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Wave Incidence at Oblique angles Sandra Cruz-Pol ECE Dept. UPRM Quiz VIERNES Uno de los problemas asignados en el prontuario Enviar contestaciones a preguntas en Word por email Wave Incidence I. Normal incidence Wave arrives at 90o from the surface II. Standing waves Oblique incidence (lossless) Wave arrives at an angle Snell’s Law and Critical angle Two types: Parallel or Perpendicular Brewster angle Oblique incidence First, we need to define: Expression for uniform plane wave traveling in any direction x Medium 2 : e2, m2 Medium 1 : e1 , m1 The Normal , an Plane of incidence Angle of incidence qr qt qi y z=0 z Oblique incidence Uniform plane wave in general form j ( k r t ) E (r , t ) Eo cos( k r t ) Re[ Eo e ] r xaˆ x yaˆ y zaˆ z the radius or position vector k k x aˆ x k y aˆ y k z aˆ z the wave number or propagation vector Where: k 2 k x2 k y2 k z2 2 me For lossless unbounded media, k = b Oblique Incidence x Medium 2 : e2, m2 Medium 1 : e1 , m1 kr kt qr qi ki qt y z=0 z Look at the direction of travel Medium 1 : e1 , m1 kr ki kiz Medium 2 : e2, m2 kt qr qi kix x qt y z=0 z Expression for Waves Ei Eio cos( kix x kiz z t ) Er Ero cos( k rx x k rz z t ) Et Eto cos( ktx x ktz z t ) qi ki where kiz kix k i k ix2 k iz2 b1 m1e 1 k ix b1 sin q i k iz b1 cos q i Tangential E must be Continuous Ei ( z 0) Er ( z 0) Et ( z 0) wi = wr = wt = w kix = krx = ktx = kx kiy = kry = kty = ky From this we know that frequency is a property of the wave. So is color. So 700nm is not always red!! Give example of bathing suit. kix krx b1 sin q i b1 sin q r Snell Law Equating, we get or kix = ktx b1 sinqi = b2 sinq t Also written as, n1 sinqi = n2 sinq t where, the index of refraction of a medium, ni , is defined as the ratio of the phase velocity in free space (c) to the phase velocity in the medium. c n1 = = er1 u1 Critical angle, qc …All is reflected n2 sin q c sin q t [q t 90 o ] n1 n2 er2 n1 e r1 When qt =90o, the refracted wave flows along the (for m1 m 2 ) surface and no energy is transmitted into medium 2. Example The value of the angle of er 2 sin 90 o incidence corresponding to sin q c = er1 this is called critical angle, qc. 4 o sin 42 = 1) = .67 ( If qi > qc, the incident wave 9 is totally reflected. sin 50 o = .77 = .67 (sin?? o ) sin 40 o = .64 = .67 (sin 73o ) Fiber optics Light can be guided with total reflections through thin dielectric rods made of glass or transparent plastic, known as optical fibers. The only power lost is due to reflections at the input and output ends and absorption by the fiber material (not perfect dielectric). Optical fibers have cylindrical fiber core with index of refraction nf, surrounded by another cylinder of lower, nc < nf , called a cladding. For total reflection: Use Snell and critical angle to derive: n2 sin q 3 sin q c n1 q 2 q 3 90o [Figure from Ulaby, 1999] Acceptance angle sin q a (n 2f nc2 ) n0 Exercise: optical fiber An optical fiber (in air) is made of fiber core with index of refraction of 1.52 and a cladding with an index of refraction of 1.49. Find the acceptance angle: 2 sin q a £ (n f - nc2 ) n0 (1.52 2 -1.49 2 ) = 1 Answer: ≤17.5o Parallel polarization It’s defined as E is || to incidence plane Eis Eio (cos q i xˆ sin q i zˆ )e jb1 x sinq i z cosqi H is Eio 1 e jb1 x sinq i z cosq i yˆ Er Et Ers Ero (cos q r xˆ sin q r zˆ )e jb1 x sinq r z cosq r H rs Ero 1 H ts Eto 2 kr e jb1 x sinq r z cosq r yˆ Ets Eto (cos q t xˆ sin q t zˆ )e e jb 2 x sinq t z cosq t yˆ x Medium 1 : e1 , m1 jb 2 x sinq t z cosq t Ei qr qi ki qt y kt z z=0 Medium 2 : e2, m2 Equating for continuity, the tangent fields Which components are tangent to the interface between two surfaces? y and x At z = 0 (interface): xˆ : Eio (cos q i )e jb1 x sinqi z cosqi Ero (cos q r )e jb1 x sinq r z cosq r Eto (cos q t )e jb 2 x sinqt z cosqt yˆ : Eio 1 e jb1 x sinqi z cosqi Ero 1 e jb1 x sinqi z cosqi Eto 2 e jb 2 x sinqt z cosqt xˆ : Eio cos q i Ero cos q r Eto cos q t yˆ : Eio 1 Ero 1 Eto 2 Reflection and Transmission Coefficients: Parallel Incidence Reflection Ero 2 cos q t 1 cos q i || Eio 2 cos q t 1 cos q i Eto 2 2 cos q i || Eio 2 cos q t 1 cos q i where || (1 || ) cos q i cos q t Reflection and Transmission Coefficients: Perpendicular Incidence Ero 2 cos q i 1 cos q t Eio 2 cos q i 1 cos q t Eto 2 2 cos q i Eio 2 cos q i 1 cos q t 1 Java Animation http://www.amanogawa.com/archive /Oblique/Oblique-2.html Exercise A plane wave in air with Ei yˆ10e j (3 x 4 z ) [V / m] Is incident upon planar surface of nonmagnetic dielectric material with er=4, on z>0, Find The incidence polarization of this wave The angle of incidence The time-domain expressions for the reflected electric and magnetic fields. The average power density carried by the wave in the dielectric medium. Brewster angle, qB Is defined as the incidence angle at which the reflection coefficient is 0 (all transmission). G || = h2 cosq t - h1 cosq B =0 h2 cosq t + h1 cosq B h2 cosq t - h1 cosq B = 0 sin q B|| = 1- (e1m 2 / e 2 m1 ) 1- (e1 / e 2 )2 * qB is known as the polarizing angle The Brewster angle does not exist for perpendicular polarization for nonmagnetic materials. Exercise A plane wave in free space with Ei (10 yˆ 5zˆ) cos(t 2 y 4 z )[V / m] Is incident upon planar surface of nonmagnetic lossless dielectric material with er=4, on z>0, Find The polarization of the incident wave The angle of incidence and transmission The reflection and transmission coefficients. The E field that is reflected and in dielectric Answers: Brewster angle 26.6 o, 12.9 o, - 0.295, 0.65, (-2.946 ŷ +1.47ẑ)cos(w t + 2y + 4z), (6.5 ŷ + 3.2 ẑ)cos(w t + 2y - 8.7z), 63.4o Reflection vs. Incidence angle. Reflection vs. incidence angle for different types of soil and parallel or perpendicular polarization. Exercise (Brewster angle) A wave in air is incident upon the flat boundary of a nonmagnetic soil medium with er=4, at qi=50o. Find reflection and transmission coefficients for both incident polarizations, and the Brewster angle. Answers: || 0.16, || 0.58 0.48, 0.52 EMAG q B|| 63.4 o Summary: wave incidence Property Reflection coefficient Transmission coefficient Relation Normal Incidence 2 1 2 1 2 2 i 2 1t R Power Transmissivity T 1 R sin q t cos q t 1 cos q i 2 cos q i 1 cos q t || 2 2 cos q t 1 cos q i 2 cos q i 1 cos q t 2 2 cos q i 2 2 cos q i || 2 cos q i 1 cos q t 2 cos q t 1 cos q i 1 R 2 n1 sin q i n2 Parallel 1 Power Reflectivity Snell’s Law: Perpendicular 2 T 1 R where n2 m r 2e r 2 || 1 || R|| || cos q i cos q t 2 T|| 1 R||