electromagnetic wave propagation lecture 15: multilayer

Electromagnetic Wave PropagationLecture 15: Multilayer film applications II

Daniel Sjoberg

Department of Electrical and Information Technology

October 12, 2010

Outline

1 Antireflection coatings at oblique incidence

2 Omnidirectional dielectric mirrors

3 Polarizing beam splitters

4 Reflection and refraction in birefringent media

5 Brewster and critical angles in birefringent media

6 Giant birefringent optics

7 Concluding remarks

Daniel Sjoberg, Department of Electrical and Information Technology

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?


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]);


Demo program


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


Outline









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.


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


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.


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.Daniel Sjoberg, Department of Electrical and Information Technology

Frequency response

nH = 3, nL = 1.38, N = 30

Calculated with multidiel.m.


Bandwidths and angle

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

Omnidirectional Not omnidirectional


Outline









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.Daniel Sjoberg, Department of Electrical and Information Technology

Wavelength dependence

na = nb = 1.55 (quartz), nH = 2.3 (zinc sulfide), nL = 1.25(cryolite), N = 5

Not many material combinations exist that satisfy

na = nb =

√2nHnL√n2H + n2L


Another solution

na = nb = 1.62 (dense flint glass), nH = 2.04 (zirconium oxide),nL = 1.385 (magnesium fluoride), N = 5

Lower reflectance for the TM polarization.


Outline









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


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);


Reflection coefficients

The reflection coefficients are

ρTM =η′TM − ηTM

η′TM + ηTM=n1n3

√n′23 −N2 sin2 θ − n′21 n

′23

√n23 −N2 sin2 θ

n1n3

√n′23 −N2 sin2 θ + n

′21 n′23

√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


Outline









Critical angles

n3 < n′3, n2 < n′2 n3 > n′3, n2 > n′2

sin θ′c =n3n

′3√

n23n′23 + n

′21 (n

′23 − n23)

sin θc =n3n

′3√

n23n′23 + n21(n

23 − n

′23 )

sin θ′c =n2n′2

sin θc =n′2n2


Brewster angles

tan θB =n3n

′3

n21

√n21 − n

′21

n23 − n′23

tan θ′B =n3n

′3

n′21

√n21 − n

′21

n23 − n′23

Requires n1 > n′1 and n3 > n′3, or n1 < n′1 and n3 < n′3.


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]


Normalized TM reflection

Normalized by reflectance at normal incidence.Daniel Sjoberg, Department of Electrical and Information Technology

Differences from the isotropic case

I The Brewster angle can be zero

I The Brewster angle may not exist

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

increase monotonically with angle)

I There may be total internal reflection for one polarization butnot for the other

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


Multilayer birefringent structures

Everything keeps generalizing, just use transverse componentsconsistently. Snel’s law is Na sin θa = Ni sin θi = Nb sin θb, andeverything is coded in (where ni = [ni,1, ni,2, ni,3])

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


Outline









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.


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.


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.


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.


GBO mirrors

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


GBO reflective polarizer

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

Mismatch only in one direction.


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 Particularly, the Brewster angle can be controlled

I Materials can be manufactured from birefringent polymers


Outline









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


Only projects and orals remaining!


electromagnetic wave propagation lecture 15: multilayer

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