electromagnetic wave propagation lecture 14: multilayer

Electromagnetic Wave PropagationLecture 14: Multilayer film applications

Daniel Sjoberg

Department of Electrical and Information Technology

October 17, 2013

Outline

1 Multilayer dielectric structures at oblique incidence

2 Perfect lens in negative-index media

3 Antireflection coatings at oblique incidence

4 Omnidirectional dielectric mirrors

5 Reflection and refraction in birefringent media

6 Giant birefringent optics

7 Concluding remarks

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Outline








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General geometry

(Fig. 8.1.1 in Orfanidis)

Most concepts generalize. Important steps:

I Use transverse field components

I The tangential wave vector is conserved

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Oblique propagation in bianisotropic materials

The formulation from lecture 4 can be used for numericalimplementation of completely general bianisotropic materials:

∂

∂z

(Et

Ht × z

)= −jωW(z) ·

(Et

Ht × z

)

W =

(0 −z × II 0

)·[(εtt ξttζtt µtt

)−A(kt)

]·(I 00 z × I

)

A(kt) =

[(0 −ω−1kt × z

ω−1kt × z 0

)−(εt ξtζt µt

)](εzz ξzzζzz µzz

)−1 [(0 −ω−1z × kt

ω−1z × kt 0

)−(εz ξzζz µz

)]In this lecture, we focus on isotropic media.

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Wave concepts

Tangential phase (Snel’s law):

na sin θa = ni sin θi = nb sin θb ⇒ cos θi =

√1− n2a sin

2 θan2i

Phase thickness:

δi = kz,iì =ω

c0niì cos θi = 2π

f

f0

niìλ0

cos θi

Transverse wave impedance and refractive index:

ηTM =kzωε

=

√µ

ε

kzω√εµ

= η cos θ = η0cos θ

n=

η0nTM

ηTE =ωµ

kz=

√µ

ε

ω√εµ

kz=

η

cos θ= η0

1

n cos θ=

η0nTE

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Propagation of split fields

Reflection and transmission coefficients:

ρT,i =nT,i−1 − nT,inT,i−1 + nT,i

, τT,i = 1 + ρT,i

Matching and propagation matrices:(ET,i+

ET,i−

)=

1

τT,i

(1 ρT,iρT,i 1

)︸︷︷︸

matching

(ejδi 00 e−jδi

)︸︷︷︸

propagation

(ET,i+1,+

ET,i+1,−

)

Reflection coefficient recursion:

ΓT,i =ρT,i + ΓT,i+1e

−2jδi

1 + ρT,iΓT,i+1e−2jδi

initialized by ΓT,M+1 = ρT,M+1.

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Propagation of total fields

Propagation matrix:(ET,i

HT,i

)=

(cos δi jηT,i sin δi

jη−1T,i sin δi cos δi

)(ET,i+1

HT,i+1

)Transverse impedance recursion:

ZT,i = ηT,iZT,i+1 + jηT,i tan δiηT,i + jZT,i+1 tan δi

, ZT,M+1 = ηT,b

Generalization of multidiel.m:

[Gamma1, Z1] = multidiel(n, L, lambda, theta, pol)

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Single dielectric slab


ΓT,1 =ρT,1 + ρT,2e

−2jδ1

1 + ρT,1ρT,2e−2jδ1=ρT,1(1− e−2jδ1)

1− ρ2T,1e−2jδ1

δ1 = 2πf

f0

n1`1λ0

cos θ1 = πf

f1, f1 =

f0

2n1`1λ0

cos θ1

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Frequency shift


I Notch frequency is shifted up

I Reflection level is increased

I Bandwidth is decreased

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Frustrated total internal reflection


kx = k0na sin θ

kza =√k20n

2a − k2x = k0na cos θ

kzb =

k0√n2b − n2a sin

2 θ if θ ≤ θc−jk0

√n2a sin

2 θ − n2b = −jαzb if θ ≥ θc11 / 50

Dependence on thickness and angle

(Figs. 8.4.3 and 8.4.4 in Orfanidis)

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Power flow

The combined field from two evanescent waves is

ET(z) =1 + ρa

1− ρ2ae−2αzbd

[e−αzbz − ρae−2αzbdeαzbz

]Ea+

HT(z) =1− ρa

1− ρ2ae−2αzbd

[e−αzbz + ρae

−2αzbdeαzbz] Ea+ηTa

The power transported in one evanescent wave is zero,

Pz =1

2Re{ET+H

∗T+} =

1

2Re

{(1 + ρa)(1− ρ∗a)|1− ρ2ae−2jαzbd|

|Ea+|2

ηTae−2αzbz

}= 0

since (1 + ρa)(1− ρ∗a) = 1− |ρa|2 + ρa − ρ∗a = 2j Im(ρa). But thecombination of two evanescent waves can carry power:

Pz =1

2Re{ETH

∗T} = · · · = (1− |Γ |2) |Ea+|

2

2ηTa

|Γ |2 = sinh2(αzbd)

sinh2(αzbd) + sin2 φa, ρa = ejφa

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Simulation

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Outline








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Slab of negative index

ε = −ε0, µ = −µ0, n =√

εµε0µ0

= −1, η =√

µε = η0.


Snel’s law:na sin θa = n sin θ ⇒ θ = −θa

Zero reflection since η = η0.

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Field solutions

The electric field is (TM polarization)

E =

E0

(x− kx

kzz)e−jkzze−jkxx, for z ≤ 0

E0

(x+ kx

kzz)ejkzze−jkxx, for 0 ≤ z ≤ d

E0

(x− kx

kzz)e−jkz(z−2d)e−jkxx, for z ≥ d

where the longitudinal wave number is

kz = sqrte(k20 − k2x) =

{√k20 − k2x, if k2x ≤ k20

−j√k2x − k20, if k20 ≤ k2x

Thus, for evanescent waves where kz = −jα, the negative mediumamplifies the transmitted wave.

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Amplification of evanescent waves


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Influence of non-ideal parameters

Transmittance below is computed as 10 log10 |T e−jkzd|2, which isnot necessarily less than one for evanescent waves.


Peaks correspond to plasmon resonances, where

Re(kx,res) ≈ 1d

∣∣∣ ε0−εε0+ε

∣∣∣. This implies a resolution of

∆x = 1/Re(kx,res).

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Simulation

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Outline








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Antireflection coatings

nb = 1.5

na = 1 Er = 0Ei

nb = 1.5 nb = 1.5

n2 = 1.38

na = 1 ErEi na = 1 Er = 0Ei

n1 = 1.22 n3 = 2.45

Suitably designed thin layers can reduce reflection from interfaces.What happens at oblique incidence?

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Two-layer coating


n = [1, 1.38, 2.45, 1.5]; L = [0.3294, 0.0453];

la0 = 550; la = linspace(400, 700, 101); pol= ’te’;

G0 = abs(multidiel(n, L, la/la0)).^2*100;

G20 = abs(multidiel(n, L, la/la0, 20, pol)).^2*100;



plot(la, [G0; G20; G30; G40]);

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Demo program

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Typical effects at oblique incidence

I Frequency shift

I Difference in polarization

I Possible to redesign for specific angle of incidence

I Higher reflection for higher angles

I Brewster angle for TM polarization

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Outline








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Dielectric mirrors


Increasing angle of incidence implies

I Frequency shift

I Polarization dependence

Omnidirectionality requires overlap of frequency bands for TE andTM for all angles.

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Necessary condition

The angles in all layers are related by Snel’s law

na sin θa = nH sin θH = nL sin θL = nb sin θb

and the phase thicknesses are

δH = 2πf

f0LH cos θH, δL = 2π

f

f0LL cos θL

where Li = ni`i/λ0 and cos θi =√

1− n2a sin2 θa/n2i . To prevent

a wave from the outside to access the Brewster angle inside themirror (which would imply transmission), we need

na <nHnL√n2H + n2L

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Bandgap

All procedures from normal incidence are applicable whenconsidering transverse field components. We look for wavespropagating with the effective wave number K, so that thereshould exist solutions(

Ei,THi,T

)=

(cos(δL) jηLT sin(δL)

jη−1LT sin(δL) cos(δL)

)·(

cos(δH) jηHT sin(δH)

jη−1HT sin(δH) cos(δH)

) (Ei+2,T

Hi+2,T

)︸︷︷︸=e−jK`(Ei,T

Hi,T)

This is an eigenvalue equation, which can be put in the form

cos(K`) =cos(δH + δL)− ρ2T cos(δH − δL)

1− ρ2T

where ρT = nHT−nLTnHT+nLT

depends on polarization.

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Bandgaps

| cos(K`)| ≤ 1 for propagating fields:

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0f/f0

2.5

2.0

1.5

1.0

0.5

0.0

0.5

1.0

1.5theta = 45, nH/nL = 1.68

TETM

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0f/f0

2.5

2.0

1.5

1.0

0.5

0.0

0.5

1.0

1.5theta = 45, nH/nL = 3.36

TETM

The bandgap increases with contrast, and when increasing theangle:

I The entire bandgap shifts up in frequency for bothpolarizations (the layers become optically thinner).

I The TM bandgap decreases, and the TE bandgap increases.

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Frequency response

nH = 3, nL = 1.38, N = 30


Calculated with multidiel.m.

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Bandwidths and angle

nH = 3, nL = 1.38 nH = 2, nL = 1.38


Omnidirectional Not omnidirectional

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Polarizing beam splitter


The TM polarization travels at the Brewster angle if

na sin θa = na1√2=

nHnL√n2H + n2L

Hence TM is transmitted and TE reflected.33 / 50

Outline








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Effective refractive index

D =

ε1 0 00 ε2 00 0 ε3

·E, B = µ0H, k = Nk0k


N =1√

cos2 θn21

+ sin2 θn23

N = n2

ηTM = η0N cos θ

n21ηTE =

η0n2 cos θ

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Snel’s law

Snel’s law is N sin θ = N ′ sin θ′ , or, more explicitly,

n1n3 sin θ√n21 sin

2 θ + n23 cos2 θ

=n′1n

′3 sin θ

′√n′21 sin2 θ′ + n

′23 cos2 θ′

(TM)

n2 sin θ = n′2 sin θ′ (TE)

Can be solved for θ as function of θ′ or vice versa. Coded in

thb = snel(na, nb, tha, pol);

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Reflection coefficients

The reflection coefficients are

ρTM =η′TM − ηTM

η′TM + ηTM=n1n3

√n′23 −N2 sin2 θ − n′1n′3

√n23 −N2 sin2 θ

n1n3

√n′23 −N2 sin2 θ + n′1n

′3

√n23 −N2 sin2 θ

ρTE =η′TE − ηTE

η′TE + ηTE=n2 cos θ −

√n′22 − n22 sin2 θ

n2 cos θ +√n′22 − n22 sin2 θ

Can be computed using routine fresnel.

I If n3 = n′3 we have ρTM =n1−n′1n1+n′1

independent of θ.

I If n = [n1, n1, n3] and n′ = [n3, n3, n1] we have

ρTE = ρTM =n1 cos θ −

√n23 − n21 sin2 θ

n1 cos θ +√n23 − n21 sin2 θ

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Examples of Fresnel coefficients


(a) n = [1.63, 1.63, 1.5] n′ = [1.63, 1.63, 1.63]

(b) n = [1.54, 1.54, 1.63] n′ = [1.5, 1.5, 1.5]

(c) n = [1.8, 1.8, 1.5] n′ = [1.5, 1.5, 1.5]

(d) n = [1.8, 1.8, 1.5] n′ = [1.56, 1.56, 1.56]

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Differences from the isotropic case

I The Brewster angle can be zero (a)

I The Brewster angle may not exist (c)

I The Brewster angle may be imaginary (both ρTE and ρTM

increase monotonically with angle) (d)

I The critical angle can be different for TE and TM (b)

I The reflection coefficient can be the same for TE and TM

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Outline








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Dielectric mirror


The layers are now birefringent with

nH = [nH1, nH2, nH3], nL = [nL1, nL2, nL3]

The code multidiel.m can be used to calculate the response.

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First design

N = 50, nH = [1.8, 1.8, 1.5], nL = [1.5, 1.5, 1.5], na = nb = 1


Relatively similar TE and TM, TM is a bit more narrowband.

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Second design

N = 30, nH = [1.8, 1.8, 1.5], nL = [1.5, 1.5, 1.8], na = nb = 1.4


Identical TE and TM, but depends on angle.

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Slight change of second design

N = 30, nH = [1.8, 1.8, 1.5], nL = [1.5, 1.5, 1.9], na = nb = 1.4


Perturbation at higher angles.

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GBO mirrors

M. F. Weber et al, Science, vol 287, p. 2451, 2000.

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GBO reflective polarizer

N = 80, nH = [1.86, 1.57, 1.57], nL = [1.57, 1.57, 1.57],na = nb = 1


Mismatch only in one direction.

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Giant birefringent optics

I The “giant” refers to the relatively large birefringence

I Commercialized by 3M

I Careful matching of refractive indices may eliminatedifferences between polarizations

I Materials can be manufactured from birefringent polymers

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Outline








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Concluding remarks

This course has dealt with the following subjects in depth:

I Propagation of electromagnetic waves in stratified media

I Material modeling in time and frequency domain, isotropic aswell as bianisotropic

I Fundamental wave properties: polarization, wave impedance,wave number

I Tangential fields and tangential wave numbers are continuousacross interfaces, and have been used consistently

I Numerical methods and examples have appeared throughoutI Lots of codes from Orfanidis and on course home page

I Several applications of multilayer structures

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Only projects and orals remaining!

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electromagnetic wave propagation lecture 14: multilayer

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