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Fresnel Equations and Light Guiding 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 1. The Fresnel equations describe reflection and transmission coefficients as a function of intensity. 2. This is the power reflection and transmission plot for an EM wave that is TE (transverse electric) polarized: T n1 = 1.0 n2 = 1.5 R θi θB [Degrees] 3. The phase matching condition for refraction is a direct result of the boundary conditions. 2 Oblique Incidence at Dielectric Interface Hr Et Er Transverse Electric Field Ht Ei Hi Hr Er Et Transverse Magnetic Field Ht Ei Hi y• 3 Partial TE Analysis i = ŷE i e−jkix x−jkiz z E o z=0 Hr Et Er r = ŷE r e−jkrx x+jkrz z E o Ht Ei Hi t = ŷE t e−jktx x−jktz z E o ωi = ωr = ωt y• Tangential E must be continuous at the boundary z = 0 for all x and for t. Eoi e−jkix x + Eor e−jkrx x = Eot e−jktx x This is possible if and only if kix = krx = ktx and ωi = ωr = ωt. The former condition is phase matching kix = krx = ktx 4 Snells Law Er n1 n2 Et Ei y• kix = krx kix = ktx n1 sin θi = n1 sin θr n1 sin θi = n2 sin θt Snells Law θ i = θr 5 TE Analysis - Set Up Hr kr Er Ei i = ŷEo ej(−kix x−kiz z) E z=0 Et r = ŷrEo ej (−kix x+kiz z) E kt t = ŷtEo ej(−kix x−kiz z) E θr θt θi ∂μH ∇×E =− ∂t = −jωμH −jk × E = 1 k×E H ωμ Ht ki Hi Medium 1 y• Medium 2 To get H, use Faradays Law i = (ẑkix − x̂kiz ) Eo ej(−kix x−kiz z) H ωμ1 kx2 + kz2 = k 2 = ω 2 μ r = (ẑkix + x̂kiz ) rEo ej(−kix x+kiz z) H ωμ1 kx = k sin θ kz = k cos θ t = (ẑktx + x̂ktz ) tEo ej(−ktx x−ktz z) H ωμ2 6 TE & TM Analysis – Solution TE solution comes directly from the boundary condition analysis ηt cos θi − ηi cos θt r= ηt cos θi + ηi cos θt 2ηt cos θi t= ηt cos θi + ηi cos θt TM solution comes from ε μ 2ηt−1 cos θi t = −1 ηt cos θi + ηi−1 cos θt ηt−1 cos θi − ηi−1 cos θt r = −1 ηt cos θi + ηi−1 cos θt Note that the TM solution provides the reflection and transmission coefficients for H, since TM is the dual of TE. 7 Fresnel Equations - Summary From Shen and Kong … just another way of writing the same results TE Polarization rTE Eor μ2 kiz − μ1 ktz = i = μ2 kiz + μ1 ktz Eo tTE Eot 2μ2 kiz = i = μ2 kiz + μ1 ktz Eo TM Polarization 8 rTM Eor 2 kiz − 1 ktz = i = 2 kiz + 1 ktz Eo tTM Eot 22 kiz = i = 2 kiz + 1 ktz Eo Reflection of Light (Optics Viewpoint … μ1 = μ2) Hr Hi y• Et Er Ht Ei TM Hi TM: r = n2 cos θi − n1 cos θt n2 cos θi + n1 cos θt E-field parallel to the plane of incidence 9 θc Reflection Coefficients E-field perpendicular to the plane of Ht incidence TE Hr n1 cos θi − n2 cos θt n1 cos θi + n2 cos θt Et Er Ei TE: r⊥ = n1 = 1.44 n2 = 1.00 |r⊥ | θB |r | Incidence Angle θi Brewsters Angle Incident ray (unpolarised) Reflected ray (TE polarised) θB + θt = 90° n1 n2 Image in the Public Domain Sir David Brewster (1781 –1868) was a Scottish scientist, inventor and writer. Rediscovered and popularized kaleidoscope in 1815. n1 sin (θB ) = n2 sin (90 − θB ) Refracted ray (slightly polarised) = n2 cos (θB ) θB = arctan @ θi = θB TE: r⊥ = n1 cos θB − n2 cos θt = 0 n1 cos θB + n2 cos θt n2 cos θB = n1 cos θt n1 sin θB = n2 sin θt n2 cos θB − n1 cos θt TM: r = =0 n2 cos θB + n1 cos θt 10 n2 n1 Total Internal Reflection Beyond the critical angle, refraction no longer occurs – thereafter, you get total internal reflection TOTAL INTERNAL REFLECTION n2sinθ2 = n1sinθ1 θcrit = sin-1(n1/n2) Image in the Public Domain – for glass (n2 = 1.5), the critical internal angle is 42° – for water, its 49° – a ray within the higher index medium cannot escape at shallower angles (look at sky from underwater…) incoming ray hugs surface n1 = 1.0 n2 = 1.5 42° 11 Snells Law Diagram Tangential field is continuous … kix = kit Refraction Total Internal Reflection kx kx Radius = k2 ktx θr θt θc kz θt θc kix Radius = k1 k 1 < k2 k 1 > k2 12 θt kz Total Internal Reflection & Evanscence Snells Law dictates n1 sin(θi) = n2 sin(θt) , or equivalently, kix = ktx . For n1 > n2 , θt = 90° at θi = sin-1(n2/n1) θC. What happens for θi > θC ? n1 > n2 kx kr θc kt ktz2 = kt2 – ktx2 < 0 ktz = ± j αtz , with αtz real. The refracted, or transmitted, wave takes the complex exponential form θt kz θc ki 1 2 exp(- j ktx x - αtz z) . This is a non-uniform plane wave that travels in the x direction and decays in the z direction. It carries no time average power into Medium 2. This phenomenon is referred to as total internal reflection. This is the similar to reflection of radio waves by the ionosphere. 13 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. 14 Frustrated Total Internal Reflection In Suburbia D k1 d An evanescent field can propagate once the field is again in a high-index material. 15 Applications of Evanescent Waves fingertip Image in the Public Domain light source camera The camera observes TIR from a fingerprint valley and blurred TIR from a fingerprint ridge. Image in the Public Domain 16 Light Propagating Through a Multimode Optical Fibre Jacket 400 μm Buffer 250 μm Cladding 125 μm Core 8 μm Single Mode Fibre Structure Image in the Public Domain The optic fiber used in undersea cables is chosen for its exceptional clarity, permitting runs of more than 100 kilometers between repeaters to minimize the number of amplifiers and the distortion they cause. Image in the Public Domain A cross-section of a submarine communications cable: Submarine communication cables crossing the Scottish shore 1. Polyethylene 2. "Mylar" tape 3. Stranded steel wires 4. Aluminum water barrier 5. Polycarbonate 6. Copper or aluminum tube 7. Petroleum jelly 8. Optical fibers Typically 69 mm in diameter and weigh around 10 kg per meter 17 Image by Jmb at http://en.wikipedia.org/ wiki/File:Submarine_Telephone_Cables_ PICT8182_1.JPG on Wikipedia. Optical Waveguides Examples Image by Apreche http://www.flickr.com/photos/ apreche/69061912/ on flickr Image by Rberteig http://www.flickr.com/photos/ rberteig/89584968/ on flickr LCD screen lit by two backlights coupled into a flat waveguide Optical fiber 18 Image by Mike Licht http://www.flickr.com/photos/notionscapital/ 2424165659/ on flickr Todays Culture Moment Global Fiber Optic Network Image in the Public Domain Image by MIT OpenCourseWare Image by Paul Keleher http://commons.wiki media.org/wiki/File:Trench_USA-fiber.jpg on Wikimedia Commons. 19 Todays Culture Moment Laying Transcontinental Cables Image in the Public Domain 20 METAL REFLECTION © Kyle Hounsell. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse Three Ways to Make a Mirror MULTILAYER REFLECTION Image is in the public domain Image is in the public domain: http://en.wikipedia.org/wiki/ File:Dielectric_laser_mirror_from_a_dye_las er.JPG TOTAL INTERNAL REFLECTION 21 Transporting Light We can transport light along the z-direction by bouncing it between two mirrors y Mirror x Mirror d z θ ..the ray moves along both y- and z-axes.. Ex (y, z) = Ae±jky y e−jβz …where ky = nko sin θ kz = nko cos θ 22 nko θ kz ky Transverse Electric (TE) Modes d −jβz +jky y −jky y Ex (y, z) = A1 e + A2 e e = am um (y)e−jβm z y x Mirror Mirror STANDING WAVE IN y-DIRECTION d/2 0 −d/2 z m=1 2 3 6 23 Perfect Conductor Waveguide y Mirror x Mirror d/2 0 −d/2 z m=1 2 3 6 2/d cos ky y if m = odd um (y) = 2/d sin ky y if m = even kym Boundary Conditions π =m d mπy 2/d cos d um (y) = mπ y 2/d sin d 24 if m = odd if m = even Transporting Light y Mirror x Mirror d/2 0 −d/2 z m=1 2 3 6 2/d cos ky y if m = odd um (y) = 2/d sin ky y if m = even kym = nko sin θm nko ky θ kz,m π =m d kz,m = nko cos θm 25 2 kz, m 2 π = k 2 − m2 2 d Transporting Light y Mirror x Mirror d/2 0 −d/2 z m=1 2 3 6 nko ky θ kz,m The solutions can be plotted along a circle of radius k=nko… ky = ko sin θ nko M kym = nko sin θm m π d kz,m = nko cos θm 3 2 1 kz,m kz,m nko = nko cos θm π =m d 26 Waveguide Mode Propagation Velocity Mirror Mirror Velocity along the direction of the guide… …steeper angles take longer to travel through the guide 27 Lowest Frequency Guided Mode Number of Modes Cut-off Frequency Frequency of light (color) Mirror Mirror 28 Solutions for a Dielectric Slab Waveguide y d 2 m=1 2 3 4 8 0 d − 2 z What does it mean to be a mode of a waveguide? 29 Slab Dielectric Waveguides ky n2 k o n1 k o M n1 ko sin θc steepest incidence angle m θc 1 θm 0 0 0 n2 k o n1 k o 30 shallowest incidence angle kz = β Comparison of Mirror Guide and Dielectric Waveguide Metal Waveguide Dielectric Waveguide θ n2 n1 n2 31 Key Takeaways n1 > n2 kx Total Internal Reflection. What happens for θi > θC ? ktz2 = kt2 – ktx2 < 0 ktz = ± j αtz , with αtz real. kr kt θc Evanescent field θt kz exp(- j ktx x - αtz z) θc ki Waveguide Modes mπy 2/d cos d um (y) = mπy 2/d sin Mirror d 1 2 x y Mirror d/2 0 −d/2 m=1 2 3 z 6 32 if m = odd if m = even 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.