Lect24 - Physics 212 Lecture 24: Polarization Physics 212...

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Unformatted text preview: Physics 212 Lecture 24: Polarization Physics 212 Lecture 24, Slide 1 Physics Music Who is the Artist? BB BB A) B) C) D) E) Nnenna Freelon Diana Krall Carmen McRae Diane Schurr Darden Purcell The BEST Local The Female Vocalist I have ever heard here !! heard She sings this Friday She at Iron Post at Happy Hour: 5pm – 7pm Happy 7pm Theme of the week: JAZZ (from classic to new & local) JAZZ Physics 212 Lecture 24, Slide 2 Physics Your Comments “Why the preparation exercises cant be Why put up now so we can start studying now and learn this stuff better” and Hour Exam 3 Prep Exercises Are now UP !! We will discuss in detail… Changes in Intensity and Changes Polarization Polarization “the difference between a the polarizer and a quarter wave plate” polarizer “Birefringent materials appear to be Birefringent breaking relativity--isn't the speed of breaking isn't light CONSTANT in ANY material?” light No Problem. No Relativity deals with relative motion Relativity Speed in materials will be different REFRACTION .. Next lecture 35 “Is this the last Is homework?” homework? NO I’m still writing them 30 25 20 “Could please tell people to Could be quiet because they get louder each lecture and now I can barely hear you” can 05 15 10 PLEASE !! 5 0 Confused Confident Physics 212 Lecture 24, Slide 3 Physics So far we have considered plane waves that look like this: From now on just draw E and remember that B is still there: Physics 212 Lecture 24, Slide 4 Physics Linear Polarization “I was a bit confused by the introduction of the "e-hat" vector (as in its purpose/usefulness)” Physics 212 Lecture 24, Slide 5 Physics Polarizers The molecular structure of a polarizer causes the component of The the E field perpendicular to the Transmission Axis to be absorbed. Physics 212 Lecture 24, Slide 6 Physics Quick ACT The molecular structure of a The polarizer causes the component of BB the E field perpendicular to the field Transmission Axis to be absorbed. Transmission Eo Suppose we have a beam traveling in the + z-direction. At t = 0 and z = 0, the electric field is aligned along the At positive x-axis and has a magnitude equal to Eo positive y B) Eocosθ C) 0 D) Eo/sinθ E) Eo/cosθ sθ A) Eosinθ co Eo What is the component of Eo along a direction in the x-y plane that makes an angle of θ with respect to the xaxis? axis? z Eo θ y Physics 212 Lecture 24, Slide 7 Physics “I can't believe your teaching us the law of "Malus"(Malice). I thought malice was to be avoided?” Physics 212 Lecture 24, Slide 8 Physics Preflight 2 Two Polarizers 60 50 40 30 20 10 0 The second polarizer is orthogonal to the first no light will come through. cos(90o) = 0 Physics 212 Lecture 24, Slide 9 Physics Two Polarizers Preflight 4 BB 80 60 40 20 0 Any non-horizontal polarizer after the first polarizer will produce polarized light AT THAT ANGLE Part of that light will make it through the horizontal polarizer Physics 212 Lecture 24, Slide 10 Physics There is no reason that φ has to be the same for Ex and Ey: has and Making φx different from φy causes circular or elliptical polarization: Making Example: π o φx − φ y = 90 = 2 o θ = 45 = π 4 E0 Ex = cos(kz − ωt ) 2 E0 Ey = sin(kz − ωt ) 2 At t=0 RCP Physics 212 Lecture 24, Slide 11 Physics Q: How do we change How the relative phase between Ex and Ey? and A: Birefringence Birefringence By picking the right thickness we can change thickness the relative phase by exactly 90o. Right hand Right rule rule This changes linear to linear to circular polarization circular polarization and is called a quarter wave plate quarter Physics 212 Lecture 24, Slide 12 Physics “talk something about intensity” NOTE: No NOTE: Intensity is lost passing through the QWP ! the BEFORE QWP: ˆj i + ˆ E = Eo sin( kz − ωt ) 2 2 2 I = cε o E 2 = cε o E x + E y AFTER QWP: Eˆ E = o i cos( kz − ωt ) + ˆ sin( kz − ωt ) j 2 [ 2 2 Eo Eo 2 sin 2 (kz − ωt ) = cε o Eo 1 = cε o + 2 2 2 2 2 I = cε o E 2 = cε o E x + E y 2 Eo = cε o cos 2 (kz − ωt ) + sin 2 (kz − ωt ) 2 2 2 Eo Eo = cε o 1 = cε o 2 2 THE SAME !! Physics 212 Lecture 24, Slide 13 Physics Right or Left ??? “red fox” got it? Right circularly polarized Do right hand rule Fingers along slow direction Cross into fast direction If thumb points in direction of propagation: RCP Physics 212 Lecture 24, Slide 14 Physics Circular Light on Linear Polarizer Q: What happens when circularly polarized light is put through a polarizer along the y (or x) axis ? polarizer (or BB A) I = 0 B) I = ½ I0 C) I = I0 I = ε 0c E 2 = ε 0 c E x2 + Xy E2 E02 = ε 0c cos 2 (kz − ωt ) 2 12 11 = ⋅ ε 0 cE02 22 Half of before Physics 212 Lecture 24, Slide 15 Physics A B Preflight 6 BB 50 Case A: Ex is absorbed I A = I 0 cos2 (45o ) IA = 1 I 20 Case B: (Ex,Ey) phase changed 40 30 20 I B = I0 10 0 Physics 212 Lecture 24, Slide 16 Physics Intensity: 2 I = ε 0 c Ex2 + E y SE A PH QW Plate CH AN G E Both Ex and Ey are still there, so intensity is the same Physics 212 Lecture 24, Slide 17 Physics 2 I = ε 0 c Ex2 + E y RB O BS A Polarizer CO MP ON EN T Ex is missing, so intensity is lower Physics 212 Lecture 24, Slide 18 Physics A B Preflight 8 BB 40 30 RCP 1/4 λ 20 10 Z 0 Physics 212 Lecture 24, Slide 19 Physics A B Preflight 10 BB 50 40 30 ½λ 20 Z Z 10 0 Physics 212 Lecture 24, Slide 20 Physics Executive Summary: Polarizers & QW Plates: Polarized Light Birefringence Circularly or Un-polarized Light RCP Ex = E0 2 cos(kx) Ey = E0 sin(kx) 2 Physics 212 Lecture 24, Slide 21 Physics Demos: What else can we put in there to change the polarization? Physics 212 Lecture 24, Slide 22 Physics Calculation Light is incident on two linear polarizers and a quarter wave plate (QWP) as shown. What is the intensity I3 in terms of I1? fast 45o y x w slo 60o I1 I2 I3 z • Conceptual Analysis • Linear Polarizers: absorbs E field component perpendicular to TA • Quarter Wave Plates: Shifts phase of E field components in fast-slow directions • Strategic Analysis • Determine state of polarization and intensity reduction after each object • Multiply individual intensity reductions to get final reduction. Physics 212 Lecture 24, Slide 23 Physics Calculation Light is incident on two linear polarizers and a quarter wave plate (QWP) as shown. fast 45o y E1 x w slo E RCP x I1 Ey 60o λ/4 BB BB I2 I3 z • What is the polarization of the light after the QWP? y y (A) LCP (B) RCP (C) x Light incident on QWP is linearly polarized at 45o to fast axis LCP or RCP? Easiest way: Right Hand Rule: (D) x (E) unpolarized Light will be circularly polarized after QWP Curl fingers of RH back to front Thumb points in dir of propagation if right hand polarized. RCP Physics 212 Lecture 24, Slide 24 Physics Calculation Light is incident on two linear polarizers and a quarter wave plate (QWP) as shown. fast 45o y E1 x w slo E RCP x I1 Ey 60o λ/4 BB I2 z I3 • What is the intensity I2 of the light after the QWP? (A) I2 = I1 BEFORE: Ex = E1 sin(kz − ωt ) 2 E E y = 1 sin( kz − ωt ) 2 (B) I2 = ½ I1 (C) I2 = ¼ I1 No absorption: Just a phase change ! I = ε 0c E 2 x +E 2 y Same before & after ! AFTER: Ex = E1 cos(kz − ωt ) 2 Ey = E1 sin(kz − ωt ) 2 Physics 212 Lecture 24, Slide 25 Physics Calculation Light is incident on two linear polarizers and a quarter wave plate (QWP) as shown. fast 45o y E1 x w slo E RCP x I1 Ey 60o E3 λ/4 BB I2 = I1 I3 z • What is the polarization of the light after the 60o polarizer? y (A) LCP (B) RCP (C) y 60o x (D) 60o x (E) unpolarized Absorption: only passes components of E parallel to TA (θ = 60o) E Ey E3 = E x sin θ + E y cos θ 3 E3 = 1 ( sin(kz − ωt + θ ) ) E3 2 E 60o E3 = 1 ( cos(kz − ωt ) sin θ + sin(kz − ωt ) cos θ ) 2 Ex Physics 212 Lecture 24, Slide 26 Physics Calculation Light is incident on two linear polarizers and a quarter wave plate (QWP) as shown. fast 45o y E1 x w slo E RCP x I1 Ey 60o E3 λ/4 I2 = I1 BB I3 I3 = ½ I1 z • What is the intensity I3 of the light after the 60o polarizer? (A) I3 = I1 Ey E3 3 E E3 = 1 2 (B) I3 = ½ I1 I ∝ E2 (C) I3 = ¼ I1 I3 = 1 I1 2 NOTE: This does not depend on θ !! 60o Ex Physics 212 Lecture 24, Slide 27 Physics Follow Up 1 Replace the 60o polarizer with another QWP as shown. fast 45o y E x w slo E RCP x I1 slow Ey fast Ey λ/4 I2 = I1 I3 E3 Ex z BB • What is the polarization of the light after the last QWP? y y (A) LCP (B) RCP Easiest way: Efast is λ/4 ahead of Eslow (C) x (D) x (E) unpolarized Brings Ex and Ey back in phase !! Physics 212 Lecture 24, Slide 28 Physics Follow Up 2 Replace the 60o polarizer with another QWP as shown. fast 45o y E x w slo E RCP x I1 slow Ey λ/4 fast Ey I2 = I1 E3 I3 = I1 Ex BB z • What is the intensity I3 of the light after the last QWP? (A) I1 BEFORE: Ex = (B) ½ I1 No absorption: Just a phase change ! E1 cos(kz − ωt ) 2 E E y = 1 sin(kz − ωt ) 2 (C) ¼ I1 Ex = Intensity = < E2 > I before E12 = 2 AFTER: I after E12 = 2 E1 cos(kz − ωt ) 2 Ey = E1 cos(kz − ωt ) 2 Physics 212 Lecture 24, Slide 29 Physics Follow Up 3 Consider light incident on two linear polarizers as shown. Suppose I2 = 1/8 I0 y x E1 60o I0 E2 I1 I1 = ½ I0 I2 = 1/8 I0 z BB • What is the possible polarization of the INPUT light? (A) LCP y (B) x 45o (C) unpolarized • After first polarizer: LP along y-axis with intensity I1 • After second polarizer: LP at 60o wrt y-axis • Intensity: I2 = I1cos2(60o) = ¼ I1 • I2 = 1/8 I0 fl I1 = ½ I0 (D) all of above Question is: What kind of light loses ½ of its intensity after passing through vertical polarizer? (E) none of above Answer: Everything except LP at θ other than 45o Physics 212 Lecture 24, Slide 30 Physics ...
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This note was uploaded on 02/09/2012 for the course PHYSICS 212 taught by Professor Mestre during the Spring '11 term at University of Illinois at Urbana–Champaign.

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