electromagnetic wave propagation lecture 12: oblique ... · 4 critical angle, brewster angle 5...

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

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Outline

1 Introduction

2 Snel’s law





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?

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Outline

1 Introduction

2 Snel’s law





7 Conclusions

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The spelling

Willebrord Snel van Royen (1580–1626)

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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

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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

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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.

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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





7 Conclusions

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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

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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

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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

)

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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.

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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′.

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Demo

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Outline

1 Introduction

2 Snel’s law





7 Conclusions

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Critical angle

Refraction Reflection


sin θ′c =n

n′sin θc =

n′

n

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Examples

Prism

Optical manhole

Optical fiber

(Figs. 7.5.2, 7.5.3, 7.5.5 in Orfanidis)

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Optical manhole — Snel’s window

http://www.uwphotographyguide.com/snells-window-underwater,photo taken using fisheye lens to cover the angle.

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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.

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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

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The Fresnel rhomb

Thus, after two reflections the TM and TE polarizations differ inphase by π/2.


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.

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Goos-Hanchen shift

The phase shift in the reflection coefficient also gives rise to theGoos-Hanchen shift (see Example 7.5.7).


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).


DTE =2 sin θ0

k0n√sin2 θ0 − sin2 θc

, DTM =(n′)2DTE

(n2 + 1) sin2 θ0 − (n′)2

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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


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◦.


Can be used to obtain linear polarization, but loses power throughpartial transmission of TE component.

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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





7 Conclusions

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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.

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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).


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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.


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

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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


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


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.


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





7 Conclusions

40 / 46

The Zenneck surface wave


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.

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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

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Outline

1 Introduction

2 Snel’s law





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.

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electromagnetic wave propagation lecture 12: oblique ... · 4 critical angle, brewster angle 5...

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