Electromagnetic Field Theory: A Problem Solving Approach Part 55

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Sinusoidal Time Variations 515 while the second has its electric field polarized in the y direction. Each solution alone is said to be linearly polarized because the electric field always points in the same direction for all time. If both field solutions are present, the direction of net electric field varies with time. In particular, let us say that the x and y components of electric field at any value of z differ in phase by angle 4: E = Re [Eoi. + E, e'i,] e• ' = Eo cos wti. + E, cos (ot + 4))i, (31) We can eliminate time as a parameter, realizing from (31) that cos wt = EE, sin w (32) (32) cos at cos 4 - EE, = (EJEI,) cos 4)- EE, sin 4 sin 4 and using the identity that sin wt + cos 2 (Ot = 1(E ) I Eo. (EJEn)2 cos 2 4)+ (E/E,) 2 sin 2 (2E.E/EýoE,0 ) cos 4 40 (33) to give us the equation of an ellipse relating E, to E,: (E +( E, 2 2EE, 2 S cos4 = sin2 4 (E,. 0 E E.E, (34) as plotted in Figure 7-11 a. As time increases the electric field vector traces out an ellipse each period so this general case of the superposition of two linear polarizations with arbitrary phase 4 is known as elliptical polarization. There are two important special cases: (a) Linear Polarization If E. and E, are in phase so that (E.E,\ 2 4 = 0, (34) reduces to E, E E,, E•0 = tanO=-• (35) Ex E.K E, E,. The electric field at all times is at a constant angle 0 to the x axis. The electric field amplitude oscillates with time along this line, as in Figure 7-1 lb. Because its direction is always along the same line, the electric field is linearly polarized. (b) Circular Polarization If both components have equal amplitudes but are 90* out of phase, E.o= Eo Eo, 4 = fir/2 516 Eletrodynamics-Fieldsand Waves E=Exoix +Eoe ioi ( Ex)2 + xo fyo -xo Yo0 Figure 7-11 (a)Two perpendicular field components with phase difference 46 have the tip of the net electric field vector tracing out an ellipse each period. (b) If both field components are in phase, the ellipse reduces to a straight line. (c) If the field components have the same magnitude but are 90* out of phase, the ellipse becomes a circle. The polarization is left circularly polarized to z-directed power flow if the electric field rotates clockwise and is (d) right circularly polarized if it rotates counterclockwise. (34) reduces to the equation of a circle: E2 +E2 = E0 (37) The tip of the electric field vector traces out a circle as time evolves over a period, as in Figure 7-11 c. For the upper (+) sign for 4 in (36), the electric field rotates clockwise while the negative sign has the electric field rotating counterclockwise. These cases are, respectively, called left and right circular polarization for waves propagating in the +z direction as found by placing the thumb of either hand in the direction of power flow. The fingers on the left hand curl in the direction of the rotating field for left circular polarization, while the fingers of the right hand curl in the direction of the rotating field for right circular polarization. Left and right circular polarizations reverse for waves traveling in the -z direction. 7-4-7 Wave Propagation in Anisotropic Media Many properties of plane waves have particular applications to optics. Because visible light has a wavelength on the order of 500 nm, even a pencil beam of light 1 mm wide7 is 2000 wavelengths wide and thus approximates a plane wave. 1 517 Sinusoidal Time Variations ,•wt= 0, 2r = Exo coswt = EO coswt (b) - ) 2 Ex 0 Wt = r Left circular polarization Right circular polarization (c) Figure 7-11 (a) Polarizers Light is produced by oscillating molecules whether in a light bulb or by the sun. This natural light is usually unpolarized as each molecule oscillates in time and direction independent of its neighbors so that even though the power flow may be in a single direction the electric field phase changes randomly with time and the source is said to be incoherent. Lasers, an acronym for "light amplification by stimulated emission of radiation," emits coherent light by having all the oscillating molecules emit in time phase. A polarizer will only pass those electric field components aligned with the polarizer's transmission axis so that the transmitted light is linearly polarized. Polarizers are made of such crystals as tourmaline, which exhibit dichroism-the selective absorption of the polarization along a crystal axis. 518 Electrodynamics-Fieldsand Waves The polarization perpendicular to this axis is transmitted. Because tourmaline polarizers are expensive, fragile, and of small size, improved low cost and sturdy sheet polarizers were developed by embedding long needlelike crystals or chainlike molecules in a plastic sheet. The electric field component in the long direction of the molecules or crystals is strongly absorbed while the perpendicular component of the electric field is passed. For an electric field of magnitude Eo at angle 4 to the transmission axis of a polarizer, the magnitude of the transmitted field is E, = Eo cos 4 (38) so that the time-average power flux density is = I Re [i(r) x A*(r)]1 =2 71 cos' 4 (39) which is known as the law of Malus. (b) Double Refraction (Birefringence) If a second polarizer, now called the analyzer, is placed parallel to the first but with its transmission axis at right angles, as in Figure 7-12, no light is transmitted. The combination is called a polariscope. However, if an anisotropic crystal is inserted between the polarizer and analyzer, light is transmitted through the analyzer. In these doubly refracting crystals, light polarized along the optic axis travels at speed c1u while light polarized perpendicular to the axis travels at a slightly different speed c,. The crystal is said to be birefringent. If linearly polarized light is incident at 450 to the axis, E(z = 0, t) = Eo(i, + i,) Re (edw) (40) the components of electric field along and perpendicular to the axis travel at different speeds: E,(z, t) = Eo Re (ei<(ct-hII)), kt = a/c1 E,(z, t) = Eo Re (eijt-h''z), k± = wo/c (41) After exiting the crystal at z = 1,the total electric field is E(z = i, t) = Eo Re [ei'(e-iLi,,+ e-i'i,)] = Eo Re [ei"(-l-')(i, + eihlr-k)'i,)] (42) Sinusoidal Time Variations 519 Crossed polarizer (analyzer) Incident field at = 0. at Elliptically polarized wave L Complex electric field vector rotates clockwise along crystal Doubly refracting (birefringent) medium Linearly polarized wave Waves polarized along this axis travel at speed cq Transmission axis Polarizer Figure 7-12 When a linearly polarized wave passes through a doubly refracting (birefringent) medium at an angle to the crystal axes, the transmitted light is elliptically polarized. which is of the form of (31) for an elliptically polarized wave where the phase difference is S= (kll- kJ)1 = ol 1 cli 1 c-L When 4 is an integer multiple of 27r, the light exiting the crystal is the same as if the crystal were not there so that it is not transmitted through the analyzer. If 45 is an odd integer multiple of 7r, the exiting light is also linearly polarized but perpendicularly to the incident light so that it is polarized in the same direction as the transmission axis of the analyzer, and thus is transmitted. Such elements are called half-wave plates at the frequency of operation. When 4 is an odd integer multiple of ur/2, the exiting light is circularly 5• 0 Electrodynamics-Fiedr s and Waves polarized and the crystal serves as a quarter-wave plate. However, only that polarization of light along the transmission axis of the analyzer is transmitted. Double refraction occurs naturally in many crystals due to their anisotropic molecular structure. Many plastics and glasses that are generally isotropic have induced birefringence when mechanically stressed. When placed within a polariscope the photoelastic stress patterns can be seen. Some liquids, notably nitrobenzene, also become birefringent when stressed by large electric fields. This phenomena is called the Kerr effect. Electro-optical measurements allow electric field mapping in the dielectric between high voltage stressed electrodes, useful in the study of high voltage conduction and breakdown phenomena. The Kerr effect is also used as a light switch in high-speed shutters. A parallel plate capacitor is placed within a polariscope so that in the absence of voltage no light is transmitted. When the voltage is increased the light is transmitted, being a maximum when 4 = w. (See problem 17.) 7-5 NORMAL INCIDENCE ONTO A PERFECT CONDUCTOR A uniform plane wave with x-directed electric field is normally incident upon a perfectly conducting plane at z = 0, as shown in Figure 7-13. The presence of the boundary gives rise to a reflected wave that propagates in the -z direction. There are no fields within the perfect conductor. The known incident fields traveling in the +z direction can be written as Ei(z, t) = Re (Ei eik1t-'i) (1) Hi(z, t)= Re (eitm-2)i(1 while the reflected fields propagating in the -z direction are similarly E,(z, t) = Re (P, ei"c+Ai.) H,(z, t) = Re ( - 7 ei +• i, (2) where in the lossless free space 710 = v4"o/eo, k = ,"ego (3) Note the minus sign difference in the spatial exponential phase factors of (1) and (2) as the waves are traveling in opposite directions. The amplitude of incident and reflected magnetic fields are given by the ratio of electric field amplitude to the wave impedance, as derived in Eq. (15) of Section I Normal Incidence onto a Perfect Conductor eo, AO 521 (170=v/-o) Er = Re(E, e j(r+kX)i. ) t)= (, 0 H, (z, H,, = -2E ?0 ,o coskzcoswt j(ot+kz) H, = Re(-L e -710 Ex(s, m Figure 7-13 A uniform plane wave normally incident upon a perfect conductor has zero electric field at the conducting surface thus requiring a reflected wave. The source of this reflected wave is the surface current at z = 0, which equals the magnetic field there. The total electric and magnetic fields are 900 out of phase in time and space. 7-3-2. The negative sign in front of the reflected magnetic field for the wave in the -z direction arises because the power flow S, = E, x H, in the reflected wave must also be in the -z direction. The total electric and magnetic fields are just the sum of the incident and reflected fields. The only unknown parameter E, can be evaluated from the boundary condition at z =0 where the tangential component of E must be continuous and thus zero along the perfect conductor: Ei+E,== 0 2=> (4) The total fields are then the sum of the incident and reflected fields E.(z, t)= Ei(z, t) + E,(z, t) = Re [Ei(e i- ~ -e+3k ' ) eic] = 2E, sin kz sin wt H,(z, t)= Hi(z, t)+H,(z, t) (e-jkz + e +L') eiw = Re = 2Ei cos kz cos ot - 7lo (5) 522 Electrodynamics-Fiedsand Waves where we take Ai = Ei to be real. The electric and magnetic fields are 90* out of phase with each other both in time and space. We note that the two oppositely traveling wave solutions combined for a standing wave solution. The total solution does not propagate but is a standing sinusoidal solution in space whose amplitude varies sinusoidally in time. A surface current flows on the perfect conductor at z = 0 due to the discontinuity in tangential component of H, 2E, K, = H,(z= O)=-cos t 11o (6) giving rise to a force per unit area on the conductor, F = 2K x oH = p0oH, (z = 0)i = 2eoE? cos 2 Wti, (7) known as the radiation pressure. The factor of 2 arises in (7) because the force on a surface current is proportional to the average value of magnetic field on each side of the interface, here being zero for z = 0+. 7-6 NORMAL INCIDENCE ONTO A DIELECTRIC 7-6-1 Lossless Dielectric We replace the perfect conductor with a lossless dielectric of permittivity e2 and permeability l2, as in Figure 7-14, with a uniform plane wave normally incident from a medium with permittivity el and permeability j1. In addition to the incident and reflected fields for z < 0, there are transmitted fields which propagate in the +z direction within the medium for z > 0: Ei(z, t) = Re [4i ei-A)i,], HE(z, t)= Re[ ei(ut-kI')i, , E,(z, t)= Re [- • ki = W ,L Al= 1 <0 ei"*+'A)i,] 1(1) E,(z, t) =Re [E ej"')i.•, H,(z, t)= Re [E itu*-k+:i], k2 = 0 = ] (12 1 It is necessary in (1) to use the appropriate wavenumber and impedance within each region. There is no wave traveling in the -z direction in the second region as we assume no boundaries or sources for z > 0. 523 Normal Incidence onto a Dielectric Selp E i = Re(Eie e 2 , P2 (2 VEIA E2 C2 i)( ki = kli Et = Re(Et ei H j(e t-kx -i i) j( 8- k2s) is) y I -i)e k, = k2 i 2 = Er = Re( re i (Q'I+kIdi) Hr kr.= -ki is = Re(-- e-lRt H =- Re(-t- e hjt-k2ziy) ) is Figure 7-14 A uniform plane wave normally incident upon a dielectric interface separating two different materials has part of its power reflected and part transmitted. The unknown quantities E, and E, can be found from the boundary conditions of continuity of tangential E and H at z = 0, 1 r2 from which we find the reflection R and transmission T field coefficients as R=-=- E. Ei where from (2) ml+?h 712+711 E, 2712 T= -X= Ei 72+ 2 11 E, 112+111 1+R=T If both mediums have the same wave impedance, II1 = there is no reflected wave. 12, is 524 7-6-2 Electrodynamics-Fieldsand Waves Time-Average Power Flow The time-average power flow in the region z <0 is = 2 Re [ .(z)A (z)] Re [E ,h •- = [1 +-Re 27 '" e+ik' - E e-ihI + ,e+i*'"][E 1 '1 [PA* ee+ 2 P iz- 1z ie-20k (5) a The last term on the right-hand side of (5) is zero as it is the difference between a number and its complex conjugate, which is pure imaginary and equals 2j times its imaginary part. Being pure imaginary, its real part is zero. Thus the time-average power flow just equals the difference in the power flows in the incident and reflected waves as found more generally in Section 7-3-2. The coupling terms between oppositely traveling waves have no time-average yielding the simple superposition of time-average powers: = 2[I •1-il 2-111] = 2n, [1 -R 2] (6) This net time-average power flows into the dielectric medium, as it also equals the transmitted power; 1I = 2712 7-6-3 2 2 I-'= 1I1 T 2n2 1 2 271 (7) [I-R'] Lossy Dielectric If medium 2 is lossy with Ohmic conductivity o, the solu- tions of (3) are still correct if we replace the permittivity the complex permittivity S,. 2=e 1+(8) so that the wave impedance in region 2 is complex: '12 = 'IA/ I ____ 62 by
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