Electromagnetic Wave PropagationLecture 12: Oblique incidence I
Daniel Sjoberg
Department of Electrical and Information Technology
October 6, 2011
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
Daniel Sjoberg, Department of Electrical and Information Technology
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
Daniel Sjoberg, Department of Electrical and Information Technology
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?
Daniel Sjoberg, Department of Electrical and Information Technology
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
Daniel Sjoberg, Department of Electrical and Information Technology
The spelling
Willebrord Snel van Royen (1580–1626)
Daniel Sjoberg, Department of Electrical and Information Technology
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
Daniel Sjoberg, Department of Electrical and Information Technology
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
Daniel Sjoberg, Department of Electrical and Information Technology
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.
Daniel Sjoberg, Department of Electrical and Information Technology
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.Daniel Sjoberg, Department of Electrical and Information Technology
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
Daniel Sjoberg, Department of Electrical and Information Technology
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
Daniel Sjoberg, Department of Electrical and Information Technology
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
Daniel Sjoberg, Department of Electrical and Information Technology
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
)
Daniel Sjoberg, Department of Electrical and Information Technology
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.
Daniel Sjoberg, Department of Electrical and Information Technology
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′.
Daniel Sjoberg, Department of Electrical and Information Technology
Demo
Daniel Sjoberg, Department of Electrical and Information Technology
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
Daniel Sjoberg, Department of Electrical and Information Technology
Critical angle
Refraction Reflection
(Fig. 7.5.1 in Orfanidis)
sin θ′c =n
n′sin θc =
n′
n
Daniel Sjoberg, Department of Electrical and Information Technology
Examples
Prism
Optical manhole
Optical fiber
(Figs. 7.5.2, 7.5.3, 7.5.5 in Orfanidis)
Daniel Sjoberg, Department of Electrical and Information Technology
Optical manhole — Snel’s window
http://www.uwphotographyguide.com/snells-window-underwater,photo taken using fisheye lens to cover the angle.
Daniel Sjoberg, Department of Electrical and Information Technology
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.
Daniel Sjoberg, Department of Electrical and Information Technology
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
Daniel Sjoberg, Department of Electrical and Information Technology
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.
Daniel Sjoberg, Department of Electrical and Information Technology
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
Daniel Sjoberg, Department of Electrical and Information Technology
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
Daniel Sjoberg, Department of Electrical and Information Technology
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
Daniel Sjoberg, Department of Electrical and Information Technology
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′
Daniel Sjoberg, Department of Electrical and Information Technology
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.
Daniel Sjoberg, Department of Electrical and Information Technology
Measuring Brewster’s angle between classes
Hastings A. Smith, Jr, The Physics Teacher, Feb 1979, p. 109.
Daniel Sjoberg, Department of Electrical and Information Technology
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
Daniel Sjoberg, Department of Electrical and Information Technology
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.Daniel Sjoberg, Department of Electrical and Information Technology
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)
Daniel Sjoberg, Department of Electrical and Information Technology
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
Daniel Sjoberg, Department of Electrical and Information Technology
Typical field distribution
Daniel Sjoberg, Department of Electrical and Information Technology
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.
Daniel Sjoberg, Department of Electrical and Information Technology
Evanescent square root
A recurring task is to take the square rootkz =
√ω2µ0ε− k2x = β − jα. 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.
Daniel Sjoberg, Department of Electrical and Information Technology
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.
Daniel Sjoberg, Department of Electrical and Information Technology
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.
Daniel Sjoberg, Department of Electrical and Information Technology
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
Daniel Sjoberg, Department of Electrical and Information Technology
The Zenneck surface wave
(Fig. 7.10.1 in Orfanidis)
Daniel Sjoberg, Department of Electrical and Information Technology
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.
Daniel Sjoberg, Department of Electrical and Information Technology
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.
Daniel Sjoberg, Department of Electrical and Information Technology
Typical field distribution, Zenneck wave
Daniel Sjoberg, Department of Electrical and Information Technology
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
Daniel Sjoberg, Department of Electrical and Information Technology
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.
Daniel Sjoberg, Department of Electrical and Information Technology
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