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Normal Incidence onto a Dielectric 525 We can easily explore the effect of losses in the low and large loss limits. (a) Low Losses If the Ohmic conductivity is small, we can neglect it in all terms except in the wavenumber k2 : -2 lim k2-1E (10) 2 /W624C1 82 The imaginary part of k 2 gives rise to a small rate of exponential decay in medium 2 as the wave propagates away from the z = 0 boundary. (b) Large Losses For large conductivities so that the displacement current is negligible in medium 2, the wavenumber and impedance in region 2 are complex: (k= 1-ij lim o,_ ,*'2 Q 2 (11) 1 +1 The fields decay within a characteristic distance equal to the skin depth 8. This is why communications to submerged submarines are difficult. For seawater, ~2 = 0= 41rX 10-7 henry/m and o-4 siemens/m so that for 1 MHz signals, 8-0.25m. However, at 100Hz the skin depth increases to 25 meters. If a submarine is within this distance from the surface, it can receive the signals. However, it is difficult to transmit these low frequencies because of the large 106 m. Note that as the free space wavelength, A-3 conductivity approaches infinity, lim -( 1o12 = 0 - I (12) (T=0 so that the field solution approaches that of normal incidence upon a perfect conductor found in Section 7-5. EXAMPLE 7-1 DIELECTRIC COATING A thin lossless dielectric with permittivity e and permeability M is coated onto the interface between two infinite halfspaces of lossless media with respective properties (E , p1r) and (e6, Ip), as shown in Figure 7-15. What coating parameters e and Ct and thickness d will allow all the time-average power 526 Electrodynamics-Fieldsand Waves = I_-AI E2 Y' "e- E2 d k2 H2 H1 Region 2 Region 1 No reflections if d -, 4 and 7 = n = 1,3,5... V71 1-2 ,where = 2•r is measured within the coating Figure 7-15 A suitable dielectric coating applied on the interface of discontinuity between differing media can eliminate reflections at a given frequency. from region 1 to be transmitted through the coating to region 2? Such coatings are applied to optical components such as lenses to minimize unwanted reflections and to maximize the transmitted light intensity. SOLUTION For all the incident power to be transmitted into region 2, there can be no reflected field in region 1, although we do have oppositely traveling waves in the coating due to the reflection at the second interface. Region 2 only has positively z-directed power flow. The fields in each region are thus of the following form: Region 1 Ex= Re [E 1 ei("-k&)ix], HI=Re Ee -k,>i, , ki = Si Normal Incidence onto a Dielectric 527 Coating E+ = Re [E+eit""-a'i,], k = o/c = H+= Re [E, e' t"+ni,] EH- = Re wE1 1 ei(+')i,] e[E Region 2 E2 = Re [P 2 ei(kt-k)ix], H2 = Re [E2 e••M ICw(62/ = k2 = *92 )i, , Continuity of tangential E and H at z = 0 and z = d requires 1 =t++L-, E, E+-E- ' + - e+ iu = E2 e-isd P+e-i P+ e-"d -- e-ikgd e+i 712 71 Each of these amplitudes in terms of E 1 is then E.= - 1+- ~ = ej'gdE+e- +L~ e+iv ] =12 ei d[t+ e-iud - _e+id] 77 Solving this last relation self-consistently requires that .+e-'( 1- Writing ~+and .. in terms of t (I+I)i e" +L 1 1+ =0 yields I 72)+e2ji1 +1) (1 _j =0 Since this relation is complex, the real and imaginary parts must separately be satisfied. For the imaginary part to be zero requires that the coating thickness d be an integral number of 528 Electrodynamics--Fieldsand Waves quarter wavelengths as measured within the coating, 2kd = nar d = nA/4, n = 1, 2,3,... The real part then requires 1+1+ ) ]1 n even n odd 1 For the upper sign where d is a multiple of half-wavelengths the only solution is P12=-1 (d=nA/4, n=2,4,6,...) which requires that media 1 and 2 be the same so that the coating serves no purpose. If regions 1 and 2 have differing wave impedances, we must use the lower sign where d is an odd integer number of quarter wavelengths so that fq2 ==1= 2 712#1 (d=nA/4, n =1,3,5,...) Thus, if the coating is a quarter wavelength thick as measured within the coating, or any odd integer multiple of this thickness with its wave impedance equal to the geometrical average of the impedances in each adjacent region, all the timeaverage power flow in region 1 passes through the coating into region 2: . 2 . , . 2 712 =2Re (, e-" +- e+·) (*E e +e+'hz( e' -'• 271 Note that for a given coating thickness d, there is no reflection only at select frequencies corresponding to wavelengths d = nA/4, n = 1,3,5,.... For a narrow band of wavelengths about these select wavelengths, reflections are small. The magnetic permeability of coatings and of the glass used in optical components are usually that of free space while the permittivities differ. The permittivity of the coating e is then picked so that and with a thickness corresponding to the central range of the wavelengths of interest (often in the visible). I _ Uniform and Nonuniform Plane Waves 7-7 529 UNIFORM AND NONUNIFORM PLANE WAVES Our analysis thus far has been limited to waves propagating in the z direction normally incident upon plane interfaces. Although our examples had the electric field polarized in the x direction., the solution procedure is the same for the ydirected electric field polarization as both polarizations are parallel to the interfaces of discontinuity. 7-7-1 Propagation at an Arbitrary Angle We now consider a uniform plane wave with power flow at an angle 0 to the z axis, as shown in Figure 7-16. The electric field is assumed to be y directed, but the magnetic field that is perpendicular to both E and S now has components in the x and z directions. The direction of the power flow, which we can call z', is related to the Cartesian coordinates as z'=x sin O+z cos 0 so that the phase factor kz' can be written as kz' = kx + k,z, k, = k sin 0 k, = k cos 0 where the wavenumber magnitude is scos0 0 i,+ sin Oi l E = Re(Ee Figure 7-16 The spatial dependence of a uniforiN plane wave at an arbitrary angle 0 can be expressed in terms of a vector wavenumber k as e-ik' ,where k is in the direction of power flow and has magnitude co/c. 530 Electrodynamics-Fieldsand Waves This allows us to write the fields as E = Re [ ee•t-1kx-h"z•i,] (4) A H= Re [E(-cos Oi, +sin Oi,) ei''-t--."-h.4 We note that the spatial dependence of the fields can be written as e-i'kL ,where the wavenumber is treated as a vector: k = ki + ki +ki, (5) with r = xix +yi, +zi, (6) so that kr= r=+k,y +ky +kz (7) The magnitude of k is as given in (3) and its direction is the same as the power flowS: IAI2 S= ExH = - (cos Oi, +sin ix,) cos2 (wt -k - r) 1EI 2k - cos 2 (wt - k r) - (8) where without loss of generality we picked the phase of f to be zero so that it is real. 7-7-2 The Complex Propagation Constant Let us generalize further by considering fields of the form E = Re [E e"' eH = Re [Hi e' '- e ' ' - ' tkr) ]= Re [E e( ''] = Re [i e(t - k e-"I] r) e-"a] (9) where y is the complex propagation vector and r is the position vector of (6): - = a + jk = y,i, + yi, + y i, (10) Y' r = yx + y,y + y,z We have previously considered uniform plane waves in lossless media where the wavenumber k is pure real and z directed with a =0 so that y is pure imaginary. The parameter a represents the decay rate of the fields even though the medium is lossless. If a is nonzero, the solutions are called nonuniform plane waves. We saw this decay in our quasi-static solutions of Laplace's equation where solutions had oscillations in one direction but decay in the perpendicular direction. We would expect this evanescence to remain at low frequencies. Uniform and Nonunifonm Plane Waves 531 The value of the assumed form of solutions in (9) is that the del (V) operator in Maxwell's equations can be replaced by the vector operator -y: V= -i, ax -+-i, -iz ay az = -Y (11) This is true because any spatial derivatives only operate on the exponential term in (9). Then the source free Maxwell's equations can be written in terms of the complex amplitudes as - x fi =ij•-Y -,tI= (12) 0 The last two relations tell us that y is perpendicular to both E and H. If we take y x the top equation and use the second equation, we have -7 x (y x •) = -jot (y X HI)= -jay (-jweE) -_= ACet (13) The double cross product can be expanded as -- x(,y X f) = -y(y •I) + (,y y)i = (,y ./)j = -_oWCE (14) The y --i term is zero from the third relation in (12). The dispersion relation is then y*y= (ar- k +2j k)= -W9Ie (15) For solution, the real and imaginary parts of (15) must be separately equal: '2 - k2 = --- P_ E (16) at k=0 When a= 0, (16) reduces to the familiar frequencywavenumber relation of Section 7-3-4. The last relation now tells us that evanescence (decay) in space as represented by a is allowed by Maxwell's equations, but must be perpendicular to propagation represented by k. 532 Electrodynamics-Fieldsand Waves We can compute the time-average power flow for fields of the form of (9) using (12) in terms of either E or H as follows: = - Re ( x ii*) - JRe(r (y*) x -'.Re k 2I WE (17) fi*) j ymp *) write the (powerflow to in terms of either E or(17) Re k ( jWR 9 ReH| - 1 H (I_*) `9Re Although both final expressions in (17) are equivalent, it is convenient to write the power flow in terms of either E or H. When E is perpendicular to both the real vectors a and 0, defined in (10) and (16), the dot product y* • E is zero. Such a mode is called transverse electric (TE), and we see in (17) that the time-average power flow is still in the direction of the wavenumber k. Similarly, when H is perpendicular to a and 13, the dot product y H* is zero and we have a transverse magnetic (TM) mode. Again, the time-average power flow in (17) is in the direction of k. The magnitude of k is related to w in (16). Note that our discussion has been limited to lossless systems. We can include Ohmic losses if we replace E by the complex permittivity E of Section 7-4-3 in (15) and (17). Then, there is always decay (a 4 0) because of Ohmic dissipation (see Problem 22). 7-7-3 Nonuniform Plane Waves We can examine nonuniform plane wave solutions with nonzero a by considering a current sheet in the z = 0 plane, which is a traveling wave in the x direction: K,(z = 0) = Ko cos (wt - kx) = Re (Koei"j- _I··I ' =' kx)) Uniform and Nonuniform Plane Waves 533 The x-directed surface current gives rise to a y-directed magnetic field. Because the system does not depend on the y coordinate, solutions are thus of the following form: ' H,=,=Re (H1e Re (H eJ 2 "), z > 0O ), z<0 "' A. (19) -- Re Re e e - 7X Ho i, e ir"e' iie -Y2 , z>O z<0 'W e"'r], where ,yand y2 are the complex propagation vectors on each side of the current sheet: yi = yliix + yiz (20) (0 7Y2 = Y2A x + Y2 The boundary condition of the discontinuity of tangential H at z = 0 equaling the surface current yields -AI e~~,X + A2 " e- ^2- = Ko e-ik (21) which tells us that the x components of the complex propagation vectors equal the trigonometric spatial dependence of the surface current: Y.l = 2x. = jk. (22) The z components ofy'1 and y2 are then determined from (15) as 2 2 21 yx + Y2 =-••2 = + -• -We)1/•2 (23) 2 If k2 W 2e6/, then y, is pure real representing evanescence in the z direction so that we generate nonuniform plane waves. When w = 0, (23) corresponds to Laplacian solutions that oscillate in the x direction but decay in the z direction. The z component of y is of opposite sign in each region, Yl = --Y2 •= +(k -_W2ey)1/2 (24) as the waves propagate or decay away from the sheet. Continuity of the tangential component of E requires YlI = H22 H 2 = -H = Ko/2 (25) If k.= 0, we re-obtain the solution of Section 7-4-1. Increasing k. generates propagating waves with power flow in the ki,+ ksi. directions. At k = w 2 , kz = 0so that the power flow is purely x directed with no spatial dependence on z. Further increasing k. converts k,to a, as the fields decay with z. y, becomes real and 534 7-8 Electrodynamics-Fieldsand Waves OBLIQUE INCIDENCE ONTO A PERFECT CONDUCTOR 7-8-1 E Field Parallel to the Interface In Figure 7-17a we show a uniform plane wave incident upon a perfect conductor with power flow at an angle 0i to the normal. The electric field is parallel to the surface with the magnetic field having both x and z components: Ei = Re [Ei ei('tL'-.x-kz)i,] j( (-cos Oii. +sin Oii.) eit Hi=Re [ - . -k. Z) - where k,i = k sin Oi ki = k cos Oi k=< 0 eA, T1 1 x 0= - H, a=- E, )P~kor OH E, Hj (b) Figure 7-17 A uniform plane wave obliquely incident upon a perfect conductor has its angle of incidence equal to the angle of reflection. (a) Electric field polarized parallel to the interface. (b) Magnetic field parallel to the interface. I

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