Chapter 14 Matrix Treatment of Polarization

Transcription

1 Chapter 4 Matri Treatment of Polarization Lecture Notes for Modern Optics based on Pedrotti & Pedrotti & Pedrotti Instructor: Naer Eradat Spring 29 5//29 Matri Treatment of Polarization

2 Polarization Polarization of an electromagnetic wave is direction of the electric field vector E. Mathematical presentation of polarized light: Jones Vectors Mathematical presentation of polarizers (optical components) : Jones Matrices Linear polarizer Phase retarder Rotators 5//29 Matri Treatment of Polarization 2

3 Mathematical presentation of polarized light Electric field of an electromagnetic wave propagating p galong z-direction: ikz ( ωt+ φ ) E Re E e = E = E E= E+ E with and comple field components ikz ( ωt+ φ ) E = E e E = Re E ikz ( ωt+ φ E ) ( E= e + E ikz ω t + φ ) iφ i φ ikz ( ω t ) ( ) ikz ω t e = E e + E e e = E e Comple amplitude vector E also contains phase State of polarization of a wave is determined b relative amplitudes and phases of of the components of E E that constitute a vector dubbed Jones vector: iφ E E e = = i φ E E e 2 ( E) ( E) Jones vector are normalized if + =. 2 Plane wave ( ) ( ) 5//29 Matri Treatment of Polarization 3

4 Special cases of Jones vector Particular forms of Jones vector: φ = φ = linearl polarized light along a line making an angle α with the ais: iφ E e Acosα cosα E = = A iφ = E Asin sin e α α cos Vertical polarization: E ( π /2) = = sin ( π / 2) cos ( ) Horizontal polarization: E = = sin ( ) cos( 6 ) Polarized at α = 6 : E = = sin ( 6 ) 2 3 Conclusion: The light presented b a Jones vactor that both o f its elements, a and b, are real (not bothzero) ( b a) - is a linearl polarized light along the angle α = tan / 5//29 Matri Treatment of Polarization 4

5 Lissajous Figures For general case of φ and φ the head of the E vector traces an ellipse rather than a straight line. The relative phase difference of the E and E, Δ φ = φ φ determines the shape of the Lissajous figure and the state of polarization of the wave. o o 5//29 Matri Treatment of Polarization 5

6 LCP and RCP Eample: consider electric field of an EM wave that has E = E = A and E leads E b ε = π / 2. Determine the state of polarization and deduce the normalized Jones vectors for this light. We write the comple amplitudes as iωt E = Acosωt E = E e E = A cos ω t E = A cos ω t i( ωt ε) π E = E E cos( ) cos E sin e = A ωt ε E = A ωt = A ωt 2 ( ω ω ) E = E + E = A cos t+ sin t = A the E vector traces out a circle of radius A iφ E = Eo = A E e A Finding the Jones vector: then E = = A iφ iπ /2 = φ, φ π / 2 = = E Ae oe i ( ( )) * 2 2 * 2 Normalization: i E E= A + ii = 2 A = A = / 2 The normalized Jones vector is: E = 2 We call this a left-circularl polarized i light or LCP since when we view this light head-on we see the E vector tip is rotating counterclockwise on a circle of radius / 2. Figure shows the E at diffrent times. If E leads E b π /2 the E would rotate clockwise. E = 2 i We have right-circularl polarized light or RCP in this case. 5//29 Matri Treatment of Polarization 6 o

7 Ellipticall polarized light Eample: consider electric field of an EM wave that has E = A and E = B where A and B are positive numbers and E leads/lags E b ε = π / 2. Determine the state of polarization and deduce the normalized Jones vectors for this light. i t E = Acosωt E E e ω = E = Acosωt i ( ωt ε ) ( ) π E E E cos cos e = B ωt ε E = B ωt = 2 E = Acosωt E = B sin ω t ( ) ( ) ( ) * 2 * 2 2 Normalization: E E = A + ib ib = A + B = A A Jones vector counterclockwise E = 2 2 iπ /2 = 2 2 ib A + B Be A + B A A Jones vector clockwise E = iπ /2 = A + B Be A + B ib Conclusion 2: the Jones vector with elements un-equal in magnitude, one of which is pure imaginar, represents an ellipticall polarized light. Figure shows the for two cases of E > E (major ais along ) and E < E E (major ais along ). 5//29 Matri Treatment of Polarization 7 o

8 Ellipticall polarized light oriented at an angle relative to ais Eample: consider electric field of an EM wave that has E = A and E = b where A and B ( ) are positive numbers and E and E have phase difference of Δ φ ± m + /2 π and Δφ ± mπ where m=, ±, ± 2,... Determine the state of polarization and deduce the normalized Jones vectors for this light. We can assume Δ Δφ = ε and φ =, φ = ε iφ E e A A A = = iφ iε = = E be bcos ibsin B ic e ε + ε + E ( )( ) ( ) ( ) * 2 * Normalization: E E = A + B+ ic B+ ic = A + B + C = o Counterclockwise rotation, general case Jones vector of an ellipticall polarized light with major ais inclined at an angle α is: A = where tan 2α = A B C B+ ic + + E E E cosε o 2 2 E C and E = A, E = B + C, ε = tan B C > counteclockwise If A > and C < clockwise Note: polarization state represented b the Jones vector does not change if it is multiplied b a constant. So we can alwas make A >. 5//29 Matri Treatment of Polarization 8 E

9 Usefulness and some properties of the Jones vectors Two properties p of the polarization vector or Jones vector : ) polarization state of a wave does not change if its Jones vector is multiplied b a constant. It onl affects the amplitude. i 2) )polarization state of a wave does not change if its Jones vector is multiplied b a constant phase factor e φ. It promotes phase of each element b φ but not the phase difference Δφ. Eample : illustrating usefulness of the Jones vectors. a) Polarization state of a superposition of two waves can be found b adding the Jones vectors: 2 i + = i Conclusion: We can generate a linearl polarized light with miing equal portions of LCP and RCP light. Eample 2: superposition of horizontall and verticall linearl polarized light: + = Conclusion : b miing equal protions of verticall and horizontall linearl polarized light we can get linearl polarized light at an angle 45. 5//29 Matri Treatment of Polarization 9

10 Summar of the polarization states and their Jones vectors 5//29 Matri Treatment of Polarization

11 Mathematical presentation of polarizers a b We can represent an optical instrument or device b its transfer matri or abcd matri: M = c d There are optical devices that affect (change) state of polarization of the light. Our goal is to represent each of these devices with a transfer matri such that multipling it with the Jones vector of the original light produces the resulting light. iφ iφ a b E e Ee i i c d = or ME = E φ φ E e E e o 5//29 Matri Treatment of Polarization

12 Linnear polarizers Linear polarizer: it selectivel removes virations in a given direction and transmits in perpendicular direction. Partial polarization: : sometimes the process of removing other polarizations is partial and not % efficient. See the figure how the output light is polarized along the transmission ais (TA). Along the TA: a b a( ) + b( ) = c d = c ( ) + d ( ) = a b a() b( ) c d = c() d( ) + = Perpendicular to TA: + = The result for linear polarizer along the ais (verticall) is: Linear polarizer, TA vertical M =. Linear polarizer, TA horizontal M = Eercise: Derive the polarization matices for the linear polarizer at 45 : a b a b and c d = c d Polarization same as the polarizer's TA Polarization perpendicular to the polarizer' sta = Linear polarizer, TA at 45 : M = 2 The most general case of the linear polarizer: 2 cos θ sinθcosθ Linear polarizer, TA at θ: M = 2 sinθcosθ sin θ 5//29 Matri Treatment of Polarization 2

13 Phase retarder The pase retarder introduces a phase difference between the orthogonal polarization components. If the speed of light in each orthogonal direction is different, there would be a cumulative phase difference Δφ between the components as light emerges from the media. Fast ais (FA): the ais along which the speed of light is faster or inde of refraction is lower. Slow ais (SA): the ais along which the speed of light is slower or inde of refraction is higher. Finding the matri for retarder : we want a matri that t will illtransform iφ i( φ+ ε) iφ i( φ+ ε) E e to E e and E e to E e ( ) iφ i φ+ ε iε a b E e E e e = = c d M Phase retarder iφ ( ) i iε φ + ε E e E e e M Quarter wave plate (QWP), a retarder with the net phase difference π /2 i /4 π π π e π ε ε = SA horizontal, let ε =, and ε = M = iπ / e iπ /4 π π π e ε ε = SA vertical, let ε =, and ε =+ M = iπ / e iπ /4 iπ /4 M = e QWP, SA vertical, e QWP, SA horizontal i M = i Half wave plate (HWP): a retarder with the net phase difference ε ε = π iπ /2 iπ /2 M = e QWP, SA vertical, = e QWP, SA horizontal M 5//29 Matri Treatment of Polarization 3

14 Phase rotator The phase rotator rotates the drection of polarization of the linearl polarized light b some angle β. a b cosθ cos( θ + β ) c d sinθ = sin ( θ + β) M cos β sin β M = sin β cos β Phase rotator 5//29 Matri Treatment of Polarization 4

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16 Eample: Production of circularl polarized light b combining a linear polarizer with a QWP iπ /4 iπ /4 e = e i 2 2 i e QWP slow ais horizontal Linearl polarized light at 45 The incident light is divided l i equall between the slow and iπ /4 iπ /4 fast ais and becomes right-circularl polarized = e i 2 2 i QWP slow ais vertical Linearl polarized light at 45 The incident light is divided equall between the slow and fast ais and becomes left-circularl polarized 5//29 Matri Treatment of Polarization 6

17 Eample: Left circularl polarized light is passing through an eighth wave plate Eighth wave plate is a phase retarder that introduces a relative phase difference of 2 π/8 or π/4 between the SA and FA. Assume ε = M iε e ε = = iε ε = π /4 iπ /4 e e Phase retarderl Eighth wave plate iπ /4 iπ /4 i3 π /4 = = e i ie e Eighth wave plate Left-circularl polarized light Ellipticall polarized light 3 π /4 e = + i 2 2 E A 2 2 = wher E B+ ic e E = A=, B= and C =, E = B + C = and 2 2 2E Eocosε tan 2α = α = E E A > and C> so the have the same sign so the ellipticall polarized light has counterclockwise rotation. 5//29 Matri Treatment of Polarization 7

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