
Typical field distribution
35 / 46
Complex waves
The generalization of evanescent waves, which are strictly definedonly in lossless media, is necessary for lossy media ε = εR − jεI. Inorder to avoid using complex angles θ, use the wavenumbers
ηTM = η cos θ =ηkzk
=kzωε, ηTE =
η
cos θ=ηk
kz=ωµ
kz
(Fig. 7.9.1 in Orfanidis)
The wave vector k = β − jαmay be complex, but must satisfy
k · k = ω2µε
The real vectors β and α need not beparallel.
36 / 46
Evanescent square root
A recurring task is to take the square rootkz =
√ω2µ0ε− k2x = βz − jαz. In order to guarantee α > 0, the
square root is defined as
kz =
{√ω2µ0(εR − jεI)− k2x if εI 6= 0
−j√k2x − ω2µ0εR if εI = 0
Thus, everything works fine for complex valued materialcoefficients, but real valued needs some extra attention forevanescent waves.
Matlab code sqrte.m in Orfanidis’ files.
37 / 46
Oblique incidence on a lossy medium
(Fig. 7.9.1 in Orfanidis)
To the left: kx = k sin θ, kz = k cos θ.To the right: k′x = kx, k′z = β′z − jα′z.
It can be shown that ρTMρTE
= β′z−jα′z−k sin θ tan θβ′z−jα′z+k sin θ tan θ
, implying ellipticpolarization of the reflected wave.
38 / 46
No standard Brewster angle for lossy media
The TM reflection coefficient is
ρTM =η′TM − ηTM
η′TM + ηTM=k′zε− kzε′
k′zε+ kzε′
which cannot be exactly zero when ε, kz and kx are real and ε′ iscomplex.
(Fig. 7.6.3 in Orfanidis)
But when kx = βx − jαx and kz = βz − jαz, we can achieveρTM = 0. This is the Zenneck surface wave.
39 / 46
Outline
1 Introduction
2 Snel’s law
3 Transverse impedance and propagation
4 Critical angle, Brewster angle
5 Evanescent and complex waves
6 Zenneck surface wave
7 Conclusions
40 / 46
The Zenneck surface wave
(Fig. 7.10.1 in Orfanidis)
41 / 46
Conditions for the Zenneck wave
The TM reflection coefficient is
ρTM =k′zε− kzε′
k′zε+ kzε′= 0 ⇒ k′zε = kzε
′
Using k2x + k2z = ω2µ0ε and (k′x)2 + (k′z)
2 = ω2µ0ε′ and kx = k′x
we find
kx = ω√µ0
√εε′√ε+ ε′
, kz = ω√µ0
ε√ε+ ε′
, k′z = ω√µ0
ε′√ε+ ε′
This results in complex wave vectors on both sides of the interface.For weakly lossy media (ε′ = εR − jεI where εI/εR � 1, we canestimate
αx|αz|
=
√ε
εR
Thus, the attenuation in the z-direction (αz) is larger than in thex-direction (αx) if εR > ε.
42 / 46
Example: air-water interface
Consider an interface between air (ε = ε0) and sea water:
ε′ = 81ε0 − jσ/ω, σ = 4S/m
The wave numbers at 1GHz and 100MHz are
f = 1GHz f = 100MHz
ε′/ε0 = 81− 72j ε′/ε0 = 81− 720jk = β − jα = 20.94 k = β − jα = 2.094k′ = β′ − jα′ = 203.76− 77.39j k′ = β′ − jα′ = 42.01− 37.54jkx = βx − jαx = 20.89− 0.064j kx = βx − jαx = 2.1− 0.001jkz = βz − jαz = 1.88 + 0.71j kz = βz − jαz = 0.06 + 0.05jk′z = β′z − jα′z = 202.97− 77.80j k′z = β′z − jα′z = 42.01− 37.59j
Thus, the attenuation in the z-direction is much larger than in thex-direction, and the wave can propagate relatively freely along theinterface.
43 / 46
Typical field distribution, Zenneck wave
44 / 46
Outline
1 Introduction
2 Snel’s law
3 Transverse impedance and propagation
4 Critical angle, Brewster angle
5 Evanescent and complex waves
6 Zenneck surface wave
7 Conclusions
45 / 46
Conclusions
I The tangential wavenumber kx is the same in both mediumsdue to phase matching.
I All standard formulas for normal incidence are valid whenconsidering tangential field components, and splitting the fieldinto TM and TE polarizations.
I The normal wavenumber kz =√ω2µε− k2x may be real or
imaginary in the lossless case, or complex in the lossy case.
I The critical angle is the largest angle of refraction, or thesmallest angle of total reflection.
I There is a phase shift at total internal reflection, which isdifferent for different polarizations.
I Complex wave vectors k = β − jα are necessary for lossymedia.
46 / 46