electromagnetic wave propagation lecture 12: oblique ... · goos-h anchen shift the phase shift in...

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

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?


Outline

1 Introduction

2 Snel’s law





7 Conclusions


The spelling

Willebrord Snel van Royen (1580–1626)


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


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


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.


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





7 Conclusions


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


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


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

)


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.


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′.


Demo


Outline

1 Introduction

2 Snel’s law





7 Conclusions


Critical angle

Refraction Reflection


sin θ′c =n

n′sin θc =

n′

n


Examples

Prism

Optical manhole

Optical fiber

(Figs. 7.5.2, 7.5.3, 7.5.5 in Orfanidis)


Optical manhole — Snel’s window

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


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.


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


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.


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


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


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′


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.


Measuring Brewster’s angle between classes

Hastings A. Smith, Jr, The Physics Teacher, Feb 1979, p. 109.


Outline

1 Introduction

2 Snel’s law





7 Conclusions


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).



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


Typical field distribution


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.


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.


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.


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.


Outline

1 Introduction

2 Snel’s law





7 Conclusions


The Zenneck surface wave



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.


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.


Typical field distribution, Zenneck wave


Outline

1 Introduction

2 Snel’s law





7 Conclusions


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.


electromagnetic wave propagation lecture 12: oblique ... · goos-h anchen shift the phase shift in...

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