Fresnel Equations and EM Power Flow

Reading - Shen and Kong – Ch. 4

Outline

• Review of Oblique Incidence • Review of Snell
s Law • Fresnel Equations • Evanescence and TIR • Brewster
s Angle • EM Power Flow

1 TRUE / FALSE

H 1. This EM wave is TE r Et (transverse electric) polarized: Er

Ei Ht

H i y• 2. Snell
s Law only works for TE polarized light.

3. Total internal reflection only occurs when light goes from a high index material to a low index material.

2 Refraction

Water Waves E&M Waves on flickr Image by NOAA Photo library Image by NOAA http://www.flickr.com/photos/ noaaphotolib/5179052751/

Waves refract at the top where the water is shallower Refraction involves a change in the direction of wave propagation due to a change in propagation speed. It involves the oblique incidence of waves on media boundaries, and hence wave propagation in at least two dimensions.

3 Refraction in Suburbia

Think of refraction as a pair of wheels on an axle going from a sidewalk onto grass. The wheel in the grass moves slower, so the direction of the wheel pair changes.

Note that the wheels move faster (longer space) on the sidewalk, and slower (shorter space) in the grass.

4 Total Internal Reflection in Suburbia

Moreover, this wheel analogy is mathematically equivalent to the refraction phenomenon. One can recover Snell
s

law from it: n1sinθ1 = n2sinθ2 .

The upper wheel hits the sidewalk and starts to go faster, which turns the axle until the upper wheel re-enters the grass and wheel pair goes straight again.

5 Oblique Incidence (3D view)

k i θi x y kiz kix z Boundary

kix = ki sin(θi)

kiz = ki cos(θi)

Identical definitions for kr and kt

6 Oblique Incidence at Dielectric Interface

Hr

Et Er Transverse Electric Field Ei Ht

Hi

Hr Et

Er Transverse H t Magnetic Field Ei

Hi y • Why do we consider only these two polarizations?

7 Partial TE Analysis E E j(ωt−k·kr) z = 0 E = Eoe Hr ωi = ωr = ωt Et Er

- i −jkixx−jkiz z Ei =ˆyE e Ht o Ei - r −jkrxx+jkrz z Er =ˆyEo e Hi • y - t −jktxx−jktz z Et =ˆyEoe

Tangential E must be continuous at the boundary z = 0 for all x and for t.

i −jkix r −jkrxx t −jktxx Eoe+ Eo e = Eoe

This is possible if and only if kix = krx = ktx and ωi = ωr = ωt. The former condition is phase matching.

8 Snell
s Law Diagram

Tangential E field is continuous … kix = ktx

Refraction Total Internal Reflection

kx kx

Radius = k2

ktx

θr θt θt θc

θt kz θc kz kix

Radius = k1

k1 k2

9 Snell
s Law

n1 n2

x y • z

kix = krx kix = ktx n1 sin θi = n1 sin θr n1 sin θi = n2 sin θt

θi = θr Snell
s Law

10 TE Analysis - Set Up

j(−k x−k z) E- =ˆyE e ix iz z = 0 i o H r - j(−kixx+kiz z) kr Er =ˆyrEoe Et k t j(−k x−k z) Er - ix iz Et =ˆytEoe θr H ∂μH- θi θt t ∇×- E- = − Ei ∂t - - ki −jk × E = −jωμH To get H, use H 1
i y • H- = k × E- Faraday s ωμ Law Medium 1 Medium 2 Eo - j(−kixx−kiz z) Hi =(ˆzkix − xkˆ iz ) e ωμ1 2 2 2 2 kx+ k z = k = ω μ� rE - o j(−kixx+kiz z) Hr =(ˆzkix +ˆxkiz) e kx = k sin θ ωμ1 k = k cos θ z tEo - j(−ktxx−ktz z) Ht =(ˆzktx +ˆxktz ) e ωμ2

11 TE Analysis – Boundary Conditions

k2 = k2 = ω2μ� Incident Wavenumber: i r

Phase Matching: kix = krx = ktx θr = θi

ki sin θi = kt sin θt Tangential E: 1+ r = t Normal μH: 1+ r = t E tE Tangential H: o (1 − r)k = o k ωμ itz ωμ itz i √ t μ ε t t μi cos θt ηi cos θt (1 − r)= t √ = t μiεi μt cos θi ηt cos θi

12 Solution Boundary conditions are:

1+ r = t z = 0 Hr η cos θ kr i t Et 1 − r = t kt Er ηt cos θi θr

θi θt Ht Ei

ki 2ηt cos θi Hi y • t = ηt cos θi + ηi cos θt Medium 1 Medium 2 η cos θ − η cos θ r = t i i t Et = tEi ηt cos θi + ηi cos θt Er = rEi

Fresnel Equations

13 Today
s Culture Moment Sir David Brewster • Scottish scientist • Studied at University of Edinburgh at age 12 • Independently discovered Fresnel lens • Editor of Edinburgh Encyclopedia and contributor to Encyclopedia Britannica (7th and 8th editions) • Inventor of the Kaleidoscope • Nominated (1849) to the National Institute of 1781 –1868 France. Fresnel Lens

Kaleidoscope

All Images are in the public domain

14 TM Case is the dual of TE

E Gauss: k·E = 0 H Gauss: k·H = 0 Faraday: k х E = ωμH Ampere: k х H = - ωεE E → H H → - E ε → μ E Gauss: k·E = 0 μ → ε H Gauss: k·H = 0 Faraday: k х E = ωμH Ampere: k х H = - ωεE

The TM solution can be recovered from the TE solution. So, consider only the TE solution in detail.

15 TE & TM Analysis – Solution

TE solution comes directly from the boundary condition analysis

2ηt cos θi ηt cos θi − ηi cos θt t = r = ηt cos θi + ηi cos θt ηt cos θi + ηi cos θt

TM solution comes from ε ↔ μ

−1 −1 −1 η cos θi − η cos θt 2η cos θi r = t i t = t η−1 cos θ + η−1 cos θ η−1 cos θ + η−1 cos θ t i i t t i i t

Note that the TM solution provides the reflection and transmission coefficients for H, since TM is the dual of TE.

16 Fresnel Equations - Summary

TE-polarization Er N − N Et 2N o i t o i rE = = tE = = Ei N + N Ei N + N o i t o i t

1 1 Ni = cos θi Nt = cos θt η1 η2

TM Polarization Hr M − M Ht 2M o i t o i rH = = tH = = Hi M + M Hi M + M o i t o i t

Mi = η1 cos θi Mt = η2 cos θt

17 Fresnel Equations - Summary

From Shen and Kong … just another way of writing the same results

TE Polarization TM Polarization

Er μ k − μ k Er � k − � k o 2 iz 1 tz o 2 iz 1 tz rTE = = rTM = = Ei μ k + μ k Ei � k + � k o 2 iz 1 tz o 2 iz 1 tz

Et 2μ k Et 2� k o 2 iz o 2 iz tTE = = tTM = = Ei μ k + μ k Ei � k + � k o 2 iz 1 tz o 2 iz 1 tz

18 Fresnel Equations

n1 = 1.0 n2 = 1.5 1.5 Hr Et t Er TM 1

tTE TE H s Ei noticefleR t tneicffieoC 0.5

Hi rTM 0 θB Hr Et rTE

Er -0.5 Ht Ei TM -1 Hi • 0 30 60 90 y Incidence Angle 19 Fresnel Equations

Total Internal Reflection n1 = 1.5 n2 = 1.0 1

Hr Et 0.8 Er 0.6

E TE Ht i 0.4 rTE

Hi 0.2

θB θC Hr Et 0 rTM

Reflection Coefficients Reflection -0.2 θ = sin-1(n / n ) Er C 2 1 H t -0.4 Ei TM 0 30 60 90 Incidence Angleθi Hi y •

20 n cos θ − n cos θ Reflection of Light 1 i 2 t TE: r⊥ = n1 cos θi + n2 cos θt (Optics Viewpoint … μ1 = μ2) n2 cos θi − n1 cos θt TM: r� = n2 cos θi + n1 cos θt Hr Et Er

θc E TE Ht i n1 =1.44 n2 =1.00 Hi

Hr Et |r⊥|

Er |r�| H θB t Coefficients Reflection Ei TM

Incidence Angle θi Hi y •

21 Brewster
s Angle

(Optics Viewpoint … μ1 = μ2)

r⊥ =0 Requires Unpolarized TE Polarized Incident Ray Reflected Ray Snell
s Law n1 = n2

r� =0 Can be

Image in the Public Domain Snell
s Law Satisfied with Sir David Brewster

θB = arctan(n2/n1)

θB + θT = 90° Slightly Polarized Reflracted Ray

Light incident on the surface is absorbed, and then reradiated by oscillating electric dipoles (Lorentz oscillators) at the interface between the two media. The dipoles that produce the transmitted (refracted) light oscillate in the polarization direction of that light. These same oscillating dipoles also generate the reflected light. However, dipoles do not radiate any energy in the direction along which they oscillate. Consequently, if the direction of the refracted light is perpendicular to the direction in which the light is predicted to be specularly reflected, the dipoles will not create any TM-polarized reflected light.

22 Energy Transport

A cos θ A cos θ i θ r i θr T⊥

TE A ni R⊥ nt

T� A cos θt

n1 =1.0 TM n =1.5 2 Cross-sectional Areas i i Si = E H cos θi R o o � θB --- t t S = E × H St = EoH o cos θt θi [Degrees] S = ErHr cos θ r o o r

23 Transmitted Power Fraction: � t 2 �2 t t (E ) cos θt S E H cos θ o μ2 N t o o t 2 t Ts = = = � = t i i 2 � E Si EoHo cos θi i 1 Ni (Eo) cos θi μ1 � 2 t μ2 (Ho) cos θt �2 Mt = � = t2 2 μ H i 1 Mi (Ho) cos θi �1 t 2 (Eo) 2 2 ≡ tE i (Eo)

t 2 (Ho) 2 2 ≡ tH i (Ho)

24 Reflected Power Fraction: � r 2 � r r (E ) 1 S E H cos θ o μ1 r o o r 2 Rs = = = � = rE i i 2 � Si EoHo cos θi (Ei ) 1 o μ1 � r 2 μ1 (Ho ) �1 = � = r2 2 μ H i 1 2 (Ho) (Er) �1 o 2 2 ≡ r E i (Eo) (Hr)2 o 2 … and from ENERGY CONSERVATION we know: 2 ≡ r H (Hi ) o TS + RS =1

25 Summary CASE I:

E-field is polarized perpendicular to the plane of incidence

… then E-field is parallel (tangential) to the surface, 1+r = t and continuity of tangential fields requires that: E E

TS + RS =1 N 2 t 2 (tE ) +(rE ) =1 Ni N 2 t 2 (1 + rE ) +(rE ) =1 Ni

Er N − N Et 2N o i t o i rE = = tE =1+rE = = Ei N + N Ei N + N o i t o i t

26 Summary CASE II:

H-field is polarized perpendicular to the plane of incidence

… then H-field is parallel (tangential) to the surface, 1+rH = tH and continuity of tangential fields requires that:

TS + RS =1

2 Mt 2 (tH ) +(rH ) =1 Mi

2 Mt 2 (1 + rH ) +(rH ) =1 Mi

Hr M − M Ht 2M o i t o i rH = = tH =1+rH = = Hi M + M Hi M + M o i t o i t

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