Waves and Polarization

Transcription

1 Waves and Polariation.nb Optics 55- James C. Want ( Modified) Waves and Polariation WP - Show that E@r, td = ÅÅÅÅÅ A Cos@kr-wtD is a solution to the wave equation. r 3-D Wave Equation: In polar coordinates: E - ÅÅÅÅÅÅÅ c E ÅÅÅÅÅÅÅÅÅÅÅ t = () E = ÅÅÅÅÅÅ r ÅÅÅÅÅÅÅÅ r i k r ÅÅÅÅÅÅÅÅ E r + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ r Sin@qD ÅÅÅÅÅÅÅÅ q i k Sin@qD ÅÅÅÅÅÅÅÅÅ E q + For spherical waves there is no dependence on q and f so we have ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ Å r Sin@qD E ÅÅÅÅÅÅÅÅÅÅÅÅ f () E@r, td = r r E@r, td Jr N. Thus r (3) i ÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ r ÅÅÅÅÅÅÅÅ E@r, td - ÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ E@r, td = r r k r c t The problem is to show that (4) E@r, td = A Cos@k r ωtd (5) r is a solution to the above equation if certain conditions concerning k, w, and c are satisfied. I will use Mathematica to do the algebra. Let e@r_, t_d = a Cos@k r ωtd; r Then, substituting Eq. (5) into Eq. (4) ields FullSimplifA r r Hr r e@r, tdl c a H c k +ω L Cos@k r t ωd c r t,t e@r, tde

2 Waves and Polariation.nb Optics 55- James C. Want ( Modified) Thus, E[r,t]= A ÅÅÅÅÅ r Cos@kr - wtd is a solution if H c k +ω L=, or k = ÅÅÅÅÅÅ w. c WP - Suppose that a reflecting surface has a complex reflection coefficient r = r o e Âd. incident electric field is given b This means that if a normal E i (x,t)=a cos(kx - wt + f) the reflected wave is given b E r Hx, tl = r o A cos(-kx - wt + f + d) Let a plane wave be incident at an angle to the plane reflecting surface. Determine E i (x,,t) and E r (x,,t). Show that the total field can be written as E tot =A H - r o ) cos[k (xcosq + sinq) - wt + f] + r o A cos[k sinq - wt + f + d/ ] cos[kx cosq - d/ ]. Interpret this result for the following four cases and find the spatial frequenc of the standing wave component: i) q =, r o =, ii) q =, r o, iii) q, r o =, iv) q, r o. θ θ Y X E i = a Cos@k Hx Cos@θD + Sin@θDL ωt +φd; E r =ρ o a Cos@k H x Cos@θD + Sin@θDL ωt +φ+δd;

3 Waves and Polariation.nb Optics 55- James C. Want ( Modified) 3 E tot = E i + E r a Cos@φ t ω+k Hx Cos@θD + Sin@θDLD + a Cos@δ+φ t ω+k H x Cos@θD + Sin@θDLD ρ o While this is the correct answer, it is not in the form we want. If we let a = k Sin[q] - w t + f + d/ and b = k x Cos[q] - d/ then E tot = a Cos@α+βD +ρ o a Cos@α βd a Cos@α+βD + a Cos@α βd ρ o Subtracting and adding (a ρ o Cos@α+βDLields E tot = Factor@a HCos@α+βD ρ o Cos@α+βDLD + TrigExpand@ρ o a H Cos@α βd + Cos@α+βDLD a Cos@α+βDH +ρ o L + a Cos@αD Cos@βD ρ o E tot = E tot ê. 8α k Sin@θD ωt +φ+δê, β k x Cos@θD δê < a Cos@φ t ω+k x Cos@θD + k Sin@θDD H +ρ o L + a CosA δ k x Cos@θDE CosA δ +φ t ω+k Sin@θDE ρ o which is the result we wanted. ü i) q =, r o = E tot ê. 8θ >, ρ o > < a CosAkx δ E CosA δ +φ t ωe Normal standing wave having eros spaced l/. ü ii) q =, r o π E tot ê. 8θ >< a Cos@k x+φ t ωd H +ρ o L + a CosAk x δ E CosA δ +φ t ωe ρ o Same standing wave pattern as in i), but reduced amplitude. Also, have travelling wave in +x direction. ü iii) q π, r o = E tot ê. 8ρ o < a CosA δ k x Cos@θDE CosA δ +φ t ω+k Sin@θDE

4 Waves and Polariation.nb Optics 55- James C. Want ( Modified) 4 We have a standing wave in the x direction. The phase of the standing wave varies as a sinusoid in the direction. ü iv) q π, r o π E tot a Cos@φ t ω+k x Cos@θD + k Sin@θDD H +ρ o L + a CosA δ k x Cos@θDE CosA δ +φ t ω+k Sin@θDE ρ o We have a superposition of case iii) with a plane wave travelling along the incident direction with reduced amplitude. WP - 3 A given argon laser has a cw output of watt. The output beam has a Gaussian amplitude distribution which is truncated at the point where the amplitude falls to /e its axial value. The resulting beam diameter is mm. a) What is the on-axis electric field? b) What would the on-axis electric field be if we had a uniform amplitude distribution across the beam? ( watt total power) ü a) The Ponting vector is given b < S >= ε o ce ; E = E o r êr o where r o = mm. The total power is given b P = < S > da = r o π ε o c E o r êr o r r θ r o r P =πε o ce o êr o r r =πε o ce o J 4 ê r N H L o

5 Waves and Polariation.nb Optics 55- James C. Want ( Modified) 5 Solving for the square of the on-axis electric field ields E o = 4 P πε o cr o H L E o = $%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% P r o πε o c H L = 3 $%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%% π 8.85 H3 8 L H L The on-axis electric field is then E o = volt m volt m ü b) ε o ce o π r = watt E o = $%%%%%%%%%%%%%% ε o c π = 3 $%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 8.85 H3 %%%%%%%%%%%%%% 8 L π volt 3 m = 5485 volt m WP - 4 I have two electric fields E = E a ` eâf e ÂHk-wtL and E = E a ` eâf e ÂHk-wtL where à and à are unit vectors in the direction of oscillation of the electric field. B breaking the electric field into components in the x and directions show that the time average irradiance is proportional to E + E + E E Hà ÿ à L Cos@f -f D. First we need to break the electric field into x and components. Let one electric field be at an angle q with respect to the x-axis and the second be at an angle q with respect to the x-axis. Then. e = i e Cos@θ D φ Hk ωtl ; ke Sin@θ D φ Hk ωtl and e = i e Cos@θ D φ Hk ωtl ; ke Sin@θ D φ Hk ωtl Next we need to add the two x-components and the two -components.

6 Waves and Polariation.nb Optics 55- James C. Want ( Modified) 6 e3 = e + e; MatrixForm@e3D i Hk t ωl+ φ Cos@θ D e + Hk t ωl+ φ Cos@θ D e k Hk t ωl+ φ Sin@θ D e + Hk t ωl+ φ Sin@θ D e The irradiance is obtained b multipling each component b its complex conugate. e4 = ComplexExpand@e3 Conugate@e3DD; MatrixForm@Simplif@e4DD i Cos@θ D e + Cos@θ D Cos@θ D Cos@φ φ D e e + Cos@θ D e k Sin@θ D e + Cos@φ φ D Sin@θ D Sin@θ D e e + Sin@θ D e The total irradiance is given b the sum of the irradiance for the x-component and the irradiance for the -component. e4d 8e + Cos@θ θ D Cos@φ φ D e e + e < Since Cos@q -q D is equal to the dot product of the two unit vectors à and à, we have the desired result. WP - 5 Prove using Jones calculus that a retarder at 45 o between vertical and horiontal quarter-wave plates is equivalent to a rotator. (Orientation of sandwich not important.) First we need to define rotation matrices and the Jones matrices for quarter-wave plates and retarders. ü Quarter-wave plate with fast axis vertical qfavjones = πê4 J N; ü Quarter-wave plate with fast axis horiontal qfahjones = πê4 J N; ü Retarder with fast axis vertical rfavjones@φ_d := φê i k φ

7 Waves and Polariation.nb Optics 55- James C. Want ( Modified) 7 ü Rotation Matrix rotjones@θ_d := J Cos@θD Sin@θD Sin@θD Cos@θD N The retarder must be rotated so it is at 45 degrees. ret45jones = rotjones@ 45 DegreeD.rfavJones@φD.rotJones@45 DegreeD; Put the retarder between the vertical and horiontal quarter-wave plates to show that we end up with a rotation matrix. rotation = FullSimplif@qfahJones.ret45Jones.qfavJonesD; MatrixForm@rotationD i Cos@ φ D Sin@ φ D k Sin@ φ D Cos@ φ D Thus, we have rotation matrix of angle f/. Now we want to rotate the quarter-wave plate, retarder, quarter-wave plate sandwich, to show that it's orientation does not change it's effect. We will rotate it an angle a. MatrixForm@FullSimplif@rotJones@ αd.rotation.rotjones@αddd i Cos@ φ D Sin@ φ D k Sin@ φ D Cos@ φ D Thus, the orientation of the sandwich is not important. WP - 6 Use Jones calculus to show that a half-wave plate converts right handed circularl polaried light into left handed circularl polaried light and the phase of the light can be changed b rotating the half-wave plate. Right handed circular polariation can be represented b rcjones = è!!! J N; A half wave plate with the fast axis horiontal can be written as rfahjones = J N; The rotation matrix can be written as

8 Waves and Polariation.nb Optics 55- James C. Want ( Modified) 8 rotjones@θ_d := J Cos@θD Sin@θD Sin@θD Cos@θD N outputpolariation = rotjones@ θd.rfahjones.rotjones@θd.rcjones; TrigToExp@TrigFactor@outputPolariationDD êê MatrixForm θ i k è!!!! θ è!!!! But this is equal to θ è!!! J N Thus, right handed circular has been changed to left handed circular and the phase is changed at twice the rotation rate of the half-wave plate. WP - 7 Derive a Mueller matrix for a half-wave plate having a vertical fast axis. Utilie our result to convert a right-handed state into a left-handed state. Verif that the same wave plate will convert a left-handed state to a right-handed state. Advancing or retarding the relative phase b p should have the same effect. Check this b deriving the matrix for a horiontal fast axis half-wave plate. ü Mueller matrix for a half-wave plate having a vertical fast axis The Mueller matrix for a quarter-wave plate having the fast axis vertical is i qfavmueller = ; k Thus, the Mueller matrix for a half-wave plate having the fast axis vertical is hfavmueller = qfavmueller. qfavmueller; hfavmueller êê MatrixForm i k The Stokes vectors for right and left handed circular polariation are given b

9 Waves and Polariation.nb Optics 55- James C. Want ( Modified) 9 i i rcstokes = ; lcstokes = ; k k Convert right to left hfavmueller. rcstokes êê MatrixForm i k Convert left to right hfavmueller. lcstokes êê MatrixForm i k ü Mueller matrix for a half-wave plate having a horiontal fast axis The Mueller matrix for a quarter-wave plate having the fast axis horiontal is i qfahmueller = ; k Thus, the Mueller matrix for a half-wave plate having the fast axis horiontal is hfahmueller = qfahmueller. qfahmueller; hfahmueller êê MatrixForm i k Thus the Mueller matrix for a half-wave plate having the fast axis horiontal is the same as the Mueller matrix for a half-wave plate having the fast axis vertical. WP - 8 a) Assume I want to rotate the polariation of a plane polaried wavefront 9 o b using onl "perfect" linear polariers.what is the maximum flux transmission I can obtain using onl two polariers? What is the minimum number of perfect polariers required to have a flux transmission greater than 95%? b) The indices of refraction for quart for the sodium ellow line are n o =.544 and n e =.553. Calculate the thickness of a quarter-wave plate made from quart.

10 Waves and Polariation.nb Optics 55- James C. Want ( Modified) ü a) From the Malus Law trans@n_d := CosA 9 n If there are two polariers, trans@d = 4 E n Determining n for a flux transmission greater than 95% For@n =, trans@nd.95, n++d; Print@n, " ", N@trans@nDDD Thus, we need 49 perfect polariers. ü b) SolveAH L t == 88t 6.36 µm<<.589 µm,te 4 Can also have a thickness of (4 m+) 6.36 mm, where m is an integer since (4) 6.36 mm is a wave plate. WP - 9 Use Jones calculus to show that two half-wave plates at angle q between them are equivalent to a rotator through angle q. The matrix for a retarder with the fast axis vertical is rfavjones@φ_d := φê i k φ The rotation matrix is given b

11 Waves and Polariation.nb Optics 55- James C. Want ( Modified) rotjones@θ_d := J Cos@θD Sin@θD Sin@θD Cos@θD N The matrix for two half-wave plates at an angle q between them is given b MatrixForm@TrigReduce@rotJones@ θd.rfavjones@πd.rotjones@θd.rfavjones@πddd Cos@ θd Sin@ θd J Sin@ θd Cos@ θd N Since this is minus the rotation matrix for a rotation of - q we have the result we want. WP - Let E x = A x cos(k-wt) and E = A cos(k-wt+f). Show that J E x N + J E N J E x N J E N Cos@φD = Sin@φD A x A A x A e x = a x Cos@k ωtd; e = a Cos@k ωt +φd; FullSimplifAJ e x N + i e a x k Sin@φD Therefore, a J e x N i e Cos@φDE a x k a J e x N + J e N J e x N J e N Cos@φD = Sin@φD a x a a x a WP - Show that in terms of the Stokes parameters the degree of polariation can be written as V = Hs + s + s 3 L ê s o Degree of Polariation = V = Polaried Light Total Amount of Light Let the total amount of light be s o For the unpolaried portion of the light

12 Waves and Polariation.nb Optics 55- James C. Want ( Modified) i k s o è!!!!!!!!!!!!!!!!!!!!!!!!!!!!! s + s + s 3 For the polaried portion of the light i k è!!!!!!!!!!!!!!!!!!!!!!!!!!!!! s + s + s 3 s s s 3 Therefore, V = è!!!!!!!!!!!!!!!!!!!!!!!!!!!!! s + s + s 3 s o WP - A linearl polaried laser source is used with a Twman-Green interferometer. A 5-5 non-polariation sensitive beamsplitter is used in the interferometer. The flat mirror in the reference arm of the interferometer is tilted slightl to give straight equi-spaced fringes in the output. A quarter-wave plate is placed in the test arm of the interferometer. The fast axis of the quarter-wave plate makes an angle q with respect to the direction of polariation of the light incident upon the quarter-wave plate. A perfect polarier is placed in the output of the interferometer. The transmission axis of the polarier is in the direction of polariation of the light coming from the reference arm. a) Calculate the irradiance of the interference pattern as a function of q. What is the fringe visibilit as a function of q? b) Assume the polarier is non-perfect in that it transmits 95% of the light having a polariation along the transmission axis of the polarier and 5% of the light having the orthogonal polariation. What is the observed fringe visibilit as a function of q? Since the light goes through the quarter-wave plate twice it will act as a half-wave plate and it will rotate the direction of polariation b an angle q. ü a) The irradiance of the interference pattern is given b i = i + i + è!!!!!!!!!! i i Cos@φD, where i = i Cos@ θd

13 Waves and Polariation.nb Optics 55- James C. Want ( Modified) 3 i = i H + Cos@ θd + Cos@ θd Cos@φDL visibilit = i max i min = i max + i min Cos@ θd + Cos@ θd ü b) i = i HH + Cos@ θd + Cos@ θd Cos@φDL H.95L + Sin@ θd H.5LL.95 H Cos@ θdl visibilit =.95 H + Cos@ θd L + H.5L Sin@ θd WP - 3 a) What are the four filters we associate with the four Stokes parameters? b) Suppose that an ideal polarier is rotated at a rate w between a similar pair of stationar crossed polariers. Give the ratio of the modulation frequenc of the emergent flux densit to the rotation frequenc of the polarier. c) I am working on a problem involving partiall polaried light. Which should I use, Jones or Mueller matrices? Explain (briefl). d) I am working on a problem using completel polaried light where I want to keep track of the phase of the beam. Should I use Jones or Mueller matrices? Explain (briefl). ü a) All filters transmit 5% of natural radiation. i) isotropic filter ii) linear horiontal polariation filter i) linear polariation at 45 filter i) right-handed circular polariation filter ü b) As the polarier rotates 36 no light will be transmitted at 4 orientations. Therefore, the ratio is 4.

14 Waves and Polariation.nb Optics 55- James C. Want ( Modified) 4 ü c) Mueller matrices should be used since Jones matrices can be used onl with completel polaried light. ü d) Jones matrices should be used since the keep track of the phase. WP - 4 A linearl polaried source is used with a Young's two-pinhole interferometer. The same amount of light is transmitted through each pinhole. A half-wave plate is placed over one pinhole. The fast axis of the half-wave plate makes an angle q with respect to the direction of polariation of the light incident upon the half-wave plate. A perfect polarier is placed in the output of the interferometer. The transmission axis of the polarier is in the direction of polariation of the light coming from the pinhole without the half-wave plate. a) What is the fringe visibilit as a function of q? b) Assume the polarier is non-perfect in that it transmits 95% of the light having a polariation along the transmission axis of the polarier and 5% of the light having the orthogonal polariation. What is the observed fringe visibilit as a function of q? The irradiance of the two-beam interference pattern is given b i = i + i + è!!!!!!!!!! i i Cos@φD; The half-wave plate rotates the direction of polariation b an angle q. The irradiance of the beam passing through the half-wave plate and the polarier is given b i = i Cos@ θd ; The irradiance of the two-beam interference pattern is then given b i = i H + Cos@ θd + Cos@ θd Cos@φDL ü a) The visibilit is given b v = i max i min = i max + i min 4 Cos@ θd H + Cos@ θd L = Cos@ θd + Cos@ θd

15 Waves and Polariation.nb Optics 55- James C. Want ( Modified) 5 ü b) The irradiance of the main interference fringe pattern is reduced b a factor of.95 and we have to add in.5 times the irradiance of the orthogonall polaried beam. i = i HH + Cos@ θd + Cos@ θd Cos@φDL.95 + Sin@ θd H.5LL v = i max i min = i max + i min.95 H Cos@ θdl.95 H + Cos@ θd L + H.5L Sin@ θd WP - 5 glass? a) Using less than 5 words define polariation angle. b) Using less than 5 words define critical angle. c) What is the approximate power reflectance at normal incidence for common glass? d) What is the approximate power reflectance at a nearl 9 degree angle of incidence for common ü a) Angle of incidence for which onl the polariation component polaried normal to the incident plane, and therefore parallel to the surface, is reflected ü b) When light passes from a medium of lower index to a medium of higher index the angle of refraction is alwas less than the angle of incidence. When the incident ras approach an angle of 9 the refracted ras approach a fixed angle called the critical angle. ü c) 4% ü d) %

16 Waves and Polariation.nb Optics 55- James C. Want ( Modified) 6 WP - 6 A quart plate in the setup shown below has its axes at 45 degrees relative to the transmission axis of the linear polarier. The indices of refraction for quart are n o =.544 and n e =.553. The laser wavelength is 633 nm. a) What is the minimum quart thickness for having no light reflected back to the laser? b) What is the minimum quart thickness (other than ero) for having the maximum amount of light reflected back to the laser? Laser Polarier Quart ü a) The quart plate needs to rotate the direction of polariation b 9 degrees. Thus, we want the quart plate to act as a half-wave plate. Since the light is going through the quart plate twice the quart plate must be a quarter-wave plate. SolveAH L t == 88t µm<<.633 µm,te 4 ü b) We want the quart plate to act as a wave-plate. Since the light is going through the quart plate twice the quart plate must be a half-wave plate. SolveAH L t == 88t µm<<.633 µm,te

17 Waves and Polariation.nb Optics 55- James C. Want ( Modified) 7 WP - 7 A 633 nm wavelength linearl polaried source is used with a Young s two-pinhole interferometer. The same amount of light is transmitted through each pinhole. A 7.58 micron thick quart plate is placed over one pinhole where the axes of the quart plate are at 45 degrees relative to the direction of polariation of the incident light. The indices of refraction for quart are no=.544 and ne=.553. If when the quart plate is removed the fringe contrast is unit, what is the fringe contrast with the quart plate in place? Calculation of retardation of quart plate in units of the wavelength n t λ = H L 7.58 µm.633 µm = 4 Therefore, one beam has circular polariation. The circularl polaried beam can be thought of as two linearl polaried beams with the amplitude of each beam is ÅÅÅÅÅÅÅÅ that of the incident beam. e = e i o Hω t+φl + è!!! k Hω tl î + i = i i o + k + è!!! Cos@φD + io e o contrast = i max i è!!!! min = = i max + i min 4 è!!! è!!! Hω tl ĵ è!!!