EECE 380 Problems Solution Manual, Fall 2011

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 Demos 8.2 351 Lesson #52 Chapter — Section: 8-2 Topics: Snell’s laws Highlights: • Reflection and refraction • Index of refraction Special Illustrations: • Example 8-4 • Technology Brief on “Lasers” (CD-ROM) Lasers Lasers are used in CD and DVD players, bar-code readers, eye surgery and multitudes of other systems and applications. A laser—acronym for light amplification by stimulated emission of radiation—is 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 Readers A 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. 353 Lessons #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 8-7 355 Lessons #57–59 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 8 Chapter 8 Section 8-1: Reflection and Transmission at Normal Incidence Problem 8.1 A plane wave in air with an electric field amplitude of 20 V/m is incident normally upon the surface of a lossless, nonmagnetic medium with εr = 25. Determine: (a) the reflection and transmission coefficients, (b) the standing-wave ratio in the air medium, and (c) the average power densities of the incident, reflected, and transmitted waves. Solution: (a) η1 = η0 = 120π (Ω), η2 = η0 √εr = 120π 5 = 24π (Ω). From Eqs. (8.8a) and (8.9), Γ = η2 −η1 η2 +η1 = 24π−120π 24π+120π = −96 144 = −0.67, τ = 1+Γ = 1−0.67 = 0.33. (b) S = 1+|Γ| 1−|Γ| = 1+0.67 1−0.67 = 5. (c) According to Eqs. (8.19) and (8.20), Si av = |Ei 0|2 2η0 = 400 2×120π = 0.52 W/m2, Sr av = |Γ|2Si av = (0.67)2 ×0.52 = 0.24 W/m2, St av = |τ|2 |Ei 0|2 2η2 = |τ|2 η1 η2 Si av = (0.33)2 × 120π 24π ×0.52 = 0.28 W/m2. Problem 8.2 A plane wave traveling in medium 1 with εr1 = 2.25 is normally incident upon medium 2 with εr2 = 4. Both media are made of nonmagnetic, non- conducting materials. If the electric field of the incident wave is given by Ei = ˆy8cos(6π×109t −30πx) (V/m), (a) obtain time-domain expressions for the electric and magnetic fields in each of the two media, and CHAPTER 8 357 (b) determine the average power densities of the incident, reflected and transmitted waves. Solution: (a) Ei = ˆy8cos(6π×109t −30πx) (V/m), η1 = η0 √εr1 = η0 √ 2.25 = η0 1.5 = 377 1.5 = 251.33 Ω, η2 = η0 √εr2 = η0 √ 4 = 377 2 = 188.5 Ω, Γ = η2 −η1 η2 +η1 = 1/2−1/1.5 1/2+1/1.5 = −0.143, τ = 1+Γ = 1−0.143 = 0.857, Er = ΓEi = −1.14 ˆycos(6π×109t +30πx) (V/m). Note that the coefficient of x is positive, denoting the fact that Er belongs to a wave traveling in −x-direction. E1 = Ei +Er = ˆy[8cos(6π×109t −30πx)−1.14cos(6π×109t +30πx)] (A/m), Hi = ˆz 8 η1 cos(6π×109t −30πx) = ˆz31.83cos(6π×109t −30πx) (mA/m), Hr = ˆz1.14 η1 cos(6π×109t +30πx) = ˆz4.54cos(6π×109t +30πx) (mA/m), H1 = Hi +Hr = ˆz[31.83cos(6π×109t −30πx)+4.54cos(6π×109t +30πx)] (mA/m). Since k1 = ω√µε1 and k2 = ω√µε2 , k2 = �ε2 ε1 k1 = � 4 2.25 30π = 40π (rad/m), E2 = Et = ˆy8τcos(6π×109t −40πx) = ˆy6.86cos(6π×109t −40πx) (V/m), H2 = Ht = ˆz 8τ η2 cos(6π×109t −40πx) = ˆz36.38cos(6π×109t −40πx) (mA/m). (b) Si av = ˆx 82 2η1 = 64 2×251.33 = ˆx127.3 (mW/m2), Sr av = −|Γ|2Si av = −ˆx(0.143)2 ×0.127 = −ˆx2.6 (mW/m2), 358 CHAPTER 8 St av = |Et 0|2 2η2 = ˆxτ2 (8)2 2η2 = ˆx (0.86)264 2×188.5 = ˆx124.7 (mW/m2). Within calculation error, Si av +Sr av = St av. Problem 8.3 A plane wave traveling in a medium with εr1 = 9 is normally incident upon a second medium with εr2 = 4. Both media are made of nonmagnetic, non- conducting materials. If the magnetic field of the incident plane wave is given by Hi = ˆz2cos(2π×109t −ky) (A/m), (a) obtain time domain expressions for the electric and magnetic fields in each of the two media, and (b) determine the average power densities of the incident, reflected and transmitted waves. Solution: (a) In medium 1, up = c √εr1 = 3×108 √ 9 = 1×108 (m/s), k1 = ω up = 2π×109 1×108 = 20π (rad/m), Hi = ˆz2cos(2π×109t −20πy) (A/m), η1 = η0 √εr1 = 377 3 = 125.67 Ω, η2 = η0 √εr2 = 377 2 = 188.5 Ω, Ei = −ˆx2η1 cos(2π×109t −20πy) = −ˆx251.34cos(2π×109t −20πy) (V/m), Γ = η2 −η1 η2 +η1 = 188.5−125.67 188.5+125.67 = 0.2, τ = 1+Γ = 1.2, Er = −ˆx251.34×0.2cos(2π×109t +20πy) = −ˆx50.27cos(2π×109t +20πy) (V/m), CHAPTER 8 359 Hr = −ˆz 50.27 η1 cos(2π×109t +20πy) = −ˆz0.4cos(2π×109t +20πy) (A/m), E1 = Ei +Er = −ˆx[25.134cos(2π×109t −20πy)+50.27cos(2π×109t +20πy)] (V/m), H1 = Hi +Hr = ˆz[2cos(2π×109t −20πy)−0.4cos(2π×109t +20πy)] (A/m). In medium 2, k2 = �ε2 ε1 k1 = � 4 9 ×20π = 40π 3 (rad/m), E2 = Et = −ˆx251.34τcos � 2π×109t − 40πy 3 � = −ˆx301.61cos � 2π×109t − 40πy 3 � (V/m), H2 = Ht = ˆz 301.61 η2 cos � 2π×109t − 40πy 3 � = ˆz1.6cos � 2π×109t − 40πy 3 � (A/m). (b) Si av = ˆy |E0|2 2η1 = ˆy (251.34)2 2×125.67 = ˆy251.34 (W/m2), Sr av = −ˆy|Γ|2(251.34) = ˆy10.05 (W/m2), St av = ˆy(251.34−10.05) = ˆy241.29 (W/m2). Problem 8.4 A 200-MHz left-hand circularly polarized plane wave with an electric field modulus of 5 V/m is normally incident in air upon a dielectric medium with εr = 4 and occupying the region defined by z ≥ 0. (a) Write an expression for the electric field phasor of the incident wave, given that the field is a positive maximum at z = 0 and t = 0. (b) Calculate the reflection and transmission coefficients. (c) Write expressions for the electric field phasors of the reflected wave, the transmitted wave, and the total field in the region z ≤ 0. 360 CHAPTER 8 (d) Determine the percentages of the incident average power reflected by the boundary and transmitted into the second medium. Solution: (a) k1 = ω c = 2π×2×108 3×108 = 4π 3 rad/m, k2 = ω up2 = ω c √εr2 = 4π 3 √ 4 = 8π 3 rad/m. LHC wave: �Ei = a0(ˆx+ ˆyejπ/2)e− jkz = a0(ˆx+ jˆy)e− jkz, Ei(z,t) = ˆxa0 cos(ωt −kz)− ˆya0 sin(ωt −kz), |Ei| = [a2 0 cos2(ωt −kz)+a2 0 sin2(ωt −kz)]1/2 = a0 = 5 (V/m). Hence, �Ei = 5(ˆx+ jˆy)e− j4πz/3 (V/m). (b) η1 = η0 = 120π (Ω), η2 = η0 √εr = η0 2 = 60π (Ω). Equations (8.8a) and (8.9) give Γ = η2 −η1 η2 +η1 = 60π−120π 60π+120π = −60 180 = −1 3 , τ = 1+Γ = 2 3 . (c) �Er = 5Γ(ˆx+ jˆy)ejk1z = −5 3(ˆx+ jˆy)ej4πz/3 (V/m), �Et = 5τ(ˆx+ jˆy)e− jk2z = 10 3 (ˆx+ jˆy)e− j8πz/3 (V/m), �E1 = �Ei + �Er = 5(ˆx+ jˆy) � e− j4πz/3 − 1 3ej4πz/3 � (V/m). (d) % of reflected power = 100×|Γ|2 = 100 9 = 11.11%, % of transmitted power = 100×|τ|2 η1 η2 = 100× �2 3 �2 × 120π 60π = 88.89%. CHAPTER 8 361 Problem 8.5 Repeat Problem 8.4 after replacing the dielectric medium with a poor conductor characterized by εr = 2.25, µr = 1, and σ = 10−4 S/m. Solution: (a) Medium 1: η1 = η0 = 120π (Ω), k1 = ω c = 2π×2×108 3×108 = 4π 3 (rad/m). Medium 2: σ2 ωε2 = 10−4 ×36π 2π×2×108 ×2.25×10−9 = 4×10−3. Hence, medium 2 is a low-loss dielectric. From Table 7-1, α2 = σ2 2 �µ2 ε2 = σ2 2 120π √εr2 = σ2 2 × 120π √ 2.25 = 10−4 2 × 120π 1.5 = 1.26×10−2 (NP/m), β2 = ω√µ2ε2 = ω√εr2 c = 2π (rad/m), η2 = �µ2 ε2 � 1+ jσ2 2ωε2 � = 120π √εr2 � 1+ j2×10−3� ≃ 120π 1.5 = 80π (Ω). LHC wave: �Ei = a0(ˆx+ jˆy)e− jk1z, |�Ei| = a0 = 5 (V/m), �Ei = 5(ˆx+ jˆy)e− j4πz/3 (V/m). (b) According to Eqs. (8.8a) and (8.9), Γ = η2 −η1 η2 +η1 = 80π−120π 80π+120π = −0.2, τ = 1+Γ = 1−0.2 = 0.8. (c) �Er = 5Γ(ˆx+ jˆy)ejk1z = −(ˆx+ jˆy)ej4πz/3 (V/m), �Et = 5τ(ˆx+ jˆy)e−α2ze− jβzz = 4(ˆx+ jˆy)e−1.26×10−2ze− j2πz (V/m), �E1 = �Ei + �Er = 5(ˆx+ jˆy)[e− j4πz/3 −0.2e j4πz/3] (V/m). 362 CHAPTER 8 (d) % of reflected power = 100|Γ|2 = 100(0.2)2 = 4%, % of transmitted power = 100|τ|2 η1 η2 = 100(0.8)2 × 120π 80π = 96%. Problem 8.6 A 50-MHz plane wave with electric field amplitude of 50 V/m is normally incident in air onto a semi-infinite, perfect dielectric medium with εr = 36. Determine (a) Γ, (b) the average power densities of the incident and reflected waves, and (c) the distance in the air medium from the boundary to the nearest minimum of the electric field intensity, |E|. Solution: (a) η1 = η0 = 120π (Ω), η2 = �µ2 ε2 = 120π √εr2 = 120π 6 = 20π (Ω), Γ = η2 −η1 η2 +η1 = 20π−120π 20π+120π = −0.71. Hence, |Γ| = 0.71 and θη = 180◦. (b) Si av = |Ei 0|2 2η1 = (50)2 2×120π = 3.32 (W/m2), Sr av = |Γ|2Si av = (0.71)2 ×3.32 = 1.67 (W/m2). (c) In medium 1 (air), λ1 = c f = 3×108 5×107 = 6 m. From Eqs. (8.16) and (8.17), lmax = θrλ1 4π = π×6 4π = 1.5 m, lmin = lmax − λ1 4 = 1.5−1.5 = 0 m (at the boundary). Problem 8.7 What is the maximum amplitude of the total electric field in the air medium of Problem 8.6, and at what nearest distance from the boundary does it occur? CHAPTER 8 363 Solution: From Problem 8.6, Γ = −0.71 and λ = 6 m. |�E1|max = (1+|Γ|)Ei 0 = (1+0.71)×50 = 85.5 V/m, lmax = θrλ1 4π = π×6 4π = 1.5 m. Problem 8.8 Repeat Problem 8.6 after replacing the dielectric medium with a conductor with εr = 1, µr = 1, and σ = 2.78×10−3 S/m. Solution: (a) Medium 1: η1 = η0 = 120π = 377 (Ω), λ1 = c f = 3×108 5×107 = 6 m, Medium 2: σ2 ωε2 = 2.78×10−3 ×36π 2π×5×107 ×10−9 = 1. Hence, Medium 2 is a quasi-conductor. From Eq. (7.70), η2 = �µ2 ε2 � 1− jε′′ 2 ε′ 2 �−1/2 = 120π � 1− j σ2 ωε2 �−1/2 = 120π(1− j1)−1/2 = 120π( √ 2)−1/2ej22.5◦ = (292.88+ j121.31) (Ω). Γ = η2 −η1 η2 +η1 = (292.88+ j121.31)−377 (292.88+ j121.31)+377 = −0.09+ j0.12 = 0.22∠114.5◦ . (b) Si av = |Ei 0|2 2η1 = 502 2×120π = 3.32 (W/m2), |Sr av| = |Γ|2Si av = (0.22)2(3.32) = 0.16 (W/m2). (c) In medium 1 (air), λ1 = c f = 3×108 5×107 = 6 m. For θr = 114.5◦ = 2 rad, Eqs. (8.16) and (8.17) give lmax = θrλ1 4π + (0)λ1 2 = 2(6) 4 +0 = 3 m, 364 CHAPTER 8 lmin = lmax − λ1 4 = 3− 6 4 = 3−1.5 = 1.5 m. Problem 8.9 The three regions shown in Fig. 8-32 (P8.9) contain perfect dielectrics. For a wave in medium 1 incident normally upon the boundary at z = −d, what combination of εr2 and d produce no reflection? Express your answers in terms of εr1, εr3 and the oscillation frequency of the wave, f. Medium 2 εr2 Medium 3 εr3 Medium 1 εr1 z = -d z = 0 z d Figure P8.9: Three dielectric regions. Solution: By analogy with the transmission-line case, there will be no reflection at z = −d if medium 2 acts as a quarter-wave transformer, which requires that d = λ2 4 and η2 = √η1η3 . The second condition may be rewritten as η0 √εr2 = � η0 √εr1 η0 √εr3 �1/2 , or εr2 = √εr1εr3 , λ2 = λ0 √εr2 = c f√εr2 = c f(εr1εr3)1/4 , and d = c 4f(εr1εr3)1/4 . CHAPTER 8 365 Problem 8.10 For the configuration shown in Fig. 8-32 (P8.9), use transmission- line equations (or the Smith chart) to calculate the input impedance at z = −d for εr1 = 1, εr2 = 9, εr3 = 4, d = 1.2 m, and f = 50 MHz. Also determine the fraction of the incident average power density reflected by the structure. Assume all media are lossless and nonmagnetic. Solution: In medium 2, λ = λ0 √εr2 = c f√εr2 = 3×108 5×107 ×3 = 2 m. Hence, β2 = 2π λ2 = π rad/m, β2d = 1.2π rad. At z = −d, the input impedance of a transmission line with load impedance ZL is given by Eq. (2.63) as Zin(−d) = Z0 �ZL + jZ0tanβ2d Z0 + jZLtanβ2d � . In the present case, Z0 = η2 = η0/√εr2 = η0/3 and ZL = η3 = η0/√εr3 = η0/2, where η0 = 120π (Ω). Hence, Zin(−d) = η2 �η3 + jη2tanβ2d η2 + jη3tanβ2d � = η0 3 � 1 2 + j �1 3 � tan1.2π 1 3 + j �1 2 � tan1.2π � = η0(0.35− j0.14). At z = −d, Γ = Zin −Z1 Zin +Z1 = η0(0.35− j0.14)−η0 η0(0.35− j0.14)+η0 = 0.49e− j162.14◦. Fraction of incident power reflected by the structure is |Γ|2 = |0.49|2 = 0.24. Problem 8.11 Repeat Problem 8.10 after interchanging εr1 and εr3. Solution: In medium 2, λ = λ0 √εr2 = c f√εr2 = 3×108 5×107 ×3 = 2 m. 366 CHAPTER 8 Hence, β2 = 2π λ2 = π rad/m, β2d = 1.2π rad. At z = −d, the input impedance of a transmission line with impedance ZL is given as Eq. (2.63), Zin(−d) = Z0 � ZL + jZ0 tanβd Z0 + jZLtanβ2d � . In the present case, Z0 = η2 = η0/√εr2 = η0/3, ZL = η3 = η0/√εr1 = η0, where η0 = 120π (Ω). Hence, Zin(−d) = η2 �η3 + jη2 tan1.2π η2 + jη3 tan1.2π � = η0 3 �1+(j/3)tan1.2π (1/3)+ jtan1.2π � = η0 �1+(j/3)tan1.2π 1+ j3tan1.2π � = (0.266− j0.337)η0 = 0.43η0∠−51.7◦ . At z = −d, Γ = Zin −Z1 Zin +Z1 = 0.43∠−51.7◦ − 1 2 0.43∠−51.7◦ + 1 2 = 0.49∠−101.1◦ . Fraction of incident power reflected by structure is |Γ|2 = 0.24. Problem 8.12 Orange light of wavelength 0.61 µm in air enters a block of glass with εr = 1.44. What color would it appear to a sensor embedded in the glass? The wavelength ranges of colors are violet (0.39 to 0.45 µm), blue (0.45 to 0.49 µm), green (0.49 to 0.58 µm), yellow (0.58 to 0.60 µm), orange (0.60 to 0.62 µm), and red (0.62 to 0.78 µm). Solution: In the glass, λ = λ0 √εr = 0.61 √ 1.44 = 0.508 µm. The light would appear green. Problem 8.13 A plane wave of unknown frequency is normally incident in air upon the surface of a perfect conductor. Using an electric-field meter, it was determined that the total electric field in the air medium is always zero when measured at a CHAPTER 8 367 distance of 2 m from the conductor surface. Moreover, no such nulls were observed at distances closer to the conductor. What is the frequency of the incident wave? Solution: The electric field of the standing wave is zero at the conductor surface, and the standing wave pattern repeats itself every λ/2. Hence, λ 2 = 2 m, or λ = 4 m, in which case f = c λ = 3×108 4 = 7.5×107 = 75 MHz. Problem 8.14 Consider a thin film of soap in air under illumination by yellow light with λ = 0.6 µm in vacuum. If the film is treated as a planar dielectric slab with εr = 1.72, surrounded on both sides by air, what film thickness would produce strong reflection of the yellow light at normal incidence? Solution: The transmission line analogue of the soap-bubble wave problem is shown in Fig. P8.14(b) where the load ZL is equal to η0, the impedance of the air medium on the other side of the bubble. That is, η0 = 377 Ω, η1 = 377 √ 1.72 = 287.5 Ω. The normalized load impedance is zL = η0 η1 = 1.31. For the reflection by the soap bubble to be the largest, Zin needs to be the most different from η0. This happens when zL is transformed through a length λ/4. Hence, L = λ 4 = λ0 4√εr = 0.6 µm 4 √ 1.72 = 0.115 µm, where λ is the wavelength of the soap bubble material. Strong reflections will also occur if the thickness is greater than L by integer multiples of nλ/2 = (0.23n) µm. Hence, in general L = (0.115+0.23n) µm, n = 0,1,2,... . According to Section 2-7.5, transforming a load ZL = 377 Ω through a λ/4 section of Z0 = 287.5 Ω ends up presenting an input impedance of Zin = Z2 0 ZL = (287.5)2 377 = 219.25 Ω. 368 CHAPTER 8 Yellow Light λ = 0.6 µm Air Soap Air εr1=1 εr3=1 εr2=1.72 L (b) Transmission-line equivalent circuit η0 = 377Ω η2 ZL = η0 = 377 Ω (a) Yellow light incident on soap bubble. Figure P8.14: Diagrams for Problem 8.14. This Zin is at the input side of the soap bubble. The reflection coefficient at that interface is Γ = Zin −η0 Zin +η0 = 219.25−377 219.25+377 = −0.27. Any other thickness would produce a reflection coeffficient with a smaller magnitude. Problem 8.15 A 5-MHz plane wave with electric field amplitude of 10 (V/m) is normally incident in air onto the plane surface of a semi-infinite conducting material with εr = 4, µr = 1, and σ = 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 Ei to be along ˆx and the incident direction to be +ˆz. With k1 = ω c = 2π×5×106 3×108 = π 30 (rad/m), CHAPTER 8 369 we have Ei = ˆx10cos � π×107t − π 30 z � (V/m), η1 = η0 = 377 Ω. From Table 7-1, ε′′ ε′ = σ ωεrε0 = 100×36π π×107 ×4×10−9 = 9×104, which makes the material a good conductor, for which α2 = � π fµσ = � π×5×106 ×4π×10−7 ×100 = 44.43 (Np/m), β2 = 44.43 (rad/m), ηc2 = (1+ j) α2 σ = (1+ j) 44.43 100 = 0.44(1+ j) Ω. According to the expression for Sav2 given in the answer to Exercise 8.3, Sav2 = ˆz|τ|2 |Ei 0|2 2 e−2α2z Re � 1 η∗c2 � . 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) = |τ|2 |Ei 0|2 2 Re � 1 η∗c2 � [1−e−2α2z1] where z1 = 2 mm. τ = 1+Γ = 1+ η2 −η1 η2 +η1 = 1+ 0.44(1+ j)−377 0.44(1+ j)+377 ≈ 0.0023(1+ j) = 3.3×10−3ej45◦. Re � 1 η∗c2 � = Re � 1 0.44(1+ j)∗ � = Re � 1 0.44(1− j) � = Re � 1+ j 0.44×2 � = 1 0.88 = 1.14, P′ = (3.3×10−3)2 102 2 ×1.14[1−e−2×44.43×2×10−3] = 1.01×10−4 (W/m2). 370 CHAPTER 8 Problem 8.16 A 0.5-MHz antenna carried by an airplane flying over the ocean surface generates a wave that approaches the water surface in the form of a normally incident plane wave with an electric-field amplitude of 3,000 (V/m). Sea water is characterized by εr = 72, µr = 1, and σ = 4 (S/m). The plane is trying to communicate a message to a submarine submerged at a depth d below the water surface. If the submarine’s receiver requires a minimum signal amplitude of 0.01 (µV/m), what is the maximum depth d to which successful communication is still possible? Solution: For sea water at 0.5 MHz, ε′′ ε′ = σ ωε = 4×36π 2π×0.5×106 ×72×10−9 = 2000. Hence, sea water is a good conductor, in which case we use the following expressions from Table 7-1: α2 = � π fµσ = � π×0.5×106 ×4π×10−7 ×4 = 2.81 (Np/m), β2 = 2.81 (rad/m), ηc2 = (1+ j) α2 σ = (1+ j) 2.81 4 = 0.7(1+ j) Ω, Γ = η2 −η1 η2 +η1 = 0.7(1+ j)−377 0.7(1+ j)+377 = (−0.9963+ j3.7×10−3), τ = 1+Γ = 5.24×10−3ej44.89◦, |Et| = |τEi 0e−α2d|. We need to find the depth z at which |E t| = 0.01 µV/m = 10−8 V/m. 10−8 = 5.24×10−3 ×3×103e−2.81d, e−2.81d = 6.36×10−10, −2.81d = ln(6.36×10−10) = −21.18, or d = 7.54 (m). CHAPTER 8 371 Sections 8-2 and 8-3: Snell’s Laws and Fiber Optics Problem 8.17 A light ray is incident on a prism at an angle θ as shown in Fig. 8-33 (P8.17). The ray is refracted at the first surface and again at the second surface. In terms of the apex angle φ of the prism and its index of refraction n, determine the smallest value of θ for which the ray will emerge from the other side. Find this minimum θ for n = 4 and φ = 60◦. θ θ3 φ n A B C θ2 Figure P8.17: Prism of Problem 8.17. Solution: For the beam to emerge at the second boundary, it is necessary that θ3 < θc, where sinθc = 1/n. From the geometry of triangle ABC, 180◦ = φ+(90◦ −θ2)+(90◦ −θ3), or θ2 = φ−θ3. At the first boundary, sinθ = nsinθ2. Hence, sinθmin = nsin(φ−θ3) = nsin � φ−sin−1 �1 n �� , or θmin = sin−1 � nsin � φ−sin−1 �1 n ��� . For n = 4 and φ = 60◦, θmin = sin−1 � 4sin(60◦ −sin−1 �1 4 �� = 20.4◦. 372 CHAPTER 8 Problem 8.18 For some types of glass, the index of refraction varies with wavelength. A prism made of a material with n = 1.71− 4 30 λ0, (λ0 in µm), where λ0 is the wavelength in vacuum, was used to disperse white light as shown in Fig. 8-34 (P8.18). The white light is incident at an angle of 50◦, the wavelength λ0 of red light is 0.7 µm and that of violet light is 0.4 µm. Determine the angular dispersion in degrees. 50° 60° Red Green Violet Angular dispersion A B C θ4 θ3 θ2 Figure P8.18: Prism of Problem 8.18. Solution: For violet, nv = 1.71− 4 30 ×0.4 = 1.66, sinθ2 = sinθ nv = sin50◦ 1.66 , or θ2 = 27.48◦. From the geometry of triangle ABC, 180◦ = 60◦ +(90◦ −θ2)+(90◦ −θ3), or θ3 = 60◦ −θ2 = 60−27.48◦ = 32.52◦, and sinθ4 = nv sinθ3 = 1.66sin32.52◦ = 0.89, or θ4 = 63.18◦. CHAPTER 8 373 For red, nr = 1.71− 4 30 ×0.7 = 1.62, θ2 = sin−1 �sin50◦ 1.62 � = 28.22◦, θ3 = 60◦ −28.22◦ = 31.78◦, θ4 = sin−1 [1.62sin31.78◦] = 58.56◦. Hence, angular dispersion = 63.18◦ −58.56◦ = 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 prism emerges from bottom prism? Neglect multiple internal reflections. 45° 45° 45° 45° 90° 90° Si St 6 5 4 3 2 1 Figure P8.19: Periscope problem. Solution: Using η = η0/n, at interfaces 1 and 4, Γa = n1 −n2 n1 +n2 = 1−1.5 1+1.5 = −0.2. At interfaces 3 and 6, Γb = −Γa = 0.2. 374 CHAPTER 8 At interfaces 2 and 5, θc = sin−1 �1 n � = sin−1 � 1 1.5 � = 41.81◦. Hence, total internal reflection takes place at those interfaces. At interfaces 1, 3, 4 and 6, the ratio of power density transmitted to that incident is (1−Γ2). Hence, St Si = (1−Γ2)4 = (1−(0.2)2)4 = 0.85. Problem 8.20 A light ray incident at 45◦ passes through two dielectric materials with the indices of refraction and thicknesses given in Fig. 8-36 (P8.20). If the ray strikes the surface of the first dielectric at a height of 2 cm, at what height will it strike the screen? 45° 2cm screen n4 = 1 n1 = 1 n3 = 1.3 n2 = 1.5 3cm 4cm 5cm 45° h4 h3 h2 h1 θ2 θ3 θ4 Figure P8.20: Light incident on a screen through a multi-layered dielectric (Problem 8.20). Solution: sinθ2 = n1 n2 sinθ1 = 1 1.5 sin45◦ = 0.47. Hence, θ2 = 28.13◦, h2 = 3 cm×tanθ2 = 3 cm×0.53 = 1.6 cm, sinθ3 = n2 n3 sinθ2 = 1.5 1.3 sin28.13◦ = 0.54. Hence, θ3 = 32.96◦, CHAPTER 8 375 h3 = 4 cm×tan32.96◦ = 2.6 cm, sinθ4 = n3 n4 sinθ3 = 0.707. Hence, θ4 = 45◦, h4 = 5 cm×tan45◦ = 5 cm. Total height = h1 +h2 +h3 +h4 = (2+1.6+2.6+5) = 11.2 cm. Problem 8.21 Figure P8.21 depicts a beaker containing a block of glass on the bottom and water over it. The glass block contains a small air bubble at an unknown depth below the water surface. When viewed from above at an angle of 60◦, the air bubble appears at a depth of 6.81 cm. What is the true depth of the air bubble? 60° 10 cm θ2 da d2 θ2 θ3 x2 x x1 dt Figure P8.21: Apparent position of the air bubble in Problem 8.21. Solution: Let da = 6.81 cm = apparent depth, dt = true depth. θ2 = sin−1 �n1 n2 sinθi � = sin−1 � 1 1.33 sin60◦ � = 40.6◦, 376 CHAPTER 8 θ3 = sin−1 �n1 n3 sinθi � = sin−1 � 1 1.6 sin60◦ � = 32.77◦, x1 = (10 cm)×tan40.6◦ = 8.58 cm, x = da cot30◦ = 6.81cot30◦ = 11.8 cm. Hence, x2 = x−x1 = 11.8−8.58 = 3.22 cm, and d2 = x2 cot32.77◦ = (3.22 cm)×cot32.77◦ = 5 cm. Hence, dt = (10+5) = 15 cm. Problem 8.22 A glass semicylinder with n = 1.5 is positioned such that its flat face is horizontal, as shown in Fig. 8-38 (P8.22). Its horizontal surface supports a drop of oil, as shown. When light is directed radially toward the oil, total internal reflection occurs if θ exceeds 53◦. What is the index of refraction of the oil? θ nglass= 1.5 noil oil drop Figure P8.22: Oil drop on the flat surface of a glass semicylinder (Problem 8.22). Solution: sinθc = n2 n1 = noil 1.5, 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 float on the surface of CHAPTER 8 377 d x x 30 cm θc θc water surface Figure P8.23: Light cone bounded by total internal reflection. the water directly above the penny, would totally obscure the penny from view. Treat the penny as a point and assume that n = 1.33 for water. Solution: θc = sin−1 � 1 1.33 � = 48.75◦, d = 2x = 2[(30 cm)tanθc] = (60 cm)×tan48.75◦ = 68.42 cm. Problem 8.24 Suppose the optical fiber of Example 8-5 is submerged in water (with n = 1.33) instead of air. Determine θa and fp in that case. Solution: With n0 = 1.33, nf = 1.52 and nc = 1.49, Eq. (8.40) gives sinθa = 1 n0 (n2 f −n2 c)1/2 = 1 1.33 � (1.52)2 −(1.49)2�1/2 = 0.23, or θa = 13.1◦. The data rate fp given by Eq. (8.45) is not a function of n0, and therefore it remains unchanged at 4.9 (Mb/s). 378 CHAPTER 8 Problem 8.25 Equation (8.45) was derived for the case where the light incident upon the sending end of the optical fiber extends over the entire acceptance cone shown in Fig. 8-12(b). Suppose the incident light is constrained to a narrower range extending between normal incidence and θ′, where θ′ < θa. (a) Obtain an expression for the maximum data rate fp in terms of θ′. (b) Evaluate fp for the fiber of Example 8-5 when θ′ = 5◦. Solution: (a) For θi = θ′, sinθ2 = 1 nf sinθ′, lmax = l cosθ2 = l � 1−sin2 θ2 = l � 1− � sinθ′ nf �2 = lnf � n2 f −(sinθ′)2 , tmax = lmax up = lmaxnf c = ln2 f c � n2 f −(sinθ′)2 , tmin = l up = l nf c , τ = ∆t = tmax −tmin = l nf c   nf � n2 f −(sinθ′)2 −1  , fp = 1 2τ = c 2lnf   nf � n2 f −(sinθ′)2 −1   −1 (bits/s). (b) For: nf = 1.52, θ′ = 5◦, l = 1 km, c = 3×108 m/s, fp = 59.88 (Mb/s). CHAPTER 8 379 Sections 8-4 and 8-5: Reflection and Transmission at Oblique Incidence Problem 8.26 A plane wave in air with �Ei = ˆy20e− j(3x+4z) (V/m), is incident upon the planar surface of a dielectric material, with εr = 4, occupying the half space z ≥ 0. Determine: (a) the polarization of the incident wave, (b) the angle of incidence, (c) the time-domain expressions for the reflected electric and magnetic fields, (d) the time-domain expressions for the transmitted electric and magnetic fields, and (e) the average power density carried by the wave in the dielectric medium. Solution: (a) �Ei = ˆy20e− j(3x+4z) V/m. Since Ei is along ˆy, which is perpendicular to the plane of incidence, the wave is perpendicularly polarized. (b) From Eq. (8.48a), the argument of the exponential is − jk1(xsinθi +zcosθi) = − j(3x+4z). Hence, k1 sinθi = 3, k1 cosθi = 4, from which we determine that tanθi = 3 4 or θi = 36.87◦, and k1 = � 32 +42 = 5 (rad/m). Also, ω = upk = ck = 3×108 ×5 = 1.5×109 (rad/s). (c) η1 = η0 = 377 Ω, η2 = η0 √εr2 = η0 2 = 188.5 Ω, θt = sin−1 �sinθi √εr2 � = sin−1 �sin36.87◦ √ 4 � = 17.46◦, 380 CHAPTER 8 Γ⊥ = η2 cosθi −η1 cosθt η2 cosθi +η1 cosθt = −0.41, τ⊥ = 1+Γ⊥ = 0.59. In accordance with Eq. (8.49a), and using the relation E r 0 = Γ⊥Ei 0, �Er = −ˆy8.2e− j(3x−4z), �Hr = (ˆx cosθi + ˆzsinθi) 8.2 η0 e− j(3x−4z), where we used the fact that θi = θr and the z-direction has been reversed. Er = Re[�Erejωt] = −ˆy8.2cos(1.5×109t −3x+4z) (V/m), Hr = (ˆx17.4+ ˆz13.06)cos(1.5×109t −3x+4z) (mA/m). (d) In medium 2, k2 = k1 �ε2 ε1 = 5 √ 4 = 20 (rad/m), and θt = sin−1 ��ε1 ε2 sinθi � = sin−1 �1 2 sin36.87◦ � = 17.46◦ and the exponent of Et and Ht is − jk2(xsinθt +zcosθt) = − j10(xsin17.46◦ +zcos17.46◦) = − j(3x+9.54z). Hence, �Et = ˆy20×0.59e− j(3x+9.54z), �Ht = (−ˆxcosθt + ˆzsinθt) 20×0.59 η2 e− j(3x+9.54z). Et = Re[�Etejωt] = ˆy11.8cos(1.5×109t −3x−9.54z) (V/m), Ht = (−ˆxcos17.46◦ + ˆzsin17.46◦) 11.8 188.5 cos(1.5×109t −3x−9.54z) = (−ˆx59.72+ ˆz18.78) cos(1.5×109t −3x−9.54z) (mA/m). (e) St av = |Et 0|2 2η2 = (11.8)2 2×188.5 = 0.36 (W/m2). CHAPTER 8 381 Problem 8.27 Repeat Problem 8.26 for a wave in air with �Hi = ˆy2×10−2e− j(8x+6z) (A/m), incident upon the planar boundary of a dielectric medium (z ≥ 0) with εr = 9. Solution: (a) �H i = ˆy2×10−2e− j(8x+6z). Since Hi is along ˆy, which is perpendicular to the plane of incidence, the wave is TM polarized, or equivalently, its electric field vector is parallel polarized (parallel to the plane of incidence). (b) From Eq. (8.65b), the argument of the exponential is − jk1(xsinθi +zcosθi) = − j(8x+6z). Hence, k1 sinθi = 8, k1 cosθi = 6, from which we determine θi = tan−1 �8 6 � = 53.13◦, k1 = � 62 +82 = 10 (rad/m). Also, ω = upk = ck = 3×108 ×10 = 3×109 (rad/s). (c) η1 = η0 = 377 Ω, η2 = η0 √εr2 = η0 3 = 125.67 Ω, θt = sin−1 �sinθi √εr2 � = sin−1 �sin53.13◦ √ 9 � = 15.47◦, Γ∥ = η2 cosθt −η1 cosθi η2 cosθt +η1 cosθi = −0.30, τ∥ = (1+Γ∥) cosθi cosθt = 0.44. In accordance with Eqs. (8.65a) to (8.65d), E i 0 = 2×10−2η1 and �Ei = (ˆxcosθi − ˆzsinθi)2×10−2η1 e− j(8x+6z) = (ˆx4.52− ˆz6.03)e− j(8x+6z). 382 CHAPTER 8 �Er is similar to �Ei except for reversal of z-components and multiplication of amplitude by Γ∥. Hence, with Γ∥ = −0.30, Er = Re[�E rejωt] = −(ˆx1.36+ ˆz1.81)cos(3×109t −8x+6z) V/m, Hr = ˆy2×10−2Γ∥ cos(3×109t −8x+6z) = −ˆy0.6×10−2 cos(3×109t −8x+6z) A/m. (d) In medium 2, k2 = k1 �ε2 ε1 = 10 √ 9 = 30 rad/m, θt = sin−1 ��ε2 ε1 sinθi � = sin−1 �1 3 sin53.13◦ � = 15.47◦, and the exponent of Et and Ht is − jk2(xsinθt +zcosθt) = − j30(xsin15.47◦ +zcos15.47◦) = − j(8x+28.91z). Hence, �Et = (ˆxcosθt − ˆzsinθt)Ei 0τ∥e− j(8x+28.91z) = (ˆx0.96− ˆz0.27)2×10−2 ×377×0.44e− j(8x+28.91z) = (ˆx3.18− ˆz0.90)e− j(8x+28.91z), �Ht = ˆy Ei 0τ∥ η2 e− j(8x+28.91z) = ˆy2.64×10−2 e− j(8x+28.91z), Et = Re{�Etejωt} = (ˆx3.18− ˆz0.90)cos(3×109t −8x−28.91z) V/m, Ht = ˆy2.64×10−2 cos(3×109t −8x−28.91z) A/m. (e) St av = |Et 0|2 2η2 = |Ht 0|2 2 η2 = (2.64×10−2)2 2 ×125.67 = 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 orthogonal CHAPTER 8 383 to the direction of propagation) and the other half of the energy is polarized along the direction orthogonal to the first polarization direction. Hence, when treating natural light incident upon a planar boundary, we can consider half of its energy to be in the form of parallel-polarized waves and the other half as perpendicularly polarized waves. Determine the fraction of the incident power reflected by the