Fresnel Equations and EM Power Flow Reading - Shen and Kong – Ch. 4 Outline • Review of Oblique Incidence • Review of Snells Law • Fresnel Equations • Evanescence and TIR • Brewsters 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. Snells 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 Snells 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) E Eoe H z = 0 = r ω = ω = ω i r t 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 Snells Law Diagram Tangential E field is continuous … kix = ktx Refraction Total Internal Reflection k k x x Radius = k2 ktx θr θt θt θc θt kz θc kz kix Radius = k1 k <k k >k 1 2 1 2 9 Snells Law n1 n2 x y • z kix = krx kix = ktx n1 sin θi = n1 sin θr n1 sin θi = n2 sin θt θ = θ i r 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 - θ θ H - - ∂μH i t 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(−k x−k z) H- =(ˆzk +ˆxk ) e tx tz t tx tz ωμ 2 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 o (1 − r)kitz = kitz ωμi ωμt √ μtεt μi cos θt ηi cos θt (1 − r)=t √ = t μ ε μ cos θ η cos θ i i t i t i 12 Solution Boundary conditions are: z = 0 1+ r = t Hr η cos θ kr i t Et 1 − r = t kt Er ηt cos θi θr θi θt Ht Ei ki H 2ηt cos θi i 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 Todays 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 ηt cos θi − η cos θt 2ηt cos θi r = i 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 M = η cos θ M = η cos θ i 1 i t 2 t 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 t = o = 2 iz t = o = 2 iz TE i TM i E μ2kiz + μ1ktz E �2kiz + �1ktz o o 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 TE Ei Ht 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 Brewsters Angle (Optics Viewpoint … μ1 = μ2) r⊥ =0 Requires Unpolarized TE Polarized Incident Ray Reflected Ray Snells Law n1 = n2 r =0 � Can be Image in the Public Domain Snells 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 --- S = Et Ht cos θ S = E × H t o o t θi [Degrees] r r Sr = E H cos θr o o 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 = = = � = tE i i 2 � Si EoHo cos θi (Ei ) 1 cos θ Ni o i μ1 � 2 μ (Ht) 2 cos θ o � t M 2 2 t = � = t 2 μ H (Hi ) 1 cos θ Mi o i �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 E H cos θi i �1 o o (Eo) μ1 � r 2 μ1 (Ho ) � 1 2 = � = r 2 μ H i 1 2 (Ho) (Er) �1 o 2 2 ≡ rE i (Eo) (Hr)2 o 2 … and from ENERGY CONSERVATION we know: 2 ≡ rH (Hi ) T + R =1 o S S 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 r = o = i t t =1+r = o = i E i E E i E Ni + Nt E Ni + Nt o o 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 = = t =1+r = = i H H i Ho Mi + Mt H Mi + Mt o 27 MIT OpenCourseWare http://ocw.mit.edu 6.007 Electromagnetic Energy: From Motors to Lasers Spring 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.