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# Lect24 - Lecture 24 Polarization Physics 20 10 0 Confused...

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Unformatted text preview: Lecture 24: Polarization Physics 212 40 30 20 10 0 Confused Avg = 3.0 Confident Physics 212 Lecture 24, Slide 1 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 2 Physics Linear Polarization Physics 212 Lecture 24, Slide 3 Physics The intensity of a wave does not depend on its polarization: I = ε 0c E 2 2 = ε 0 c Ex2 + E y 2 = ε 0 c ⎡ Ex2 + E y ⎤ ⎣ ⎦ = ε 0 cE02 sin 2 ( kz − ωt − φ ) (sin 2 θ + cos 2 θ ) 12 So I = 1 1 ε 0 cE02 Just like before 2 Physics 212 Lecture 24, Slide 4 Physics Polarizers the E field perpendicular to the Transmission Axis to be absorbed. Animation Physics 212 Lecture 24, Slide 5 Physics The molecular structure of a polarizer causes the component of Physics 212 Lecture 24, Slide 6 Physics Preflight 2 Two Polarizers BB 70 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 7 Physics Preflight 3 Two Polarizers BB 80 60 40 20 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 0 Physics 212 Lecture 24, Slide 8 Physics There is no reason that φ has to be the same for Ex and Ey: Making φx different from φy causes circular or elliptical polarization: Example: π φx − φ y = 90o = 2 θ = 45o = π 4 At t=0 Ex = Ey = E0 2 E0 2 cos( kz − ωt ) sin(kz − ωt ) RCP Physics 212 Lecture 24, Slide 9 Physics Q: How do we change the relative phase between Ex and Ey? A: Birefringence By picking the right thickness we can change the relative phase by exactly 90o. This changes linear to circular polarization and is called a Physics 212 Lecture 24, Slide 10 Physics Right hand rule quarter wave plate quarter This modification does not change the intensity either: I = ε 0c E 2 2 = ε 0 c Ex2 + E y Right Handed Circular Polarization (at t = 0) E02 = ε 0c cos 2 (kz − ωt ) + sin 2 ( kz − ωt ) 2 1 So I = 1 ε 0 cE02 Just like before 2 Physics 212 Lecture 24, Slide 11 Physics Right or Left ??? 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 12 Physics Right or Left ??? How would we make this situation give Left Circularly Polarized light? Answer: Reverse the slow and fast axes of the quarter wave plate Physics 212 Lecture 24, Slide 13 Physics Circular Light on Linear Polarizer Q: What happens when circularly polarized light is put through a polarizer along the y (or x) axis ? BB A) I = 0 B) I = ½ I0 C) I = I0 I = ε 0c E 2 = ε 0 c Ex2 + Xy E2 E02 = ε 0c cos 2 ( kz − ωt ) 2 11 = ⋅ ε 0 cE02 22 Half of before Physics 212 Lecture 24, Slide 14 Physics 12 A B Preflight 6 BB Case A: Ex is absorbed I A = I 0 cos 2 (45o ) IA = 1 I 20 Case B: (Ex,Ey) phase changed 50 40 30 20 MORE… I B = I0 10 0 Physics 212 Lecture 24, Slide 15 Physics Intensity: 2 I = ε 0 c ⎡ Ex2 + E y ⎤ ⎣ ⎦ PH SE A QW Plate CH AN GE Both Ex and Ey are still there, so intensity is the same Physics 212 Lecture 24, Slide 16 Physics 2 I = ε 0 c ⎡ Ex2 + E y ⎤ ⎣ ⎦ BS A RB O Polarizer CO MP ON EN T Ex is missing, so intensity is lower Physics 212 Lecture 24, Slide 17 Physics A B Preflight 8 BB 50 40 RCP 1/4 λ Z 30 20 10 0 Physics 212 Lecture 24, Slide 18 Physics A B Preflight 10 BB 50 40 30 ½λ Z Z 20 10 0 Physics 212 Lecture 24, Slide 19 Physics Executive Summary: Polarizers & QW Plates: Polarized Light Circularly or Un-polarized Light Birefringence RCP Ex = E0 2 cos(kx) Ey = E0 sin(kx) 2 Physics 212 Lecture 24, Slide 20 Physics Demos: What else can we put in there to change the polarization? Physics 212 Lecture 24, Slide 21 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? y x 45o fast sl o w 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 22 Physics Calculation Light is incident on two linear polarizers and a quarter wave plate (QWP) as shown. y x 45o fast E1 s lo w RCP Ex E y 60o I1 λ/4 I2 BB BB I3 y z • What is the polarization of the light after the QWP? (A) LCP (B) RCP (C) y x (D) x (E) unpolarized Light incident on QWP is linearly polarized at 45o to fast axis LCP or RCP? Easiest way: Right Hand Rule: Light will be circularly polarized after QWP RCP Curl fingers of RH back to front Thumb points in dir of propagation if right hand polarized. Physics 212 Lecture 24, Slide 23 Physics Calculation Light is incident on two linear polarizers and a quarter wave plate (QWP) as shown. y x 45o fast E1 s lo w RCP Ex E y 60o I1 λ/4 I2 BB I3 z • What is the intensity I2 of the light after the QWP? (A) I2 = I1 BEFORE: Ex = Ey = E1 sin(kz − ωt ) 2 E1 sin(kz − ωt ) 2 (B) I2 = ½ I1 (C) I2 = ¼ I1 AFTER: Ex = Ey = E1 cos(kz − ωt ) 2 E1 sin(kz − ωt ) 2 No absorption: Just a phase change ! I = ε 0c ⎡ E ⎣ 2 x +E ⎤ ⎦ 2 y Same before & after ! Physics 212 Lecture 24, Slide 24 Physics Calculation Light is incident on two linear polarizers and a quarter wave plate (QWP) as shown. y x 45o fast E1 s lo w RCP Ex E y 60o I1 E3 λ/4 I2 = I1 BB I3 y z • What is the polarization of the light after the 60o polarizer? (A) LCP (B) RCP (C) y 60 o x (D) 60o x (E) unpolarized Absorption: only passes components of E parallel to TA (θ = 60o) E Ey E3 = Ex sin θ + E y cos θ E3 = 1 ( sin( kz − ωt + θ ) ) 3 E3 2 E 60o E3 = 1 ( cos(kz − ωt ) sin θ + sin(kz − ωt ) cos θ ) 2 Ex Physics 212 Lecture 24, Slide 25 Physics Calculation Light is incident on two linear polarizers and a quarter wave plate (QWP) as shown. y x 45o fast E1 s lo w RCP Ex E y 60o I1 E3 λ/4 I2 = I1 I3 I3 = ½ I1 BB z • What is the intensity I3 of the light after the 60o polarizer? (A) I3 = I1 E E3 = 1 2 (B) I3 = ½ I1 (C) I3 = ¼ I1 I3 = 1 I1 2 Ey 60o I ∝ E2 E3 3 NOTE: This does not depend on θ !! Ex Physics 212 Lecture 24, Slide 26 Physics Follow Up 1 Replace the 60o polarizer with another QWP as shown. y x 45o fast E w sl o E xE y RCP slow fast Ey E3 I1 λ/4 Ex z I2 = I1 I3 BB • What is the polarization of the light after the last QWP? (A) LCP (B) RCP (C) y y x (D) x (E) unpolarized Efast Easiest way: is λ/4 ahead of Eslow Brings Ex and Ey back in phase !! Physics 212 Lecture 24, Slide 27 Physics Follow Up 2 Replace the 60o polarizer with another QWP as shown. y x 45o fast E w sl o E xE y RCP slow fast Ey E3 I1 λ/4 Ex z I2 = I1 I3 = I1 BB • What is the intensity I3 of the light after the last QWP? (A) I1 BEFORE: Ex = E1 cos(kz − ωt ) 2 I before E12 = 2 (B) ½ I1 (C) ¼ I1 AFTER: Ex = I after E12 = 2 Ey = No absorption: Just a phase change ! Intensity = < E2 > E1 cos(kz − ωt ) 2 E1 cos(kz − ωt ) 2 E E y = 1 sin(kz − ωt ) 2 Physics 212 Lecture 24, Slide 28 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 (B) y x o (C) unpolarized (D) all of above (E) none of above 45 • 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 Question is: What kind of light loses ½ of its intensity after passing through vertical polarizer? Answer: Everything except LP at θ other than 45o Physics 212 Lecture 24, Slide 29 Physics ...
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