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350 Chapter 8:Reflection, Transmission, and Waveguides Lessons #50 and 51 Chapter Section:8-1 Topics:Normal incidence Highlights: Analogy to transmission line Reflection and transmission coefficient Special Illustrations: Example 8-1 CD-ROM Modules 8.1-8.5 CD-ROM Demos8.2 351Lesson #52 Chapter Section:8-2 Topics:Snells laws Highlights: Reflection and refraction Index of refraction Special Illustrations: Example 8-4 Technology Brief on Lasers (CD-ROM) LasersLasers are used in CD and DVD players, bar-code readers, eye surgery and multitudes of other systems and applications. A laseracronym for light amplification by stimulated emission of radiationis a source of monochromatic (single wavelength), coherent (uniform wavefront), narrow-beam light, in contrast with other sources of light (such as the sun or a light bulb) which usually encompass waves of many different wavelengths with random phase (incoherent). A laser source generating microwaves is called a maser. The first maser was built in 1953 by Charles Townes and the first laser was constructed in 1960 by Theodore Maiman. 352 Lesson #53 Chapter Section:8-3 Topics:Fiber optics Highlights: Structure of an optical fiber Dispersion Special Illustrations: Example 8-5 Technology Brief on Bar-Code Reader (CD-ROM) Bar Code ReadersA bar code consists of a sequence of parallel bars of certain widths, usually printed in black against a white background, configured to represent a particular binary code of information about a product and its manufacturer.Laser scanners can read the code and transfer the information to a computer, a cash register, or a display screen. For both stationary scanners built into checkout counters at grocery stores and handheld units that can be pointed at the bar-coded object like a gun, the basic operation of a bar-code reader is the same. 353Lessons #54 and 55 Chapter Section:8-4 Topics:Oblique incidence Highlights: Parallel and perpendicular polarizations Brewster angle Total internal reflection Special Illustrations: Example 8-6 and 8-7 CD-ROM Demos 8.4-8.6 354 Lesson #56 Chapter Section:8-5 Topics:Reflectivity and transmissivity Highlights: Power relations Special Illustrations: Example 8-7 355Lessons #5759 Chapter Section:8-6 to 8-10 Topics:Waveguides Highlights: TE and TM modes Cutoff frequency Phase and group velocities Special Illustrations: Examples 8-8, 8-9, and 8-10 Lesson #60 Chapter Section:8-11 Topics:Cavity Resonators Highlights: Resonant frequency Q factor Applications 356 CHAPTER 8Chapter 8Section 8-1: Reection and Transmission at Normal IncidenceProblem 8.1 Aplanewaveinairwithanelectriceldamplitudeof20V/m isincident normally upon the surface of a lossless, nonmagnetic medium with r = 25.Determine:(a) the reection and transmission coefcients,(b) the standing-wave ratio in the air medium, and(c) the average power densities of the incident, reected, and transmitted waves.Solution:(a)1 = 0 = 120 (), 2 =0r= 1205= 24 ().From Eqs. (8.8a) and (8.9), = 212 +1= 2412024+120= 96144=0.67, = 1+ = 10.67 = 0.33.(b)S = 1+||1||= 1+0.6710.67= 5.(c) According to Eqs. (8.19) and (8.20),Siav = |Ei0|220=4002120= 0.52 W/m2,Srav =||2Siav = (0.67)20.52 = 0.24 W/m2,Stav =||2|Ei0|222=||212Siav = (0.33)212024 0.52 = 0.28 W/m2.Problem8.2 Aplanewavetravelinginmedium1withr1= 2.25isnormallyincident upon medium 2 with r2 = 4. Both media are made of nonmagnetic, non-conducting materials. If the electric eld of the incident wave is given byEi= y8cos(6109t 30x) (V/m),(a) obtain time-domain expressions for the electric and magnetic elds in each ofthe two media, andCHAPTER 8 357(b) determine the average power densities of the incident, reected and transmittedwaves.Solution:(a)Ei= y8cos(6109t 30x) (V/m),1 =0r1=02.25=01.5= 3771.5= 251.33 ,2 =0r2=04= 3772= 188.5 , = 212 +1= 1/21/1.51/2+1/1.5=0.143, = 1+ = 10.143 = 0.857,Er= Ei=1.14 ycos(6109t +30x) (V/m).Note that the coefcient of x is positive, denoting the fact that Erbelongs to a wavetraveling in x-direction.E1 = Ei+Er= y[8cos(6109t 30x) 1.14cos(6109t +30x)] (A/m),Hi= z 81cos(6109t 30x) = z31.83cos(6109t 30x) (mA/m),Hr= z1.141cos(6109t +30x) = z4.54cos(6109t +30x) (mA/m),H1 = Hi+Hr= z[31.83cos(6109t 30x) +4.54cos(6109t +30x)] (mA/m).Since k1 =1and k2 =2,k2 =

21k1 =

42.25 30 = 40 (rad/m),E2 = Et= y8cos(6109t 40x) = y6.86cos(6109t 40x) (V/m),H2 = Ht= z 82cos(6109t 40x) = z36.38cos(6109t 40x) (mA/m).(b)Siav = x8221=642251.33= x127.3 (mW/m2),Srav =||2Siav = x(0.143)20.127 = x2.6 (mW/m2),358 CHAPTER 8Stav = |Et0|222= x2 (8)222= x (0.86)2642188.5= x124.7 (mW/m2).Within calculation error, Siav +Srav = Stav.Problem 8.3 A plane wave traveling in a medium with r1 = 9 is normally incidentupon a secondmedium with r2= 4. Both media are made of nonmagnetic, non-conducting materials. If the magnetic eld of the incident plane wave is given byHi= z2cos(2109t ky) (A/m),(a) obtain time domain expressions for the electric and magnetic elds in each ofthe two media, and(b) determine the average power densities of the incident, reected and transmittedwaves.Solution:(a) In medium 1,up =cr1= 31089= 1108(m/s),k1 =up= 21091108= 20 (rad/m),Hi= z2cos(2109t 20y) (A/m),1 =0r1= 3773= 125.67 ,2 =0r2= 3772= 188.5 ,Ei= x21cos(2109t 20y)= x251.34cos(2109t 20y) (V/m), = 212 +1= 188.5125.67188.5+125.67= 0.2, = 1+ = 1.2,Er= x251.340.2cos(2109t +20y)= x50.27cos(2109t +20y) (V/m),CHAPTER 8 359Hr= z 50.271cos(2109t +20y)= z0.4cos(2109t +20y) (A/m),E1 = Ei+Er= x[25.134cos(2109t 20y) +50.27cos(2109t +20y)] (V/m),H1 = Hi+Hr= z[2cos(2109t 20y) 0.4cos(2109t +20y)] (A/m).In medium 2,k2 =

21k1 =

4920 = 403(rad/m),E2 = Et= x251.34cos

2109t 40y3

= x301.61cos

2109t 40y3

(V/m),H2 = Ht= z 301.612cos

2109t 40y3

= z1.6cos

2109t 40y3

(A/m).(b)Siav = y|E0|221= y(251.34)22125.67= y251.34 (W/m2),Srav = y||2(251.34) = y10.05 (W/m2),Stav = y(251.3410.05) = y241.29 (W/m2).Problem 8.4 A 200-MHz left-hand circularly polarized plane wave with an electriceld modulus of 5 V/m is normally incidentin air upon a dielectricmedium withr = 4 and occupying the region dened by z 0.(a) Write an expression for the electric eld phasor of the incident wave, given thatthe eld is a positive maximum at z = 0 and t = 0.(b) Calculate the reection and transmission coefcients.(c) Writeexpressions for theelectriceldphasors of thereectedwave, thetransmitted wave, and the total eld in the region z 0.360 CHAPTER 8(d) Determinethepercentages of theincident averagepower reectedbytheboundary and transmitted into the second medium.Solution:(a)k1 = c= 221083108= 43rad/m,k2 =up2= cr2 = 434 = 83rad/m.LHC wave:Ei= a0( x+ yej/2)ejkz= a0( x+ j y)ejkz,Ei(z, t) = xa0cos(t kz) ya0sin(t kz),|Ei| = [a20cos2(t kz) +a20sin2(t kz)]1/2= a0 = 5 (V/m).Hence,Ei= 5( x+ j y)ej4z/3(V/m).(b)1 = 0 = 120 (), 2 =0r= 02= 60 ().Equations (8.8a) and (8.9) give = 212 +1= 6012060+120= 60180=13 , = 1+ = 23 .(c)Er= 5( x+ j y)ejk1z=53( x+ j y)ej4z/3(V/m),Et= 5( x+ j y)ejk2z= 103( x+ j y)ej8z/3(V/m),E1 = Ei+Er= 5( x+ j y)ej4z/313ej4z/3

(V/m).(d)% of reected power = 100||2= 1009= 11.11%,% of transmitted power = 100||212= 100

23

212060= 88.89%.CHAPTER 8 361Problem 8.5 Repeat Problem 8.4 after replacing the dielectric medium with a poorconductor characterized by r = 2.25, r = 1, and = 104S/m.Solution:(a) Medium 1:1 = 0 = 120 (), k1 = c= 221083108= 43(rad/m).Medium 2:22=10436221082.25109= 4103.Hence, medium 2 is a low-loss dielectric. From Table 7-1,2 = 22

22= 22120r2= 22 1202.25= 10421201.5= 1.26102(NP/m),2 = 22 = r2c= 2 (rad/m),2 =

22

1+j222

= 120r2

1+ j2103

1201.5= 80 ().LHC wave:Ei= a0( x+ j y)ejk1z,|Ei| = a0 = 5 (V/m),Ei= 5( x+ j y)ej4z/3(V/m).(b) According to Eqs. (8.8a) and (8.9), = 212 +1= 8012080+120=0.2, = 1+ = 10.2 = 0.8.(c)Er= 5( x+ j y)ejk1z=( x+ j y)ej4z/3(V/m),Et= 5( x+ j y)e2zejzz= 4( x+ j y)e1.26102zej2z(V/m),E1 = Ei+Er= 5( x+ j y)[ej4z/30.2ej4z/3] (V/m).362 CHAPTER 8(d)% of reected power = 100||2= 100(0.2)2= 4%,% of transmitted power = 100||212= 100(0.8)212080= 96%.Problem 8.6 A50-MHzplanewavewithelectriceldamplitudeof50V/misnormally incident in air onto a semi-innite, perfect dielectric medium with r = 36.Determine (a) , (b) the average power densities of the incident and reected waves,and (c) the distance in the air medium from the boundary to the nearest minimum ofthe electric eld intensity, |E|.Solution:(a)1 = 0 = 120 (), 2 =

22= 120r2= 1206= 20 (), = 212 +1= 2012020+120=0.71.Hence, || = 0.71 and = 180.(b)Siav = |Ei0|221=(50)22120= 3.32 (W/m2),Srav =||2Siav = (0.71)23.32 = 1.67 (W/m2).(c) In medium 1 (air),1 =cf= 31085107= 6 m.From Eqs. (8.16) and (8.17),lmax = r14= 64= 1.5 m,lmin = lmax14= 1.51.5 = 0 m (at the boundary).Problem 8.7 What is the maximum amplitude of the total electric eld in the airmediumofProblem8.6, andat what nearest distancefromtheboundarydoesitoccur?CHAPTER 8 363Solution: From Problem 8.6, =0.71 and = 6 m.|E1|max = (1+||)Ei0 = (1+0.71) 50 = 85.5 V/m,lmax = r14= 64= 1.5 m.Problem8.8 Repeat Problem8.6after replacingthedielectricmediumwithaconductor with r = 1, r = 1, and = 2.78103S/m.Solution:(a) Medium 1:1 =0 = 120 = 377 (), 1 =cf= 31085107= 6 m,Medium 2:22=2.781033625107109= 1.Hence, Medium 2 is a quasi-conductor. From Eq. (7.70),2 =

22

1 j

2

2

1/2= 120

1 j 22

1/2= 120(1 j1)1/2= 120(2)1/2ej22.5= (292.88+ j121.31) (). = 212 +1=(292.88+ j121.31) 377(292.88+ j121.31) +377=0.09+ j0.12 = 0.22114.5.(b)Siav = |Ei0|221=5022120= 3.32 (W/m2),|Srav| =||2Siav = (0.22)2(3.32) = 0.16 (W/m2).(c) In medium 1 (air),1 =cf= 31085107= 6 m.For r = 114.5 = 2 rad, Eqs. (8.16) and (8.17) givelmax = r14+ (0)12= 2(6)4+0 = 3 m,364 CHAPTER 8lmin = lmax14= 364= 31.5 = 1.5 m.Problem8.9 The three regions shown in Fig. 8-32 (P8.9) contain perfectdielectrics. For a wave in medium 1 incident normally upon the boundary at z =d,what combination of r2 and d produce no reection? Express your answers in termsof r1, r3 and the oscillation frequency of the wave, f .Medium 2r2Medium 3r3Medium 1r1z = -d z = 0zdFigure P8.9: Three dielectric regions.Solution: By analogy with the transmission-line case, there will be no reection atz =d if medium 2 acts as a quarter-wave transformer, which requires thatd = 24and2 =13.The second condition may be rewritten as0r2=0r10r3

1/2, or r2 =r1r3 ,2 =0r2=cfr2=cf (r1r3)1/4 ,andd =c4f (r1r3)1/4 .CHAPTER 8 365Problem 8.10 For the conguration shown in Fig. 8-32 (P8.9), use transmission-line equations (or the Smith chart) to calculatethe input impedance at z = dforr1= 1, r2= 9, r3= 4, d = 1.2 m, andf = 50 MHz. Also determine the fractionof the incident average power density reected by the structure. Assume all mediaare lossless and nonmagnetic.Solution: In medium 2, =0r2=cfr2=310851073= 2 m.Hence,2 = 22= rad/m, 2d = 1.2 rad.At z = d, the input impedance of a transmissionline with load impedance ZLisgiven by Eq. (2.63) asZin(d) = Z0

ZL + jZ0tan2dZ0 + jZLtan2d

.In the presentcase, Z0= 2= 0/r2= 0/3 and ZL= 3= 0/r3= 0/2,where 0 = 120 (). Hence,Zin(d) = 2

3 + j2tan2d2 + j3tan2d

= 03

12 + j

13

tan1.213 + j

12

tan1.2

= 0(0.35 j0.14).At z =d, = ZinZ1Zin +Z1= 0(0.35 j0.14) 00(0.35 j0.14) +0= 0.49ej162.14.Fraction of incident power reected by the structure is ||2=|0.49|2= 0.24.Problem 8.11 Repeat Problem 8.10 after interchanging r1 and r3.Solution: In medium 2, =0r2=cfr2=310851073= 2 m.366 CHAPTER 8Hence,2 = 22= rad/m, 2d = 1.2 rad.At z =d, the input impedance of a transmission line with impedance ZL is given asEq. (2.63),Zin(d) = Z0

ZL + jZ0tandZ0 + jZLtan2d

.In the present case, Z0 = 2 = 0/r2= 0/3, ZL = 3 = 0/r1= 0, where0 = 120 (). Hence,Zin(d) = 2

3 + j2tan1.22 + j3tan1.2

= 03

1+( j/3)tan1.2(1/3) + j tan1.2

= 0

1+( j/3)tan1.21+ j3tan1.2

= (0.266 j0.337)0 = 0.43051.7.At z =d, = ZinZ1Zin +Z1=0.4351.7 120.4351.7+ 12= 0.49101.1 .Fraction of incident power reected by structure is ||2= 0.24.Problem 8.12 Orange light of wavelength0.61 m in air enters a block of glasswith r = 1.44. What color would it appear to a sensor embedded in the glass?Thewavelengthrangesof colorsare violet(0.39to 0.45 m), blue(0.45 to 0.49m),green (0.49 to 0.58 m), yellow (0.58 to 0.60 m), orange (0.60 to 0.62 m), and red(0.62 to 0.78 m).Solution: In the glass, =0r=0.611.44= 0.508 m.The light would appear green.Problem 8.13 A plane wave of unknown frequency is normally incident in air uponthe surface of a perfect conductor. Using an electric-eld meter, it was determinedthat thetotal electriceldintheairmediumisalwayszerowhenmeasuredat aCHAPTER 8 367distance of 2 m from the conductor surface. Moreover, no such nulls were observedat distances closer to the conductor. What is the frequency of the incident wave?Solution: The electric eld of the standing wave is zero at the conductor surface,and the standing wave pattern repeats itself every /2. Hence,2= 2 m, or = 4 m,in which casef=c= 31084= 7.5107= 75 MHz.Problem 8.14 Consider a thin lm of soap in air under illumination by yellow lightwith = 0.6 m in vacuum. If the lm is treated as a planardielectricslab withr = 1.72, surrounded on both sides by air, what lm thickness would produce strongreection of the yellow light at normal incidence?Solution:The transmission line analogue of the soap-bubble wave problem is shownin Fig. P8.14(b) where the load ZL is equal to 0, the impedance of the air mediumon the other side of the bubble. That is,0 = 377 , 1 =3771.72= 287.5 .The normalized load impedance iszL = 01= 1.31.Forthereectionbythesoapbubbletobethelargest, Zinneedstobethemostdifferent from0. This happens when zL is transformed through a length /4. Hence,L = 4=04r=0.6 m41.72= 0.115 m,where is the wavelength of the soap bubble material. Strong reections will alsooccur if the thickness is greater than L by integer multiples of n/2 = (0.23n) m.Hence, in generalL = (0.115+0.23n) m, n = 0, 1, 2, . . . .According to Section 2-7.5, transforming a load ZL = 377 through a /4 sectionof Z0 = 287.5 ends up presenting an input impedance ofZin = Z20ZL=(287.5)2377= 219.25 .368 CHAPTER 8Yellow Light = 0.6 mAir Soap Airr1=1 r3=1 r2=1.72L(b) Transmission-line equivalent circuit0 = 3772 ZL = 0 = 377 (a) Yellow light incident on soap bubble.Figure P8.14: Diagrams for Problem 8.14.ThisZinisattheinputsideofthesoapbubble. The reectioncoefcient atthatinterface is = Zin0Zin +0= 219.25377219.25+377=0.27.Any other thickness would produce a reection coeffcient with a smaller magnitude.Problem 8.15 A 5-MHz plane wave with electric eld amplitude of 10 (V/m) isnormally incident in air onto the plane surface of a semi-innite conducting materialwith r= 4, r= 1, and = 100 (S/m). Determine the average power dissipated(lost) per unit cross-sectional area in a 2-mm penetration of the conducting medium.Solution: For convenience, let us choose Eito be along x and the incident directionto be + z. Withk1 = c= 251063108=30(rad/m),CHAPTER 8 369we haveEi= x10cos

107t 30 z

(V/m),1 = 0 = 377 .From Table 7-1,

=r0=100361074109= 9104,which makes the material a good conductor, for which2 =

f =

51064107100 = 44.43 (Np/m),2 = 44.43 (rad/m),c2 = (1+ j) 2= (1+ j) 44.43100= 0.44(1+ j) .According to the expression for Sav2 given in the answer to Exercise 8.3,Sav2 = z||2|Ei0|22e22zRe

1c2

.The power lost is equal to the difference between Sav2 at z = 0 and Sav2 at z = 2 mm.Thus,P

= power lost per unit cross-sectional area= Sav2(0) Sav2(z = 2 mm)=||2|Ei0|22Re

1c2

[1e22z1]where z1 = 2 mm. = 1+= 1+ 212 +1= 1+ 0.44(1+ j) 3770.44(1+ j) +377 0.0023(1+ j) = 3.3103ej45.Re

1c2

= Re

10.44(1+ j)

= Re

10.44(1 j)

= Re

1+ j0.442

=10.88= 1.14,P

= (3.3103)210221.14[1e244.432103] = 1.01104(W/m2).370 CHAPTER 8Problem 8.16 A0.5-MHzantennacarriedbyanairplaneyingovertheoceansurface generates a wave that approaches the water surface in the form of a normallyincidentplane wave with an electric-eldamplitude of 3,000 (V/m). Sea water ischaracterized by r =72, r =1, and =4 (S/m). The plane is trying to communicatea message to a submarine submerged at a depth dbelow the water surface. If thesubmarines receiver requires a minimum signal amplitude of 0.01 (V/m), what isthe maximum depth d to which successful communication is still possible?Solution: For sea water at 0.5 MHz,

==43620.510672109= 2000.Hence, sea water is a good conductor, in which case we use the following expressionsfrom Table 7-1:2 =

f =

0.510641074 = 2.81 (Np/m),2 = 2.81 (rad/m),c2= (1+ j) 2= (1+ j) 2.814= 0.7(1+ j) , = 212 +1= 0.7(1+ j) 3770.7(1+ j) +377= (0.9963+ j3.7103), = 1+ = 5.24103ej44.89,|Et| =|Ei0e2d|.We need to nd the depth z at which |Et| = 0.01 V/m = 108V/m.108= 5.241033103e2.81d,e2.81d= 6.361010,2.81d = ln(6.361010) =21.18,ord = 7.54 (m).CHAPTER 8 371Sections 8-2 and 8-3: Snells Laws and Fiber OpticsProblem 8.17 A lightray is incidenton a prism at an angle as shown in Fig.8-33 (P8.17). The ray is refracted at the rst surface and again at the second surface.In terms of the apex angle of the prism and its index of refractionn, determinethe smallest value of for which the ray will emerge from the other side. Find thisminimum for n = 4 and = 60.3nABC2Figure P8.17: Prism of Problem 8.17.Solution: For the beam to emerge at the second boundary, it is necessary that3 < c,where sinc = 1/n. From the geometry of triangle ABC,180 = +(902) +(903),or 2 = 3. At the rst boundary, sin = nsin2. Hence,sinmin = nsin(3) = nsin

sin1

1n

,ormin = sin1nsin

sin1

1n

.For n = 4 and = 60,min = sin14sin(60sin1

14

= 20.4.372 CHAPTER 8Problem8.18 For some types of glass, the index of refraction varies withwavelength. A prism made of a material withn = 1.71430 0, (0 in m),where 0 is the wavelength in vacuum, was used to disperse white light as shown inFig. 8-34 (P8.18). The white light is incident at an angle of 50, the wavelength 0 ofred light is 0.7 m and that of violet light is 0.4 m. Determine the angular dispersionin degrees.5060RedGreenVioletAngular dispersionABC432Figure P8.18: Prism of Problem 8.18.Solution: For violet,nv = 1.714300.4 = 1.66, sin2 = sinnv= sin501.66,or2 = 27.48.From the geometry of triangle ABC,180 = 60 +(902) +(903),or3 = 602 = 6027.48 = 32.52,andsin4 = nvsin3 = 1.66sin32.52 = 0.89,or4 = 63.18.CHAPTER 8 373For red,nr = 1.714300.7 = 1.62,2 = sin1sin501.62

= 28.22,3 = 6028.22 = 31.78,4 = sin1[1.62sin31.78] = 58.56.Hence, angular dispersion = 63.1858.56 = 4.62.Problem 8.19 The two prisms in Fig. 8-35 (P8.19) are made of glass with n = 1.5.What fraction of the power density carried by the ray incident upon the top prismemerges from bottom prism? Neglect multiple internal reections.454545459090SiSt654321Figure P8.19: Periscope problem.Solution: Using = 0/n, at interfaces 1 and 4,a = n1n2n1 +n2= 11.51+1.5=0.2.At interfaces 3 and 6,b =a = 0.2.374 CHAPTER 8At interfaces 2 and 5,c = sin1

1n

= sin1

11.5

= 41.81.Hence, total internal reection takes place at those interfaces. At interfaces 1, 3, 4and 6, the ratio of power density transmitted to that incident is (12). Hence,StSi= (12)4= (1(0.2)2)4= 0.85.Problem 8.20 A light ray incident at 45passes through two dielectric materialswith the indices of refraction and thicknesses given in Fig. 8-36 (P8.20). If the raystrikes the surface of the rst dielectric at a height of 2 cm, at what height will it strikethe screen?452cmscreenn4 = 1n1 = 1n3 = 1.3 n2 = 1.53cm 4cm 5cm45h4h3h2h1234Figure P8.20: Light incident on a screen through a multi-layered dielectric (Problem8.20).Solution:sin2 = n1n2sin1 =11.5 sin45 = 0.47.Hence,2 = 28.13,h2 = 3 cmtan2 = 3 cm0.53 = 1.6 cm,sin3 = n2n3sin2 = 1.51.3 sin28.13 = 0.54.Hence,3 = 32.96,CHAPTER 8 375h3 = 4 cmtan32.96 = 2.6 cm,sin4 = n3n4sin3 = 0.707.Hence,4 = 45,h4 = 5 cmtan45 = 5 cm.Total height = h1 +h2 +h3 +h4 = (2+1.6+2.6+5) = 11.2 cm.Problem 8.21 FigureP8.21depictsabeakercontainingablockofglassonthebottom and water over it. The glass block contains a small air bubble at an unknowndepth below the water surface. When viewed from above at an angle of 60, the airbubble appears at a depth of 6.81 cm. What is the true depth of the air bubble?6010 cm2dad223x2xx1dtFigure P8.21: Apparent position of the air bubble in Problem 8.21.Solution: Letda = 6.81 cm = apparent depth,dt = true depth.2 = sin1n1n2sini

= sin111.33 sin60

= 40.6,376 CHAPTER 83 = sin1n1n3sini

= sin111.6 sin60

= 32.77,x1 = (10 cm) tan40.6 = 8.58 cm,x = dacot 30 = 6.81cot 30 = 11.8 cm.Hence,x2 = x x1 = 11.88.58 = 3.22 cm,andd2 = x2cot 32.77 = (3.22 cm) cot 32.77 = 5 cm.Hence, dt = (10+5) = 15 cm.Problem 8.22 A glass semicylinder with n = 1.5 is positioned such that its at faceis horizontal, as shown in Fig. 8-38 (P8.22). Its horizontal surface supports a drop ofoil, as shown. When light is directed radially toward the oil, total internal reectionoccurs if exceeds 53. What is the index of refraction of the oil?nglass= 1.5noiloil dropFigure P8.22: Oil drop on the at surface of a glass semicylinder (Problem 8.22).Solution:sinc = n2n1= noil1.5,noil = 1.5sin53 = 1.2.Problem 8.23 A penny lies at the bottom of a water fountain at a depth of 30 cm.Determine the diameter of a piece of paper which, if placed to oat on the surface ofCHAPTER 8 377dx x30 cmccwater surfaceFigure P8.23: Light cone bounded by total internal reection.the water directly above the penny, would totally obscure the penny from view. Treatthe penny as a point and assume that n = 1.33 for water.Solution:c = sin111.33

= 48.75,d = 2x = 2[(30 cm)tanc] = (60 cm) tan48.75 = 68.42 cm.Problem8.24 Suppose the optical ber of Example 8-5 is submerged in water (withn = 1.33) instead of air. Determine a andfp in that case.Solution: With n0 = 1.33, nf = 1.52 and nc = 1.49, Eq. (8.40) givessina =1n0(n2fn2c)1/2=11.33

(1.52)2(1.49)2

1/2= 0.23,ora = 13.1.The data ratefp given by Eq. (8.45) is not a function of n0, and therefore it remainsunchanged at 4.9 (Mb/s).378 CHAPTER 8Problem 8.25 Equation(8.45)was derivedfor thecase wherethe lightincidentuponthe sendingendoftheopticalber extendsovertheentireacceptanceconeshown in Fig. 8-12(b). Suppose the incident light is constrained to a narrower rangeextending between normal incidence and

, where

< a.(a) Obtain an expression for the maximum data ratefp in terms of

.(b) Evaluatefp for the ber of Example 8-5 when

= 5.Solution:(a) For i =

,sin2 =1nfsin

,lmax =lcos2=l

1sin22=l

1

sin

nf

2=lnf

n2f (sin

)2,tmax = lmaxup= lmaxnfc=ln2fc

n2f (sin

)2,tmin =lup= l nfc, = t =tmaxtmin = l nfc

nf

n2f(sin

)21,fp =12=c2lnf

nf

n2f (sin

)211(bits/s).(b) For:nf = 1.52,

= 5,l = 1 km,c = 3108m/s,fp = 59.88 (Mb/s).CHAPTER 8 379Sections 8-4 and 8-5: Reection and Transmission at Oblique IncidenceProblem 8.26 A plane wave in air withEi= y20ej(3x+4z)(V/m),is incident upon the planar surface of a dielectric material, with r = 4, occupying thehalf space z 0. Determine:(a) the polarization of the incident wave,(b) the angle of incidence,(c) the time-domain expressions for the reected electric and magnetic elds,(d) the time-domain expressions for the transmitted electric and magnetic elds,and(e) the average power density carried by the wave in the dielectric medium.Solution:(a) Ei= y20ej(3x+4z)V/m.Since Eiis along y, which is perpendicular to the plane of incidence, the wave isperpendicularly polarized.(b) From Eq. (8.48a), the argument of the exponential isjk1(xsini +zcosi) =j(3x +4z).Hence,k1sini = 3, k1cosi = 4,from which we determine thattani = 34or i = 36.87,andk1 =

32+42= 5 (rad/m).Also, = upk = ck = 31085 = 1.5109(rad/s).(c)1 = 0 = 377 ,2 =0r2= 02= 188.5 ,t = sin1sinir2

= sin1sin36.874

= 17.46,380 CHAPTER 8 = 2cos i1cost2cos i +1cost=0.41, = 1+ = 0.59.In accordance with Eq. (8.49a), and using the relation Er0 = Ei0,Er= y8.2ej(3x4z),Hr= ( x cosi + zsini) 8.20ej(3x4z),where we used the fact that i = r and the z-direction has been reversed.Er= Re[Erejt] = y8.2cos(1.5109t 3x +4z) (V/m),Hr= ( x17.4+ z13.06)cos(1.5109t 3x +4z) (mA/m).(d) In medium 2,k2 = k1

21= 54 = 20 (rad/m),andt = sin112sini

= sin112 sin36.87

= 17.46and the exponent of Etand Htisjk2(xsint +zcost) =j10(xsin17.46 +zcos17.46) =j(3x +9.54z).Hence,Et= y200.59ej(3x+9.54z),Ht= ( xcost + zsint) 200.592ej(3x+9.54z).Et= Re[Etejt] = y11.8cos(1.5109t 3x 9.54z) (V/m),Ht= ( xcos17.46 + zsin17.46)11.8188.5 cos(1.5109t 3x 9.54z)= ( x59.72+ z18.78) cos(1.5109t 3x 9.54z) (mA/m).(e)Stav = |Et0|222=(11.8)22188.5= 0.36 (W/m2).CHAPTER 8 381Problem 8.27 Repeat Problem 8.26 for a wave in air withHi= y2102ej(8x+6z)(A/m),incident upon the planar boundary of a dielectric medium (z 0) with r = 9.Solution:(a) Hi= y2102ej(8x+6z).Since Hiis along y, which is perpendicular to the plane of incidence, the wave isTM polarized, or equivalently, its electric eld vector is parallel polarized (parallel tothe plane of incidence).(b) From Eq. (8.65b), the argument of the exponential isjk1(xsini +zcosi) =j(8x +6z).Hence,k1sini = 8, k1cosi = 6,from which we determinei = tan1

86

= 53.13,k1 =

62+82= 10 (rad/m).Also, = upk = ck = 310810 = 3109(rad/s).(c)1 = 0 = 377 ,2 =0r2= 03= 125.67 ,t = sin1sinir2

= sin1sin53.139

= 15.47,

= 2cost1cosi2cost +1cosi=0.30,

= (1+

) cosicost= 0.44.In accordance with Eqs. (8.65a) to (8.65d), Ei0 = 21021 andEi= ( xcos i zsini)21021ej(8x+6z)= ( x4.52 z6.03)ej(8x+6z).382 CHAPTER 8Eris similar toEiexcept for reversal of z-components and multiplication of amplitudeby

. Hence, with

=0.30,Er= Re[Erejt] =( x1.36+ z1.81)cos(3109t 8x +6z) V/m,Hr= y2102

cos(3109t 8x +6z)= y0.6102cos(3109t 8x +6z) A/m.(d) In medium 2,k2 = k1

21= 109 = 30 rad/m,t = sin121sini

= sin113 sin53.13

= 15.47,and the exponent of Etand Htisjk2(xsint +zcost) =j30(xsin15.47 +zcos15.47) =j(8x +28.91z).Hence,Et= ( xcost zsint)Ei0

ej(8x+28.91z)= ( x0.96 z0.27)21023770.44ej(8x+28.91z)= ( x3.18 z0.90)ej(8x+28.91z),Ht= y Ei0

2ej(8x+28.91z)= y2.64102ej(8x+28.91z),Et= Re{Etejt}= ( x3.18 z0.90)cos(3109t 8x 28.91z) V/m,Ht= y2.64102cos(3109t 8x 28.91z) A/m.(e)Stav = |Et0|222= |Ht0|222 =(2.64102)22125.67 = 44 mW/m2.Problem 8.28 Natural light is randomly polarized, which means that, on average,half the light energy is polarized along any given direction (in the plane orthogonalCHAPTER 8 383to the direction of propagation) and the other half of the energy is polarized along thedirection orthogonal to the rst polarization direction. Hence, when treating naturallight incidentupon a planar boundary, we can considerhalf of its energyto be inthe form of parallel-polarized waves and the other half as perpendicularly polarizedwaves. Determine the fraction of the incident power reected by the planar surfaceof a piece of glass with n = 1.5 when illuminated by natural light at 70.Solution: Assume the incident power is 1 W. Hence:Incident power with parallel polarization = 0.5 W,Incident power with perpendicular polarization = 0.5 W.2/1 = (n2/n1)2= n2= 1.52= 2.25. Equations (8.60) and (8.68) give = cos 70

2.25sin270cos 70 +

2.25sin270=0.55,

= 2.25cos 70 +

2.25sin2702.25cos 70 +

2.25sin270= 0.21.Reected power with parallel polarization = 0.5(

)2= 0.5(0.21)2= 22 mW,Reected power with perpendicular polarization = 0.5()2= 0.5(0.55)2= 151.3 mW.Total reected power = 22+151.3 = 173.3 mW, or 17.33%..Problem 8.29 A parallel polarized plane wave is incident from air onto a dielectricmedium with r = 9 at the Brewster angle. What is the refraction angle?1r1 = 1r2 = 92Figure P8.29: Geometry of Problem 8.29.Solution: For nonmagnetic materials, Eq. (8.72) gives1 = B = tan1

21= tan13 = 71.57.384 CHAPTER 8Butsin2 = sin1r2= sin13= sin71.573= 0.32,or 2 = 18.44.Problem 8.30 A perpendicularly polarized wave in air is obliquely incident uponaplanarglass-airinterfaceat anincidenceangleof30. Thewavefrequencyis600THz (1 THz= 1012Hz), whichcorrespondsto greenlight, andtheindexofrefraction of the glass is 1.6. If the electric eld amplitude of the incident wave is 50V/m, determine(a) the reection and transmission coefcients, and(b) the instantaneous expressions for E and H in the glass medium.Solution:(a)Fornonmagneticmaterials, (2/1) = (n2/n1)2. UsingthisrelationinEq.(8.60) gives =cos i

(n2/n1)2sin2icos i +

(n2/n1)2sin2i=cos30

(1.6)2sin230cos30 +

(1.6)2sin230=0.27, = 1+ = 10.27 = 0.73.(b) In the glass medium,sint = sinin2= sin301.6= 0.31,or t = 18.21.2 =

22= 0n2= 1201.6= 75 = 235.62 (),k2 =up= 2fc/n= 2f nc= 260010121.63108= 6.4106rad/m,Et0 = Ei0 = 0.7350 = 36.5 V/m.From Eqs. (8.49c) and (8.49d),Et = yEt0ejk2(xsint+zcost),Ht = ( xcos t + zsint) Et02ejk2(xsint+zcost),CHAPTER 8 385and the corresponding instantaneous expressions are:Et(x, z, t) = y36.5cos(t k2xsintk2zcost) (V/m),Ht(x, z, t) = ( xcost zcost)0.16cos(t k2xsintk2zcos t) (A/m),with = 21015rad/s and k2 = 6.4106rad/m.Problem 8.31 Show that the reection coefcient can be written in the form = sin(ti)sin(t +i).Solution: From Eq. (8.58a), = 2cos i1cost2cos i +1cost=(2/1)cos icost(2/1)cos i +cost.Using Snells law for refraction given by Eq. (8.31), we have21= sintsini,we have = sintcos icostsinisintcos i +costsini= sin(ti)sin(t +i) .Problem 8.32 Show that for nonmagnetic media, the reection coefcient

canbe written in the form

= tan(ti)tan(t +i) .Solution: From Eq. (8.66a),

is given by

= 2cos t1cosi2cos t +1cosi=(2/1)cos tcosi(2/1)cos t +cosi.For nonmagnetic media, 1 = 2 = 0 and21=

12= n1n2.386 CHAPTER 8Snells law of refraction issintsini= n1n2.Hence,

=sintsinicos tcosisintsinicos t +cosi= sintcostsinicos isintcost +sinicos i.To show that the expression for

is the same as

= tan(ti)tan(t +i) ,we shall proceed with the latter and show that it is equal to the former.tan(ti)tan(t +i)= sin(ti)cos(t +i)cos(ti)sin(t +i) .Using the identities (from Appendix C):2sinxcos y = sin(x +y) +sin(x y),and if we let x =ti and y =t +i in the numerator, while letting x =t +i andy =ti in the denominator, thentan(ti)tan(t +i)= sin(2t) +sin(2i)sin(2t) +sin(2i).But sin2 = 2sincos , and sin() =sin, hence,tan(ti)tan(t +i)= sintcostsinicos isintcost +sinicos i,which is the intended result.Problem 8.33 A parallel polarized beam of light with an electric eld amplitude of10 (V/m) is incident in air on polystyrene with r = 1 and r = 2.6. If the incidenceangle at the airpolystyrene planar boundary is 50, determine(a) the reectivity and transmissivity, and(b) the power carried by the incident, reected, and transmitted beams if the spoton the boundary illuminated by the incident beam is 1 m2in area.CHAPTER 8 387Solution:(a) From Eq. (8.68),

=(2/1)cosi +

(2/1) sin2i(2/1)cos i +

(2/1) sin2i= 2.6cos 50 +

2.6sin2502.6cos 50 +

2.6sin250=0.08,R

=|

|2= (0.08)2= 6.4103,T

= 1R

= 0.9936.(b)Pi

=|Ei0|221Acosi =(10)22120cos50 = 85 mW,Pr

= R

Pi

= (6.4103) 0.085 = 0.55 mW,Pt

= T

Pi

= 0.99360.085 = 84.45 mW.Sections 8-6 to 8-11Problem 8.34 Derive Eq. (8.89b).Solution:We start with Eqs. (8.88a and e), ezy+ j ey =jhx,jhxhzx= j ey.To eliminate hx, we multiply the top equation by and the bottom equation by ,and then we add them together. The result is: ezy+ j2 eyhzx=j2 ey.388 CHAPTER 8Multiplying all terms by ejzto convert eyto Ey (and similarly for the other eldcomponents), and then solving for Ey leads toEy =1j(22)

Ezy+Hzx

=jk2c

Ezy+Hzx

,where we used the relationk2c= 2 2.Problem 8.35 A hollow rectangular waveguide is to be used to transmit signals ata carrier frequency of 6 GHz. Choose its dimensions so that the cutoff frequency ofthe dominant TE mode is lower than the carrier by 25% and that of the next mode isat least 25% higher than the carrier.Solution:For m = 1 and n = 0 (TE10 mode) and up0 = c (hollow guide), Eq. (8.106) reducestof10 =c2a .Denote the carrier frequency asf0 = 6 GHz. Settingf10 = 0.75f0 = 0.756 GHz = 4.5 GHz,we havea =c2f10=310824.5109= 3.33 cm.If b is chosen such that a > b >a2, the second mode will be TE01, followed by TE20at f20 = 9 GHz. For TE01,f01 =c2b .Settingf01 = 1.25f0 = 7.5 GHz, we getb =c2f01=310827.5109= 2 cm.CHAPTER 8 389Problem 8.36 A TE wave propagating in a dielectric-lled waveguide of unknownpermittivity has dimensions a = 5 cm and b =3 cm. If the x-component of its electriceld is given byEx =36cos(40x)sin(100y) sin(2.41010t 52.9z), (V/m)determine:(a) the mode number,(b) r of the material in the guide,(c) the cutoff frequency, and(d) the expression for Hy.Solution:(a) Comparison of the given expression with Eq. (8.110a) reveals thatma= 40, hence m = 2nb= 100, hence n = 3.Mode is TE23.(b) From sin(t z), we deduce that = 2.41010rad/s, = 52.9 rad/m.Using Eq. (8.105) to solve for r, we haver =c222+

ma

2+

nb

2

= 2.25.(c)up0=cr= 31082.25= 2108m/s.f23 =up02

2a

2+

3b

2= 10.77 GHz.390 CHAPTER 8(d)ZTE =ExHy=

1( f23/f )2= 377r

1

10.7712

2= 569.9 .Hence,Hy =ExZTE=0.063cos(40x)sin(100y)sin(2.41010t 52.9z) (A/m).Problem 8.37 A waveguide lled with a material whose r = 2.25 has dimensionsa = 2 cm and b = 1.4 cm. If the guide is to transmit 10.5-GHz signals, what possiblemodes can be used for the transmission?Solution:ApplicationofEq. (8.106)withup0= c/r= 3 108/2.25= 2 108m/s,gives:f10 = 5 GHz (TE only)f01 = 7.14 GHz (TE only)f11 = 8.72 GHz (TE or TM)f20 = 10 GHz (TE only)f21 = 12.28 GHz (TE or TM)f12 = 15.1 GHz (TE or TM).Hence, any one of the rst four modes can be used to transmit 10.5-GHz signals.Problem 8.38 Forarectangular waveguideoperatingintheTE10mode, obtainexpressions for the surface charge density s and surface current density Js on eachof the four walls of the guide.Solution:CHAPTER 8 391For TE10, the expressions for E and H are given by Eq. (8.110) with m = 1 andn = 0,Ex = 0,Ey =jH0k2casin

xa

ejz,Ez = 0,Hx =jH0k2casin

xa

ejz,Hy = 0,Hz = H0cos

xa

ejz.The applicable boundary conditions are given in Table 6-2. At the boundary betweena dielectric (medium 1) and a conductor (medium 2),s = n2D1 = 1 n2E1,Js = n2H1,where E1 and H1 are the elds inside the guide, 1 is the permittivity of the materiallling the guide, and n2 is the normal to the guide wall, pointing away from the wall(inwardly). In view of the coordinate system dened for the guide, n2 = x for sidewall at x = 0, n2 = x for wall at x = a, etc.n2^n2^n2^n2^ybxa42310(a) At side wall 1 at x = 0, n2 = x. Hence,s = 1 x yEy|x=0 = 0Js = x( x Hx + z Hz)|x=0= y Hz|x=0= yH0ejz.392 CHAPTER 8(b) At side wall 2 at x = a, n2 = x. Hence,s = 0Js = yH0ejz.(c) At bottom surface at y = 0, n2 = y. Hence,s = 1 y yEy|y=0=jH0k2casin

xa

ejzJs = y( x Hx + z Hz)= H0 xcos

xa

z j k2ca sin

xa

ejz.(d) At top surface at y b, n2 = y. Hence,s =jH0k2casin

xa

ejzJs = H0 xcos

xa

+ z j k2ca sin

xa

ejz.Problem 8.39 A waveguide, with dimensions a = 1 cm and b = 0.7 cm, is to beused at 20 GHz. Determine the wave impedance for the dominant mode when(a) the guide is empty, and(b) the guide is lled with polyethylene (whose r = 2.25).Solution:For the TE10 mode,f10 =up02a=c2ar.When empty,f10 =31082102= 15 GHz.When lled with polyethylene, f10 = 10 GHz.According to Eq. (8.111),ZTE =

1( f10/f )2=0r

1( f10/f )2 .CHAPTER 8 393When empty,ZTE =377

1(15/20)2= 570 .When lled,ZTE =3772.25

1(10/20)2= 290 .Problem8.40 Anarrowrectangular pulse superimposed on a carrier with afrequency of 9.5 GHz was used to excite all possible modes in a hollow guide witha = 3 cm and b = 2.0 cm. If the guide is 100 m in length, how long will it take eachof the excited modes to arrive at the receiving end?Solution:With a = 3 cm, b = 2 cm, and up0= c = 3108m/s, application of Eq. (8.106)leads to:f10 = 5 GHzf01 = 7.5 GHzf11 = 9.01 GHzf20 = 10 GHzHence, the pulse with a 9.5-GHz carrier can excite the top three modes. Their groupvelocities can be calculated with the help of Eq. (8.114),ug = c

1( fmn/f )2,which gives:ug =

0.85c = 2.55108m/s, for TE100.61c = 1.84108m/s, for TE010.32c = 0.95108m/s, for TE11 and TM11Travel time associated with these modes is:T=dug= 100ug=

0.39 s, for TE100.54 s, for TE011.05 s, for TE11 and TM11.394 CHAPTER 8Problem 8.41 If the zigzag angle

is 42 for the TE10 mode, what would it be forthe TE20 mode?Solution:For TE10, the derivation that started with Eq. (8.116) led to

10 = tan1

a

, TE10 mode.Had the derivation been for n = 2 (instead of n = 1), the x-dependence would haveinvolved a phase factor(2x/a) (instead of(x/a)). The sequence of steps wouldhave led to

20 = tan1

2a

, TE20 mode.Given that

10 = 42, it follows thata= tan42 = 0.90Hence,

20 = tan1(20.9) = 60.9.Problem 8.42 Measurement of the TE101 frequency response of an air-lled cubiccavity revealed that its Q is 4802. If its volume is 64 mm3, what material are its sidesmade of?Solution:According to Eq. (8.121), the TE101 resonant frequency of a cubic cavity is givenbyf101 = 31082a=310824103= 53.0 GHz.Its Q is given byQ =a3s= 4802,which gives s = 2.78107m. Applyings =1f1010c,and solving for c leads toc6.2107S/m.CHAPTER 8 395According to Appendix B, the material is silver.Problem 8.43 A hollow cavity made of aluminum has dimensions a = 4 cm andd = 3 cm. Calculate Q of the TE101 mode for(a) b = 2 cm, and(b) b = 3 cm.Solution:For the TE101 mode, f101 is independent of b,f101 = c2

1a

2+

1d

2= 31082

14102

2+

13102

2= 6.25 GHz.For aluminum with c = 3.5107S/m (Appendix B),s =1f1010c= 1.08106m.(a) For a = 4 cm, b = 2 cm and d = 3 cm,Q =1sabd(a2+d2)[a3(d +2b) +d3(a+2b)]= 8367.(b) For a = 4 cm, b = 3 cm, and d = 3 cm,Q = 9850.Problem8.44 A50-MHz right-hand circularly polarizedplane wave withanelectric eld modulus of 30 V/m is normally incident in air upon a dielectric mediumwith r = 9 and occupying the region dened by z 0.(a) Write an expression for the electric eld phasor of the incident wave, given thatthe eld is a positive maximum at z = 0 and t = 0.(b) Calculate the reection and transmission coefcients.396 CHAPTER 8(c) Writeexpressions for theelectriceldphasors of thereectedwave, thetransmitted wave, and the total eld in the region z 0.(d) Determinethepercentages of theincident averagepower reectedbytheboundary and transmitted into the second medium.Solution:(a)k1 = c= 2501063108= 3rad/m,k2 = cr2 = 39 = rad/m.From (7.57), RHC wave traveling in +z direction:Ei= a0( x+ yej/2)ejk1z= a0( x j y)ejk1zEi(z, t) = Re

Eiejt

= Re

a0( xej(tk1z)+ yej(tk1z/2))

= xa0cos(t k1z) + ya0cos(t k1z /2)= xa0cos(t k1z) + ya0sin(t k1z)|Ei| =

a20cos2(t k1z) +a20sin2(t k1z)

1/2= a0 = 30 V/m.Hence,Ei= 30(x0 jy0)ejz/3(V/m).(b)1 = 0 = 120 (), 2 =0r2= 1209= 40 (). = 212 +1= 4012040+120=0.5 = 1+ = 10.5 = 0.5.CHAPTER 8 397(c)Er= a0( x j y)ejk1z=0.530( x j y)ejk1z=15( x j y)ejz/3(V/m).Et= a0( x j y)ejk2z= 15( x j y)ejz(V/m).E1 = Ei+Er= 30( x j y)ejz/315( x j y)ejz/3= 15( x j y)[2ejz/3ejz/3] (V/m).(d)% of reected power = 100||2= 100(0.5)2= 25%% of transmitted power = 100||212= 100(0.5)212040= 75%.Problem 8.45 Consider a at 5-mm-thick slab of glass with r = 2.56.(a) If a beam of green light (0 = 0.52 m) is normally incident upon one of thesides of the slab, what percentage of the incident power is reected back by theglass?(b) To eliminate reections, it is desired to add a thin layer of antireection coatingmaterial on each side of the glass. If you are at liberty to specify the thicknessof theantireectionmaterialas well as its relativepermittivity, whatwouldthese specications be?Solution:398 CHAPTER 8ZL = 0Z2 = gAZ1 = 05 mmAir Glassr = 2.56Green LightAir5 mmZi(a) Representing the wave propagation process by an equivalent transmission linemodel, the input impedance at the left-hand side of the air-glass interface is (from2.63):Zi = Z0

ZL + jZ0tanlZ0 + jZLtanl

For the glass,Z0 = g =0r=02.56=01.6ZL = 0l = 2 l = 20rl =20.521062.56 5103= 30769.23.Subtracting the maximumpossible multiples of 2, namely 30768, leaves aremainder ofl = 1.23 rad.Hence,Zi =01.6

0 + j(0/1.6)tan1.23(0/1.6) + j0tan1.23

=

1.6+ j tan1.231+ j1.6tan1.23

1201.6=

1.6+ j0.8821+ j1.41

1201.6= 24925.8 = (224.2 j108.4) .CHAPTER 8 399With Zi now representing the input impedance of the glass, the reection coefcientat point A is: = Zi0Zi +0= 224.2 j108.4120224.2 j108.4+120= 187.34144.6610.8910.2= 0.3067154.8 .% of reected power =||2100 = 9.4%.(b) To avoid reections, we can add a quarter-wave transformer on each side of theglass.Air Glassr = 2.56Air5 mm d dAntireflection coatingAntireflection coatingAntireflection coatingThis requires that d be:d = 4 +2n, n = 0, 1, 2, . . .where is the wavelength in that material; i.e., =0/rc, where rc is the relativepermittivity of the coating material. It is also required that c of the coating materialbe:2c = 0g.Thus20rc= 00r,orrc =r =2.56 = 1.6.Hence, =0rc= 0.52 m1.6= 0.411 m,d = 4 +2n= (0.103+0.822n) (m), n = 0, 1, 2, . . .400 CHAPTER 8Problem8.46 Aparallel-polarizedplanewaveisincident fromair at ananglei = 30 onto a pair of dielectric layers as shown in the gure.(a) Determine the angles of transmission 2, 3, and 4.(b) Determine the lateral distance d.d5 cm5 cmi243AirAirr = 1r = 2.25r = 1r = 6.25Solution:(a) Application of Snells law of refraction given by (8.31) leads to:sin2 = sin1

r1r2= sin30

16.25= 0.22 = 11.54.Similarly,sin3 = sin2

r2r3= sin11.54

6.252.25= 0.333 = 19.48.And,sin4 = sin3

r3r4= sin19.48

2.251= 0.54 = 30.As expected, the exit ray back into air will be at the same angle as i.CHAPTER 8 401(b)d = (5 cm)tan2 +(5 cm)tan3= 5tan11.54 +5tan19.48 = 2.79 cm.Problem 8.47 A plane wave in air withEi= ( x2 y4 z6)ej(2x+3z)(V/m)is incident upon the planar surface of a dielectric material, with r = 2.25, occupyingthe half-space z 0. Determine(a) The incidence angle i.(b) The frequency of the wave.(c) The eld Erof the reected wave.(d) The eld Etof the wave transmitted into the dielectric medium.(e) The average power density carried by the wave into the dielectric medium.Solution:xzi2(a) From the exponential of the given expression, it is clear that the wave directionoftravel isinthexzplane. Bycomparisonwiththeexpressionsin(8.48a)for402 CHAPTER 8perpendicularpolarizationor (8.65a)for parallelpolarization, bothof which havethe same phase factor, we conclude that:k1sini = 2,k1cosi = 3.Hence,k1 =

22+32= 3.6 (rad/m)i = tan1(2/3) = 33.7.Also,k2 = k1r2 = 3.62.25 = 5.4 (rad/m)2 = sin1sini

12.25= 21.7.(b)k1 = 2fcf = k1c2= 3.631082= 172 MHz.(c) In order to determine the electric eld of the reected wave, we rst have todetermine the polarization of the wave. The vector argument in the given expressionfor Eiindicatesthat theincident waveisamixtureofparallel andperpendicularpolarizationcomponents. Perpendicularpolarizationhas a y-componentonly(see8.46a), whereas parallel polarization has only x and z components (see 8.65a). Hence,we shall decompose the incident wave accordingly:Ei= Ei +Ei

withEi = y4ej(2x+3z)(V/m)Ei

= ( x2 z6)ej(2x+3z)(V/m)From the above expressions, we deduce:Ei0 =4 V/mEi0 =

22+62= 6.32 V/m.CHAPTER 8 403Next, we calculate and for each of the two polarizations: =cos i

(2/1) sin2icos i +

(2/1) sin2iUsing i = 33.7 and 2/1 = 2.25/1 = 2.25 leads to: =0.25 = 1+ = 0.75.Similarly, =(2/1)cos i +

(2/1) sin2i(2/1)cos i +

(2/1) sin2i=0.15,

= (1+

)cos icos t= (10.15)cos 33.7cos 21.7= 0.76.The electric elds of the reected and transmitted waves for the two polarizations aregiven by (8.49a), (8.49c), (8.65c), and (8.65e):Er = yEr0ejk1(xsinrzcosr)Et = yEt0ejk2(xsint+zcost)Er

= ( xcosr + zsinr)Er0ejk1(xsinrzcosr)Et

= ( xcost zsint)Et0ejk2(xsint+zcost)Based on our earlier calculations:r = i = 33.7t = 21.7k1 = 3.6 rad/m, k2 = 5.4 rad/m,Er0 = Ei0 = (0.25) (4) = 1 V/m.Et0 = Ei0 = 0.75(4) =3 V/m.Er0 =

Ei0 = (0.15) 6.32 =0.95 V/m.Et0 =

Ei0 = 0.766.32 = 4.8 V/m.Using the above values, we have:Er= Er +Er

= ( x0.79+ y z0.53)ej(2x3z)(V/m).404 CHAPTER 8(d)Et= Et +Et

= ( x4.46 y3 z1.78)ej(2x+5z)(V/m).(e)St= |Et0|222|Et0|2= (4.46)2+32+(1.78)2= 32.062 =0r2= 3771.5= 251.3 St=32.062251.3= 63.8 (mW/m2).

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