Bài tập Trường điện từ (Có lời giải)

egion. Calculate the wavelength in centimeters and the attenuation in nepers per meter if the wave is propagating in a non-magnetic material for which a) ′r = 1 and  ′′ r = 0: In a non-magnetic material, we would have: α = ω √ µ00′r 2   √ 1 + ( ′′r ′r )2 − 1  1/2 and β = ω √ µ00′r 2   √ 1 + ( ′′r ′r )2 + 1  1/2 With the given values of ′r and  ′′ r , it is clear that β = ω √ µ00 = ω/c, and so λ = 2π/β = 2πc/ω = 3× 1010/1010 = 3 cm. It is also clear that α = 0. b) ′r = 1.04 and  ′′ r = 9.00×10−4: In this case ′′r/′r << 1, and so β .= ω √ ′r/c = 2.13 cm −1. Thus λ = 2π/β = 2.95 cm. Then α .= ω′′ 2 √ µ ′ = ω′′r 2 √ µ00√ ′r = ω 2c ′′r√ ′r = 2π × 1010 2× 3× 108 (9.00× 10−4)√ 1.04 = 9.24× 10−2 Np/m 7 2.15c) ′r = 2.5 and  ′′ r = 7.2: Using the above formulas, we obtain β = 2π × 1010√2.5 (3× 1010)√2   √ 1 + ( 7.2 2.5 )2 + 1  1/2 = 4.71 cm−1 and so λ = 2π/β = 1.33 cm. Then α = 2π × 1010√2.5 (3× 108)√2   √ 1 + ( 7.2 2.5 )2 − 1  1/2 = 335 Np/m 12.16. The power factor of a capacitor is defined as the cosine of the impedance phase angle, and its Q is ωCR, where R is the parallel resistance. Assume an idealized parallel plate capacitor having a dielecric characterized by σ, ′, and µr. Find both the power factor and Q in terms of the loss tangent: First, the impedance will be: Z = R ( 1 jωC ) R + ( 1 jωC ) = R 1− jRωC 1 + (RωC)2 = R 1− jQ 1 + Q2 Now R = d/(σA) and C = ′A/d, and so Q = ω′/σ = 1/l.t. Then the power factor is P.F = cos[tan−1(−Q)] = 1/ √ 1 + Q2. 12.17. Let η = 250 + j30 Ω and jk = 0.2 + j2 m−1 for a uniform plane wave propagating in the az direction in a dielectric having some finite conductivity. If |Es| = 400 V/m at z = 0, find: a) at z = 0 and z = 60 cm: Assume x-polarization for the electric field. Then = 1 2 Re {Es ×H∗s} = 1 2 Re { 400e−αze−jβzax × 400 η∗ e−αzejβzay } = 1 2 (400)2e−2αzRe { 1 η∗ } az = 8.0× 104e−2(0.2)zRe { 1 250− j30 } az = 315 e−2(0.2)z az W/m2 Evaluating at z = 0, obtain (z = 0) = 315az W/m2, and at z = 60 cm, Pz,av(z = 0.6) = 315e−2(0.2)(0.6)az = 248az W/m2. b) the average ohmic power dissipation in watts per cubic meter at z = 60 cm: At this point a flaw becomes evident in the problem statement, since solving this part in two different ways gives results that are not the same. I will demonstrate: In the first method, we use Poynting’s theorem in point form (first equation at the top of p. 366), which we modify for the case of time-average fields to read: −∇· = where the right hand side is the average power dissipation per volume. Note that the additional right-hand-side terms in Poynting’s theorem that describe changes in energy 8 stored in the fields will both be zero in steady state. We apply our equation to the result of part a: = −∇· = − d dz 315 e−2(0.2)z = (0.4)(315)e−2(0.2)z = 126e−0.4z W/m3 At z = 60 cm, this becomes = 99.1 W/m3. In the second method, we solve for the conductivity and evaluate = σ . We use jk = jω √ µ′ √ 1− j(′′/′) and η = √ µ ′ 1√ 1− j(′′/′) We take the ratio, jk η = jω′ [ 1− j ( ′′ ′ )] = jω′ + ω′′ Identifying σ = ω′′, we find σ = Re { jk η } = Re { 0.2 + j2 250 + j30 } = 1.74× 10−3 S/m Now we find the dissipated power per volume: σ = 1.74× 10−3 ( 1 2 ) ( 400e−0.2z )2 At z = 60 cm, this evaluates as 109 W/m3. One can show that consistency between the two methods requires that Re { 1 η∗ } = σ 2α This relation does not hold using the numbers as given in the problem statement and the value of σ found above. Note that in Problem 12.13, where all values are worked out, the relation does hold and consistent results are obtained using both methods. 12.18. Given, a 100MHz uniform plane wave in a medium known to be a good dielectric. The phasor electric field is Es = 4e−0.5ze−j20zax V/m. Not stated in the problem is the permeabil- ity, which we take to be µ0. Also, the specified distance in part f should be 10m, not 1km. Determine: a) ′: As a first step, it is useful to see just how much of a good dielectric we have. We use the good dielectric approximations, Eqs. (60a) and (60b), with σ = ω′′. Using these, we take the ratio, β/α, to find β α = 20 0.5 = ω √ µ′ [ 1 + (1/8)(′′/′)2 ] (ω′′/2) √ µ/′ = 2 ( ′ ′′ ) + 1 4 ( ′′ ′ ) This becomes the quadratic equation:( ′′ ′ )2 − 160 ( ′′ ′ ) + 8 = 0 9 12.18a (continued) The solution to the quadratic is (′′/′) = 0.05, which means that we can neglect the second term in Eq. (60b), so that β .= ω √ µ′ = (ω/c) √ ′r. With the given frequency of 100 MHz, and with µ = µ0, we find √ ′r = 20(3/2π) = 9.55, so that  ′ r = 91.3, and finally ′ = ′r0 = 8.1× 10−10 F/m. b) ′′: Using Eq. (60a), the set up is α = 0.5 = ω′′ 2 √ µ ′ ⇒ ′′ = 2(0.5) 2π × 108 √ ′ µ = 10−8 2π(377) √ 91.3 = 4.0× 10−11 F/m c) η: Using Eq. (62b), we find η .= √ µ ′ [ 1 + j 1 2 ( ′′ ′ )] = 377√ 91.3 (1 + j.025) = (39.5 + j0.99) ohms d) Hs: This will be a y-directed field, and will be Hs = Es η ay = 4 (39.5 + j0.99) e−0.5ze−j20z ay = 0.101e−0.5ze−j20ze−j0.025 ay A/m e) : Using the given field and the result of part d, obtain = 1 2 Re{Es ×H∗s} = (0.101)(4) 2 e−2(0.5)z cos(0.025)az = 0.202e−z az W/m2 f) the power in watts that is incident on a rectangular surface measuring 20m x 30m at z = 10m (not 1km): At 10m, the power density is = 0.202e−10 = 9.2×10−6 W/m2. The incident power on the given area is then P = 9.2× 10−6 × (20)(30) = 5.5 mW. 12.19. Perfectly-conducting cylinders with radii of 8 mm and 20 mm are coaxial. The region between the cylinders is filled with a perfect dielectric for which  = 10−9/4π F/m and µr = 1. If E in this region is (500/ρ) cos(ωt− 4z)aρ V/m, find: a) ω, with the help of Maxwell’s equations in cylindrical coordinates: We use the two curl equations, beginning with ∇×E = −∂B/∂t, where in this case, ∇×E = ∂Eρ ∂z aφ = 2000 ρ sin(ωt− 4z)aφ = −∂Bφ ∂t aφ So Bφ = ∫ 2000 ρ sin(ωt− 4z)dt = 2000 ωρ cos(ωt− 4z) T Then Hφ = Bφ µ0 = 2000 (4π × 10−7)ωρ cos(ωt− 4z) A/m We next use ∇×H = ∂D/∂t, where in this case ∇×H = −∂Hφ ∂z aρ + 1 ρ ∂(ρHφ) ∂ρ az where the second term on the right hand side becomes zero when substituting our Hφ. So ∇×H = −∂Hφ ∂z aρ = − 8000(4π × 10−7)ωρ sin(ωt− 4z)aρ = ∂Dρ ∂t aρ And Dρ = ∫ − 8000 (4π × 10−7)ωρ sin(ωt− 4z)dt = 8000 (4π × 10−7)ω2ρ cos(ωt− 4z) C/m 2 10 12.19a. (continued) Finally, using the given , Eρ = Dρ  = 8000 (10−16)ω2ρ cos(ωt− 4z) V/m This must be the same as the given field, so we require 8000 (10−16)ω2ρ = 500 ρ ⇒ ω = 4× 108 rad/s b) H(ρ, z, t): From part a, we have H(ρ, z, t) = 2000 (4π × 10−7)ωρ cos(ωt− 4z)aφ = 4.0 ρ cos(4× 108t− 4z)aφ A/m c) S(ρ, φ, z): This will be S(ρ, φ, z) = E×H = 500 ρ cos(4× 108t− 4z)aρ × 4.0 ρ cos(4× 108t− 4z)aφ = 2.0× 10−3 ρ2 cos2(4× 108t− 4z)az W/m2 d) the average power passing through every cross-section 8 < ρ < 20 mm, 0 < φ < 2π. Using the result of part c, we find = (1.0× 103)/ρ2az W/m2. The power through the given cross-section is now P = ∫ 2π 0 ∫ .020 .008 1.0× 103 ρ2 ρ dρ dφ = 2π × 103 ln ( 20 8 ) = 5.7 kW 12.20. If Es = (60/r) sin θ e−j2r aθ V/m, and Hs = (1/4πr) sin θ e−j2r aφ A/m in free space, find the average power passing outward through the surface r = 106, 0 < θ < π/3, and 0 < φ < 2π. = 1 2 Re {Es ×H∗s} = 15 sin2 θ 2πr2 ar W/m2 Then, the requested power will be Φ = ∫ 2π 0 ∫ π/3 0 15 sin2 θ 2πr2 ar · ar r2 sin θdθdφ = 15 ∫ π/3 0 sin3 θ dθ = 15 ( −1 3 cos θ(sin2 θ + 2) ) ∣∣∣π/3 0 = 25 8 = 3.13 W Note that the radial distance at the surface, r = 106 m, makes no difference, since the power density dimishes as 1/r2. 11 12.21. The cylindrical shell, 1 cm ¡ ρ ¡ 1.2 cm, is composed of a conducting material for which σ = 106 S/m. The external and internal regions are non-conducting. Let Hφ = 2000 A/m at ρ = 1.2 cm. a) Find H everywhere: Use Ampere’s circuital law, which states:∮ H · dL = 2πρ(2000) = 2π(1.2× 10−2)(2000) = 48π A = Iencl Then in this case J = I Area az = 48 (1.44− 1.00)× 10−4 az = 1.09× 10 6 az A/m2 With this result we again use Ampere’s circuital law to find H everywhere within the shell as a function of ρ (in meters): Hφ1(ρ) = 1 2πρ ∫ 2π 0 ∫ ρ .01 1.09× 106 ρ dρ dφ = 54.5 ρ (104ρ2 − 1) A/m (.01 < ρ < .012) Outside the shell, we would have Hφ2(ρ) = 48π 2πρ = 24/ρ A/m (ρ > .012) Inside the shell (ρ < .01 m), Hφ = 0 since there is no enclosed current. b) Find E everywhere: We use E = J σ = 1.09× 106 106 az = 1.09az V/m which is valid, presumeably, outside as well as inside the shell. c) Find S everywhere: Use P = E×H = 1.09az × 54.5 ρ (104ρ2 − 1)aφ = −59.4 ρ (104ρ2 − 1)aρ W/m2 (.01 < ρ < .012 m) Outside the shell, S = 1.09az × 24 ρ aφ = −26 ρ aρ W/m2 (ρ > .012 m) 12 12.22. The inner and outer dimensions of a copper coaxial transmission line are 2 and 7 mm, respec- tively. Both conductors have thicknesses much greater than δ. The dielectric is lossless and the operating frequency is 400 MHz. Calculate the resistance per meter length of the: a) inner conductor: First δ = 1√ πfµσ = 1√ π(4× 108)(4π × 10−7)(5.8× 107) = 3.3× 10 −6m = 3.3µm Now, using (70) with a unit length, we find Rin = 1 2πaσδ = 1 2π(2× 10−3)(5.8× 107)(3.3× 10−6) = 0.42 ohms/m b) outer conductor: Again, (70) applies but with a different conductor radius. Thus Rout = a b Rin = 2 7 (0.42) = 0.12 ohms/m c) transmission line: Since the two resistances found above are in series, the line resistance is their sum, or R = Rin + Rout = 0.54 ohms/m. 12.23. A hollow tubular conductor is constructed from a type of brass having a conductivity of 1.2 × 107 S/m. The inner and outer radii are 9 mm and 10 mm respectively. Calculate the resistance per meter length at a frequency of a) dc: In this case the current density is uniform over the entire tube cross-section. We write: R(dc) = L σA = 1 (1.2× 107)π(.012 − .0092) = 1.4× 10 −3 Ω/m b) 20 MHz: Now the skin effect will limit the effective cross-section. At 20 MHz, the skin depth is δ(20MHz) = [πfµ0σ]−1/2 = [π(20× 106)(4π × 10−7)(1.2× 107)]−1/2 = 3.25× 10−5 m This is much less than the outer radius of the tube. Therefore we can approximate the resistance using the formula: R(20MHz) = L σA = 1 2πbδ = 1 (1.2× 107)(2π(.01))(3.25× 10−5) = 4.1× 10 −2 Ω/m c) 2 GHz: Using the same formula as in part b, we find the skin depth at 2 GHz to be δ = 3.25× 10−6 m. The resistance (using the other formula) is R(2GHz) = 4.1× 10−1 Ω/m. 13 12.24a. Most microwave ovens operate at 2.45 GHz. Assume that σ = 1.2 × 106 S/m and µr = 500 for the stainless steel interior, and find the depth of penetration: δ = 1√ πfµσ = 1√ π(2.45× 109)(4π × 10−7)(1.2× 106) = 9.28× 10 −6m = 9.28µm b) Let Es = 50 0◦ V/m at the surface of the conductor, and plot a curve of the amplitude of Es vs. the angle of Es as the field propagates into the stainless steel: Since the conductivity is high, we use (62) to write α .= β .= √ πfµσ = 1/δ. So, assuming that the direction into the conductor is z, the depth-dependent field is written as Es(z) = 50e−αze−jβz = 50e−z/δe−jz/δ = 50 exp(−z/9.28)︸ ︷︷ ︸ amplitude exp(−j z/9.28︸ ︷︷ ︸ angle ) where z is in microns. Therefore, the plot of amplitude versus angle is simply a plot of e−x versus x, where x = z/9.28; the starting amplitude is 50 and the 1/e amplitude (at z = 9.28 µm) is 18.4. 12.25. A good conductor is planar in form and carries a uniform plane wave that has a wavelength of 0.3 mm and a velocity of 3× 105 m/s. Assuming the conductor is non-magnetic, determine the frequency and the conductivity: First, we use f = v λ = 3× 105 3× 10−4 = 10 9 Hz = 1 GHz Next, for a good conductor, δ = λ 2π = 1√ πfµσ ⇒ σ = 4π λ2fµ = 4π (9× 10−8)(109)(4π × 10−7) = 1.1× 10 5 S/m 12.26. The dimensions of a certain coaxial transmission line are a = 0.8mm and b = 4mm. The outer conductor thickness is 0.6mm, and all conductors have σ = 1.6× 107 S/m. a) Find R, the resistance per unit length, at an operating frequency of 2.4 GHz: First δ = 1√ πfµσ = 1√ π(2.4× 108)(4π × 10−7)(1.6× 107) = 2.57× 10 −6m = 2.57µm Then, using (70) with a unit length, we find Rin = 1 2πaσδ = 1 2π(0.8× 10−3)(1.6× 107)(2.57× 10−6) = 4.84 ohms/m The outer conductor resistance is then found from the inner through Rout = a b Rin = 0.8 4 (4.84) = 0.97 ohms/m The net resistance per length is then the sum, R = Rin + Rout = 5.81 ohms/m. 14 12.26b. Use information from Secs. 6.4 and 9.10 to find C and L, the capacitance and inductance per unit length, respectively. The coax is air-filled. From those sections, we find (in free space) C = 2π0 ln(b/a) = 2π(8.854× 10−12) ln(4/.8) = 3.46× 10−11 F/m L = µ0 2π ln(b/a) = 4π × 10−7 2π ln(4/.8) = 3.22× 10−7 H/m c) Find α and β if α+ jβ = √ jωC(R + jωL): Taking real and imaginary parts of the given expression, we find α = Re {√ jωC(R + jωL) } = ω √ LC√ 2   √ 1 + ( R ωL )2 − 1  1/2 and β = Im {√ jωC(R + jωL) } = ω √ LC√ 2   √ 1 + ( R ωL )2 + 1  1/2 These can be found by writing out α = Re {√ jωC(R + jωL) } = (1/2) √ jωC(R + jωL)+ c.c., where c.c denotes the complex conjugate. The result is squared, terms collected, and the square root taken. Now, using the values of R, C, and L found in parts a and b, we find α = 3.0× 10−2 Np/m and β = 50.3 rad/m. 12.27. The planar surface at z = 0 is a brass-Teflon interface. Use data available in Appendix C to evaluate the following ratios for a uniform plane wave having ω = 4× 1010 rad/s: a) αTef/αbrass: From the appendix we find ′′/′ = .0003 for Teflon, making the material a good dielectric. Also, for Teflon, ′r = 2.1. For brass, we find σ = 1.5× 107 S/m, making brass a good conductor at the stated frequency. For a good dielectric (Teflon) we use the approximations: α .= σ 2 √ µ ′ = ( ′′ ′ ) ( 1 2 ) ω √ µ′ = 1 2 ( ′′ ′ ) ω c √ ′r β .= ω √ µ′ [ 1 + 1 8 ( ′′ ′ )] .= ω √ µ′ = ω c √ ′r For brass (good conductor) we have α .= β .= √ πfµσbrass = √ π ( 1 2π ) (4× 1010)(4π × 10−7)(1.5× 107) = 6.14× 105 m−1 Now αTef αbrass = 1/2 (′′/′) (ω/c) √ ′r√ πfµσbrass = (1/2)(.0003)(4× 1010/3× 108)√2.1 6.14× 105 = 4.7× 10 −8 b) λTef λbrass = (2π/βTef) (2π/βbrass) = βbrass βTef = c √ πfµσbrass ω √ ′r Tef = (3× 108)(6.14× 105) (4× 1010)√2.1 = 3.2× 10 3 15 12.27. (continued) c) vTef vbrass = (ω/βTef) (ω/βbrass) = βbrass βTef = 3.2× 103 as before 12.28. A uniform plane wave in free space has electric field given by Es = 10e−jβxaz + 15e−jβxay V/m. a) Describe the wave polarization: Since the two components have a fixed phase difference (in this case zero) with respect to time and position, the wave has linear polarization, with the field vector in the yz plane at angle φ = tan−1(10/15) = 33.7◦ to the y axis. b) Find Hs: With propagation in forward x, we would have Hs = −10 377 e−jβxay + 15 377 e−jβxaz A/m = −26.5e−jβxay + 39.8e−jβxaz mA/m c) determine the average power density in the wave in W/m2: Use Pavg = 1 2 Re {Es ×H∗s} = 1 2 [ (10)2 377 ax + (15)2 377 ax ] = 0.43ax W/m2 or Pavg = 0.43 W/m2 12.29. Consider a left-circularly polarized wave in free space that propagates in the forward z direc- tion. The electric field is given by the appropriate form of Eq. (100). a) Determine the magnetic field phasor, Hs: We begin, using (100), with Es = E0(ax + jay)e−jβz. We find the two components of Hs separately, using the two components of Es. Specifically, the x component of Es is associated with a y component of Hs, and the y component of Es is associated with a negative x component of Hs. The result is Hs = E0 η0 (ay − jax) e−jβz b) Determine an expression for the average power density in the wave in W/m2 by direct application of Eq. (77): We have Pz,avg = 1 2 Re(Es ×H∗s) = 1 2 Re ( E0(ax + jay)e−jβz × E0 η0 (ay − jax)e+jβz ) = E20 η0 az W/m2 (assuming E0 is real) 16 12.30. The electric field of a uniform plane wave in free space is given by Es = 10(az + jax)e−j50y. Determine: a) f : From the given field, we identify β = 50 = ω/c (in free space), so that f = ω/2π = 50c/2π = 2.39 GHz. b) Hs: Each of the two components of Es must pair with a magnetic field vector, such that the cross product of electric with magnetic field gives a vector in the positive y direction. The overall magnitude is the electric field magnitude divided by the free space intrinsic impedance. Thus Hs = 10 377 (ax − jaz) e−j50y c) = 1 2 Re{Es ×H∗s} = 50 377 [(az × ax)− (ax × az)] = 100377 ay = 0.27ay W/m 2 d) Describe the polarization of the wave: This can be seen by writing the electric field in real instantaneous form, and then evaluating the result at y = 0: E(0, t) = 10 [cos(ωt)az − sin(ωt)ax] At t = 0, the field is entirely along z, and then acquires an increasing negative x com- ponent as t increases. The field therefore rotates clockwise in the y = 0 plane when looking back toward the plane from positive y. Since the wave propagates in the positive y direction and has equal x and z amplitudes, we identify the polarization as left circular. 12.31. A linearly-polarized uniform plane wave, propagating in the forward z direction, is input to a lossless anisotropic material, in which the dielectric constant encountered by waves polarized along y (ry) differs from that seen by waves polarized along x (rx). Suppose rx = 2.15, ry = 2.10, and the wave electric field at input is polarized at 45◦ to the positive x and y axes. Assume free space wavelength λ. a) Determine the shortest length of the material such that the wave as it emerges from the output end is circularly polarized: With the input field at 45◦, the x and y components are of equal magnitude, and circular polarization will result if the phase difference between the components is π/2. Our requirement over length L is thus βxL− βyL = π/2, or L = π 2(βx − βy) = πc 2ω( √ rx −√ry) With the given values, we find, L = (58.3)πc 2ω = 58.3 λ 4 = 14.6λ b) Will the output wave be right- or left-circularly-polarized? With the dielectric constant greater for x-polarized waves, the x component will lag the y component in time at the out- put. The field can thus be written as E = E0(ay−jax), which is left circular polarization. 17 12.32. Suppose that the length of the medium of Problem 12.31 is made to be twice that as determined in the problem. Describe the polarization of the output wave in this case: With the length doubled, a phase shift of π radians develops between the two components. At the input, we can write the field as Es(0) = E0(ax + ay). After propagating through length L, we would have, Es(L) = E0[e−jβxLax + e−jβyLay] = E0e−jβxL[ax + e−j(βy−βx)Lay] where (βy−βx)L = −π (since βx > βy), and so Es(L) = E0e−jβxL[ax−ay]. With the reversal of the y component, the wave polarization is rotated by 90◦, but is still linear polarization. 12.33. Given a wave for which Es = 15e−jβzax + 18e−jβzejφay V/m, propagating in a medium characterized by complex intrinsic impedance, η. a) Find Hs: With the wave propagating in the forward z direction, we find: Hs = 1 η [−18ejφax + 15ay] e−jβz A/m b) Determine the average power density in W/m2: We find Pz,avg = 1 2 Re {Es ×H∗s} = 1 2 Re { (15)2 η∗ + (18)2 η∗ } = 275 Re { 1 η∗ } W/m2 12.34. Given the general elliptically-polarized wave as per Eq. (93): Es = [Ex0ax + Ey0ejφay]e−jβz a) Show, using methods similar to those of Example 12.7, that a linearly polarized wave results when superimposing the given field and a phase-shifted field of the form: Es = [Ex0ax + Ey0e−jφay]e−jβzejδ where δ is a constant: Adding the two fields gives Es,tot = [ Ex0 ( 1 + ejδ ) ax + Ey0 ( ejφ + e−jφejδ ) ay ] e−jβz =  Ex0ejδ/2 (e−jδ/2 + ejδ/2)︸ ︷︷ ︸ 2 cos(δ/2) ax + Ey0ejδ/2 ( e−jδ/2ejφ + e−jφejδ/2 ) ︸ ︷︷ ︸ 2 cos(φ−δ/2) ay   e−jβz This simplifies to Es,tot = 2 [Ex0 cos(δ/2)ax + Ey0 cos(φ− δ/2)ay] ejδ/2e−jβz, which is linearly polarized. b) Find δ in terms of φ such that the resultant wave is polarized along x: By inspecting the part a result, we achieve a zero y component when 2φ− δ = π (or odd multiples of π). 18 CHAPTER 13 13.1. A uniform plane wave in air, E+x1 = E + x10 cos(10 10t − βz) V/m, is normally-incident on a copper surface at z = 0. What percentage of the incident power density is transmitted into the copper? We need to find the reflection coefficient. The intrinsic impedance of copper (a good conductor) is ηc = √ jωµ σ = (1 + j) √ ωµ 2σ = (1 + j) √ 1010(4π × 107) 2(5.8× 107) = (1 + j)(.0104) Note that the accuracy here is questionable, since we know the conductivity to only two significant figures. We nevertheless proceed: Using η0 = 376.7288 ohms, we write Γ = ηc − η0 ηc + η0 = .0104− 376.7288 + j.0104 .0104 + 376.7288 + j.0104 = −.9999 + j.0001 Now |Γ|2 = .9999, and so the transmitted power fraction is 1− |Γ|2 = .0001, or about 0.01% is transmitted. 13.2. The plane z = 0 defines the boundary between two dielectrics. For z < 0, r1 = 5, ′′r1 = 0, and µ1 = µ0. For z > 0, ′r2 = 3,  ′′ r2 = 0, and µ2 = µ0. Let E + x1 = 200 cos(ωt− 15z) V/m and find a) ω: We have β = ω √ µ0′1 = ω √ ′r1/c = 15. So ω = 15c/ √ ′r1 = 15 × (3 × 108)/ √ 5 = 2.0× 109 s−1. b) : First we need η1 = √ µ0/′1 = η0/ √ ′r1 = 377/ √ 5 = 169 ohms. Next we apply Eq. (76), Chapter 12, to evaluate the Poynting vector (with no loss and consequently with no phase difference between electric and magnetic fields). We find = (1/2)|E1|2/η1 az = (1/2)(200)2/169az = 119az W/m2. c) : First, we need to evaluate the reflection coefficient: Γ = η2 − η1 η2 + η1 = η0/ √ ′r2 − η0/ √ ′r1 η0/ √ ′r2 + η0/ √ ′r1 = √ ′r1 − √ ′r2√ ′r1 + √ ′r2 = √ 5−√3√ 5 + √ 3 = 0.13 Then = −|Γ|2 = −(0.13)2(119)az = −2.0az W/m2. d) : This will be the remaining power, propagating in the forward z direction, or = 117az W/m 2. 13.3. A uniform plane wave in region 1 is normally-incident on the planar boundary separating regions 1 and 2. If ′′1 =  ′′ 2 = 0, while  ′ r1 = µ 3 r1 and  ′ r2 = µ 3 r2, find the ratio  ′ r2/ ′ r1 if 20% of the energy in the incident wave is reflected at the boundary. There are two possible answers. First, since |Γ|2 = .20, and since both permittivities and permeabilities are real, Γ = ±0.447. we then set up Γ = ±0.447 = η2 − η1 η2 + η1 = η0 √ (µr2/′r2)− η0 √ (µr1/′r1) η0 √ (µr2/′r2) + η0 √ (µr1/′r1) = √ (µr2/µ3r2)− √ (µr1/µ3r1)√ (µr2/µ3r2) + √ (µr1/µ3r1) = µr1 − µr2 µr1 + µr2 1 13.3. (continued) Therefore µr2 µr1 = 1∓ 0.447 1± 0.447 = (0.382, 2.62) ⇒ ′r2 ′r1 = ( µr2 µr1 )3 = (0.056, 17.9) 13.4. A 10-MHz uniform plane wave having an initial average power density of 5W/m2 is normally- incident from free space onto the surface of a lossy material in which ′′2/ ′ 2 = 0.05,  ′ r2 = 5, and µ2 = µ0. Calculate the distance into the lossy medium at which the transmitted wave power density is down by 10dB from the initial 5W/m2: First, since ′′2/ ′ 2 = 0.05 << 1, we recognize region 2 as a good dielectric. Its intrinsic impedance is therefore approximated well by Eq. (62b), Chapter 12: η2 = √ µ0 ′2 [ 1 + j 1 2 ′′2 ′2 ] = 377√ 5 [1 + j0.025] The reflection coefficient encountered by the incident wave from region 1 is therefore Γ = η2 − η1 η2 + η1 = (377/ √ 5)[1 + j.025]− 377 (377/ √ 5)[1 + j.025] + 377 = (1−√5) + j.025 (1 + √ 5) + j.025 = −0.383 + j0.011 The fraction of the incident power that is reflected is then |Γ|2 = 0.147, and thus the fraction of the power that is transmitted into region 2 is 1− |Γ|2 = 0.853. Still using the good dielectric approximation, the attenuation coefficient in region 2 is found from Eq. (60a), Chapter 12: α .= ω′′2 2 √ µ0 ′2 = (2π × 107)(0.05× 5× 8.854× 10−12) 377 2 √ 5 = 2.34× 10−2 Np/m Now, the power that propagates into region 2 is expressed in terms of the incident power through (z) = 5(1− |Γ|2)e−2αz = 5(.853)e−2(2.34×10−2)z = 0.5 W/m2 in which the last equality indicates a factor of ten reduction from the incident power, as occurs for a 10 dB loss. Solve for z to obtain z = ln(8.53) 2(2.34× 10−2) = 45.8 m 13.5. The region z < 0 is characterized by ′r = µr = 1 and  ′′ r = 0. The total E field here is given as the sum of the two uniform plane waves, Es = 150e−j10z ax + (50 20◦)ej10z ax V/m. a) What is the operating frequency? In free space, β = k0 = 10 = ω/c = ω/3× 108. Thus, ω = 3× 109 s−1, or f = ω/2π = 4.7× 108 Hz. b) Specify the intrinsic impedance of the region z > 0 that would provide the appropriate reflected wave: Use Γ = Er Einc = 50ej20 ◦ 150 = 1 3 ej20 ◦ = 0.31 + j0.11 = η − η0 η + η0 2 13.5 (continued) Now η = η0 ( 1 + Γ 1− Γ ) = 377 ( 1 + 0.31 + j0.11 1− 0.31− j0.31 ) = 691 + j177 Ω c) At what value of z (−10 cm < z < 0) is the total electric field intensity a maximum amplitude? We found the phase of the reflection coefficient to be φ = 20◦ = .349rad, and we use zmax = −φ 2β = −.349 20 = −0.017 m = −1.7 cm 13.6. Region 1, z 0, are described by the following parameters: ′1 = 100 pF/m, µ1 = 25 µH/m, ′′1 = 0,  ′ 2 = 200 pF/m, µ2 = 50 µH/m, and  ′′ 2/ ′ 2 = 0.5. If E+1 = 5e −α1z cos(4× 109t− β1z)ax V/m, find: a) α1: As ′′1 = 0, there is no loss mechanism that is modeled (see Eq. (44), Chapter 12), and so α1 = 0. b) β1: Since region 1 is lossless, the phase constant for the uniform plane wave will be β1 = ω √ µ1′1 = (4× 109) √ (25× 10−6)(100× 10−12) = 200 rad/m c) : To find the power density, we need the intrinsic impedance of region 1, given by η1 = √ µ1 ′1 = √ 25× 10−6 100× 10−12 = 500 ohms Then the incident power density will be = 1 2η1 |E1|2 az = 5 2 2(500) az = 25az mW/m2 d) : To find the reflected power, we need the intrinsic impedance of region 2. This is found using Eq. (48), Chapter 12: η2 = √ µ2 ′2 1√ 1− j(′′2/′2) = √ 50× 10−6 200× 10−12 1√ 1− j0.5 = 460 + j109 ohms Then the reflection coefficient at the 1-2 boundary is Γ = η2 − η1 η2 + η1 = 460 + j109− 500 460 + j109 + 500 = −0.028 + j0.117 The reflected power fraction is then |Γ|2 = 1.44× 10−2. Therefore = − |Γ|2 = −0.36az mW/m2. 3 13.6e) : We first need the attenuation coefficient in region 2. This is given by Eq. (44) in Chapter 12, which in our case becomes α2 = ω √ µ2′2 2   √ 1 + ( ′′2 ′2 )2 − 1  1/2 = (4× 109) [ (50× 10−6)(200× 10−12) 2 ]1/2 [√ 1 + 0.25− 1]1/2 = 97.2 Np/m Now =< S + 1 > (1− |Γ|2) e−2α2z = 25(0.986)e−2(97.2)z az = 24.7e−194z az mW/m2 Note the approximately 1 cm penetration depth. 13.7. The semi-infinite regions z 1 m are free space. For 0 < z < 1 m, ′r = 4, µr = 1, and ′′r = 0. A uniform plane wave with ω = 4 × 108 rad/s is travelling in the az direction toward the interface at z = 0. a) Find the standing wave ratio in each of the three regions: First we find the phase constant in the middle region, β2 = ω √ ′r c = 2(4× 108) 3× 108 = 2.67 rad/m Then, with the middle layer thickness of 1 m, β2d = 2.67 rad. Also, the intrinsic impedance of the middle layer is η2 = η0/ √ ′r = η0/2. We now find the input impedance: ηin = η2 [ η0 cos(β2d) + jη2 sin(β2d) η2 cos(β2d) + jη0 sin(β2d) ] = 377 2 [ 2 cos(2.67) + j sin(2.67) cos(2.67) + j2 sin(2.67) ] = 231 + j141 Now, at the first interface, Γ12 = ηin − η0 ηin + η0 = 231 + j141− 377 231 + j141 + 377 = −.176 + j.273 = .325  123◦ The standing wave ratio measured in region 1 is thus s1 = 1 + |Γ12| 1− |Γ12| = 1 + 0.325 1− 0.325 = 1.96 In region 2 the standing wave ratio is found by considering the reflection coefficient for waves incident from region 2 on the second interface: Γ23 = η0 − η0/2 η0 + η0/2 = 1− 1/2 1 + 1/2 = 1 3 Then s2 = 1 + 1/3 1− 1/3 = 2 Finally, s3 = 1, since no reflected waves exist in region 3. 4 13.7b. Find the location of the maximum |E| for z < 0 that is nearest to z = 0. We note that the phase of Γ12 is φ = 123◦ = 2.15 rad. Thus zmax = −φ 2β = −2.15 2(4/3) = −.81 m 13.8. A wave starts at point a, propagates 100m through a lossy dielectric for which α = 0.5 Np/m, reflects at normal incidence at a boundary at which Γ = 0.3 + j0.4, and then returns to point a. Calculate the ratio of the final power to the incident power after this round trip: Final power, Pf , and incident power, Pi, are related through Pf = Pie−2αL|Γ|2e−2αL ⇒ Pf Pi = |0.3 + j0.4|2e−4(0.5)100 = 3.5× 10−88(!) Try measuring that. 13.9. Region 1, z 0, are both perfect dielectrics (µ = µ0, ′′ = 0). A uniform plane wave traveling in the az direction has a radian frequency of 3× 1010 rad/s. Its wavelengths in the two regions are λ1 = 5 cm and λ2 = 3 cm. What percentage of the energy incident on the boundary is a) reflected; We first note that ′r1 = ( 2πc λ1ω )2 and ′r2 = ( 2πc λ2ω )2 Therefore ′r1/ ′ r2 = (λ2/λ1) 2. Then with µ = µ0 in both regions, we find Γ = η2 − η1 η2 + η1 = η0 √ 1/′r2 − η0 √ 1/′r1 η0 √ 1/′r2 + η0 √ 1/′r1 = √ ′r1/ ′ r2 − 1√ ′r1/ ′ r2 + 1 = (λ2/λ1)− 1 (λ2/λ1) + 1 = λ2 − λ1 λ2 + λ1 = 3− 5 3 + 5 = −1 4 The fraction of the incident energy that is reflected is then |Γ|2 = 1/16 = 6.25× 10−2. b) transmitted? We use part a and find the transmitted fraction to be 1− |Γ|2 = 15/16 = 0.938. c) What is the standing wave ratio in region 1? Use s = 1 + |Γ| 1− |Γ| = 1 + 1/4 1− 1/4 = 5 3 = 1.67 5 13.10. In Fig. 13.1, let region 2 be free space, while µr1 = 1, ′′r1 = 0, and  ′ r1 is unknown. Find  ′ r1 if a) the amplitude of E−1 is one-half that of E + 1 : Since region 2 is free space, the reflection coefficient is Γ = |E−1 | |E+1 | = η0 − η1 η0 + η1 = η0 − η0/ √ ′r1 η0 + η0/ √ ′r1 = √ ′r1 − 1√ ′r1 + 1 = 1 2 ⇒ ′r1 = 9 . b) is one-half of < S + 1 >: This time |Γ|2 = ∣∣∣∣∣ √ ′r1 − 1√ ′r1 + 1 ∣∣∣∣∣ 2 = 1 2 ⇒ ′r1 = 34 c) |E1|min is one-half |E1|max: Use |E1|max |E1|min = s = 1 + |Γ| 1− |Γ| = 2 ⇒ |Γ| = Γ = 1 3 = √ ′r1 − 1√ ′r1 + 1 ⇒ ′r1 = 4 13.11. A 150 MHz uniform plane wave in normally-incident from air onto a material whose intrinsic impedance is unknown. Measurements yield a standing wave ratio of 3 and the appearance of an electric field minimum at 0.3 wavelengths in front of the interface. Determine the impedance of the unknown material: First, the field minimum is used to find the phase of the reflection coefficient, where zmin = − 12β (φ + π) = −0.3λ ⇒ φ = 0.2π where β = 2π/λ has been used. Next, |Γ| = s− 1 s + 1 = 3− 1 3 + 1 = 1 2 So we now have Γ = 0.5ej0.2π = ηu − η0 ηu + η0 We solve for ηu to find ηu = η0(1.70 + j1.33) = 641 + j501 Ω 13.12. A 50MHz uniform plane wave is normally incident from air onto the surface of a calm ocean. For seawater, σ = 4 S/m, and ′r = 78. a) Determine the fractions of the incident power that are reflected and transmitted: First we find the loss tangent: σ ω′ = 4 2π(50× 106)(78)(8.854× 10−12) = 18.4 This value is sufficiently greater than 1 to enable seawater to be considered a good con- ductor at 50MHz. Then, using the approximation (Eq. 65, Chapter 11), the intrinsic impedance is ηs = √ πfµ/σ(1 + j), and the reflection coefficient becomes Γ = √ πfµ/σ (1 + j)− η0√ πfµ/σ (1 + j) + η0 6 13.12 (continued) where √ πfµ/σ = √ π(50× 106)(4π × 10−7)/4 = 7.0. The fraction of the power reflected is Pr Pi = |Γ|2 = [ √ πfµ/σ − η0]2 + πfµ/σ [ √ πfµ/σ + η0]2 + πfµ/σ = [7.0− 377]2 + 49.0 [7.0 + 377]2 + 49.0 = 0.93 The transmitted fraction is then Pt Pi = 1− |Γ|2 = 1− 0.93 = 0.07 b) Qualitatively, how will these answers change (if at all) as the frequency is increased? Within the limits of our good conductor approximation (loss tangent greater than about ten), the reflected power fraction, using the formula derived in part a, is found to decrease with increasing frequency. The transmitted power fraction thus increases. 13.13. A right-circularly-polarized plane wave is normally incident from air onto a semi-infinite slab of plexiglas (′r = 3.45,  ′′ r = 0). Calculate the fractions of the incident power that are reflected and transmitted. Also, describe the polarizations of the reflected and transmitted waves. First, the impedance of the plexiglas will be η = η0/ √ 3.45 = 203 Ω. Then Γ = 203− 377 203 + 377 = −0.30 The reflected power fraction is thus |Γ|2 = 0.09. The total electric field in the plane of the interface must rotate in the same direction as the incident field, in order to continu- ally satisfy the boundary condition of tangential electric field continuity across the interface. Therefore, the reflected wave will have to be left circularly polarized in order to make this happen. The transmitted power fraction is now 1− |Γ|2 = 0.91. The transmitted field will be right circularly polarized (as the incident field) for the same reasons. 13.14. A left-circularly-polarized plane wave is normally-incident onto the surface of a perfect con- ductor. a) Construct the superposition of the incident and reflected waves in phasor form: Assume positive z travel for the incident electric field. Then, with reflection coefficient, Γ = −1, the incident and reflected fields will add to give the total field: Etot = Ei + Er = E0(ax + jay)e−jβz − E0(ax + jay)e+jβz = E0  (e−jβz − ejβz)︸ ︷︷ ︸ −2j sin(βz) ax + j ( e−jβz − ejβz)︸ ︷︷ ︸ −2j sin(βz) ay   = 2E0 sin(βz) [ay − jax] b) Determine the real instantaneous form of the result of part a: E(z, t) = Re { Etotejωt } = 2E0 sin(βz) [cos(ωt)ay + sin(ωt)ax] c) Describe the wave that is formed: This is a standing wave exhibiting circular polarization in time. At each location along the z axis, the field vector rotates clockwise in the xy plane, and has amplitude (constant with time) given by 2E0 sin(βz). 7 13.15. Consider these regions in which ′′ = 0: region 1, z < 0, µ1 = 4µH/m and ′1 = 10 pF/m; region 2, 0 6 cm, µ3 = µ1 and  ′ 3 =  ′ 1. a) What is the lowest frequency at which a uniform plane wave incident from region 1 onto the boundary at z = 0 will have no reflection? This frequency gives the condition β2d = π, where d = 6 cm, and β2 = ω √ µ2′2 Therefore β2d = π ⇒ ω = π (.06) √ µ2′2 ⇒ f = 1 0.12 √ (2× 10−6)(25× 10−12) = 1.2 GHz b) If f = 50 MHz, what will the standing wave ratio be in region 1? At the given frequency, β2 = (2π×5×107) √ (2× 10−6)(25× 10−12) = 2.22 rad/m. Thus β2d = 2.22(.06) = 0.133. The intrinsic impedance of regions 1 and 3 is η1 = η3 = √ (4× 10−6)/(10−11) = 632 Ω. The input impedance at the first interface is now ηin = 283 [ 632 cos(.133) + j283 sin(.133) 283 cos(.133) + j632 sin(.133) ] = 589− j138 = 605 − .23 The reflection coefficient is now Γ = ηin − η1 ηin + η1 = 589− j138− 632 589− j138 + 632 = .12  − 1.7 The standing wave ratio is now s = 1 + |Γ| 1− |Γ| = 1 + .12 1− .12 = 1.27 13.16. A uniform plane wave in air is normally-incident onto a lossless dielectric plate of thickness λ/8, and of intrinsic impedance η = 260 Ω. Determine the standing wave ratio in front of the plate. Also find the fraction of the incident power that is transmitted to the other side of the plate: With the a thickness of λ/8, we have βd = π/4, and so cos(βd) = sin(βd) = 1 √ 2. The input impedance thus becomes ηin = 260 [ 377 + j260 260 + j377 ] = 243− j92 Ω The reflection coefficient is then Γ = (243− j92)− 377 (243− j92) + 377 = −0.19− j0.18 = 0.26  − 2.4rad Therefore s = 1 + .26 1− .26 = 1.7 and 1− |Γ| 2 = 1− (.26)2 = 0.93 8 13.17. Repeat Problem 13.16 for the cases in which the frequency is a) doubled: If this is true, then d = λ/4, and thus ηin = (260)2/377 = 179. The reflection coefficient becomes Γ = 179− 377 179 + 377 = −0.36 ⇒ s = 1 + .36 1− .36 = 2.13 Then 1− |Γ|2 = 1− (.36)2 = 0.87. b) quadrupled: Now, d = λ/2, and so we have a half-wave section surrounded by air. Trans- mission will be total, and so s = 1 and 1− |Γ|2 = 1. 13.18. A uniform plane wave is normally-incident onto a slab of glass (n = 1.45) whose back surface is in contact with a perfect conductor. Determine the reflective phase shift at the front surface of the glass if the glass thickness is: (a) λ/2; (b) λ/4; (c) λ/8. With region 3 being a perfect conductor, η3 = 0, and Eq. (36) gives the input impedance to the structure as ηin = jη2 tanβ . The reflection coefficient is then Γ = ηin − η0 ηin + η0 = jη2 tanβ − η0 jη2 tanβ + η0 = η22 tan 2 β − η20 + j2η0η2 tanβ η22 tan 2 β + η20 = Γr + jΓi where the last equality occurs by multiplying the numerator and denominator of the middle term by the complex conjugate of its denominator. The reflective phase is now φ = tan−1 ( Γi Γr ) = tan−1 [ 2η2η0 tanβ η22 tan 2 β − η20 ] = tan−1 [ (2.90) tanβ tanβ − 2.10 ] where η2 = η0/1.45 has been used. We can now evaluate the phase shift for the three given cases. First, when = λ/2, β = π, and thus φ(λ/2) = 0. Next, when = λ/4, β = π/2, and φ(λ/4)→ tan−1 [2.90] = 71◦ as → λ/4. Finally, when = λ/8, β = π/4, and φ(λ/8) = tan−1 [ 2.90 1− 2.10 ] = −69.2◦ (or 291◦) 9 13.19. You are given four slabs of lossless dielectric, all with the same intrinsic impedance, η, known to be different from that of free space. The thickness of each slab is λ/4, where λ is the wavelength as measured in the slab material. The slabs are to be positioned parallel to one another, and the combination lies in the path of a uniform plane wave, normally-incident. The slabs are to be arranged such that the air spaces between them are either zero, one-quarter wavelength, or one-half wavelength in thickness. Specify an arrangement of slabs and air spaces such that a) the wave is totally transmitted through the stack: In this case, we look for a combination of half-wave sections. Let the inter-slab distances be d1, d2, and d3 (from left to right). Two possibilities are i.) d1 = d2 = d3 = 0, thus creating a single section of thickness λ, or ii.) d1 = d3 = 0, d2 = λ/2, thus yielding two half-wave sections separated by a half- wavelength. b) the stack presents the highest reflectivity to the incident wave: The best choice here is to make d1 = d2 = d3 = λ/4. Thus every thickness is one-quarter wavelength. The impedances transform as follows: First, the input impedance at the front surface of the last slab (slab 4) is ηin,1 = η2/η0. We transform this back to the back surface of slab 3, moving through a distance of λ/4 in free space: ηin,2 = η20/ηin,1 = η 3 0/η 2. We next transform this impedance to the front surface of slab 3, producing ηin,3 = η2/ηin,2 = η4/η30 . We continue in this manner until reaching the front surface of slab 1, where we find ηin,7 = η8/η70 . Assuming η < η0, the ratio ηn/ηn−10 becomes smaller as n increases (as the number of slabs increases). The reflection coefficient for waves incident on the front slab thus gets close to unity, and approaches 1 as the number of slabs approaches infinity. 13.20. The 50MHz plane wave of Problem 13.12 is incident onto the ocean surface at an angle to the normal of 60◦. Determine the fractions of the incident power that are reflected and transmitted for a) s polarization: To review Problem 12, we first we find the loss tangent: σ ω′ = 4 2π(50× 106)(78)(8.854× 10−12) = 18.4 This value is sufficiently greater than 1 to enable seawater to be considered a good con- ductor at 50MHz. Then, using the approximation (Eq. 65, Chapter 11), and with µ = µ0, the intrinsic impedance is ηs = √ πfµ/σ(1 + j) = 7.0(1 + j). Next we need the angle of refraction, which means that we need to know the refractive index of seawater at 50MHz. For a uniform plane wave in a good conductor, the phase constant is β = nsea ω c .= √ πfµσ ⇒ nsea .= c √ µσ 4πf = 26.8 Then, using Snell’s law, the angle of refraction is found: sin θ2 = nsea n1 sin θ1 = 26.8 sin(60◦) ⇒ θ2 = 1.9◦ This angle is small enough so that cos θ2 .= 1. Therefore, for s polarization, Γs .= ηs2 − ηs1 ηs2 + ηs1 = 7.0(1 + j)− 377/ cos 60◦ 7.0(1 + j) + 377/ cos 60◦ = −0.98 + j0.018 = 0.98  179◦ 10 3.20a (continued) The fraction of the power reflected is now |Γs|2 = 0.96. The fraction transmitted is then 0.04. b) p polarization: Again, with the refracted angle close to zero, the relection coefficient for p polarization is Γp .= ηp2 − ηp1 ηp2 + ηp1 = 7.0(1 + j)− 377 cos 60◦ 7.0(1 + j) + 377 cos 60◦ = −0.93 + j0.069 = 0.93  176◦ The fraction of the power reflected is now |Γp|2 = 0.86. The fraction transmitted is then 0.14. 13.21. A right-circularly polarized plane wave in air is incident at Brewster’s angle onto a semi-infinite slab of plexiglas (′r = 3.45,  ′′ r = 0, µ = µ0). a) Determine the fractions of the incident power that are reflected and transmitted: In plexiglas, Brewster’s angle is θB = θ1 = tan−1(′r2/ ′ r1) = tan −1( √ 3.45) = 61.7◦. Then the angle of refraction is θ2 = 90◦−θB (see Example 13.9), or θ2 = 28.3◦. With incidence at Brewster’s angle, all p-polarized power will be transmitted — only s-polarized power will be reflected. This is found through Γs = η2s − η1s η2s + η1s = .614η0 − 2.11η0 .614η0 + 2.11η0 = −0.549 where η1s = η1 sec θ1 = η0 sec(61.7◦) = 2.11η0, and η2s = η2 sec θ2 = (η0/ √ 3.45) sec(28.3◦) = 0.614η0. Now, the reflected power fraction is |Γ|2 = (−.549)2 = .302. Since the wave is circularly-polarized, the s-polarized compo- nent represents one-half the total incident wave power, and so the fraction of the total power that is reflected is .302/2 = 0.15, or 15%. The fraction of the incident power that is transmitted is then the remainder, or 85%. b) Describe the polarizations of the reflected and transmitted waves: Since all the p-polarized component is transmitted, the reflected wave will be entirely s-polarized (linear). The transmitted wave, while having all the incident p-polarized power, will have a reduced s-component, and so this wave will be right-elliptically polarized. 13.22. A dielectric waveguide is shown in Fig. 13.16 with refractive indices as labeled. Incident light enters the guide at angle φ from the front surface normal as shown. Once inside, the light totally reflects at the upper n1 − n2 interface, where n1 > n2. All subsequent reflections from the upper an lower boundaries will be total as well, and so the light is confined to the guide. Express, in terms of n1 and n2, the maximum value of φ such that total confinement will occur, with n0 = 1. The quantity sinφ is known as the numerical aperture of the guide. From the illustration we see that φ1 maximizes when θ1 is at its minimum value. This minimum will be the critical angle for the n1 − n2 interface, where sin θc = sin θ1 = n2/n1. Let the refracted angle to the right of the vertical interface (not shown) be φ2, where n0 sinφ1 = n1 sinφ2. Then we see that φ2 +θ1 = 90◦, and so sin θ1 = cosφ2. Now, the numerical aperture becomes sinφ1max = n1 n0 sinφ2 = n1 cos θ1 = n1 √ 1− sin2 θ1 = n1 √ 1− (n2/n1)2 = √ n21 − n22 Finally, φ1max = sin−1 (√ n21 − n22 ) is the numerical aperture angle. 11 13.23. Suppose that φ1 in Fig. 13.16 is Brewster’s angle, and that θ1 is the critical angle. Find n0 in terms of n1 and n2: With the incoming ray at Brewster’s angle, the refracted angle of this ray (measured from the inside normal to the front surface) will be 90◦ − φ1. Therefore, φ1 = θ1, and thus sinφ1 = sin θ1. Thus sinφ1 = n1√ n20 + n 2 1 = sin θ1 = n2 n1 ⇒ n0 = (n1/n2) √ n21 − n22 Alternatively, we could have used the result of Problem 13.22, in which it was found that sinφ1 = (1/n0) √ n21 − n22, which we then set equal to sin θ1 = n2/n1 to get the same result. 13.24. A Brewster prism is designed to pass p-polarized light without any reflective loss. The prism of Fig. 13.17 is made of glass (n = 1.45), and is in air. Considering the light path shown, determine the apex angle, α: With entrance and exit rays at Brewster’s angle (to eliminate reflective loss), the interior ray must be horizontal, or parallel to the bottom surface of the prism. From the geometry, the angle between the interior ray and the normal to the prism surfaces that it intersects is α/2. Since this angle is also Brewster’s angle, we may write: α = 2 sin−1 ( 1√ 1 + n2 ) = 2 sin−1 ( 1√ 1 + (1.45)2 ) = 1.21 rad = 69.2◦ 13.25. In the Brewster prism of Fig. 13.17, determine for s-polarized light the fraction of the incident power that is transmitted through the prism: We use Γs = (ηs2 − ηs1)/(ηs2 + ηs1), where ηs2 = η2 cos(θB2) = η2 n/ √ 1 + n2 = η0 n2 √ 1 + n2 and ηs1 = η1 cos(θB1) = η1 1/ √ 1 + n2 = η0 √ 1 + n2 Thus, at the first interface, Γ = (1−n2)/(1+n2). At the second interface, Γ will be equal but of opposite sign to the above value. The power transmission coefficient through each interface is 1− |Γ|2, so that for both interfaces, we have, with n = 1.45: Ptr Pinc = ( 1− |Γ|2)2 = [ 1− ( n2 − 1 n2 + 1 )2]2 = 0.76 12 13.26. Show how a single block of glass can be used to turn a p-polarized beam of iight through 180◦, with the light suffering, in principle, zero reflective loss. The light is incident from air, and the returning beam (also in air) may be displaced sideways from the incident beam. Specify all pertinent angles and use n = 1.45 for glass. More than one design is possible here. The prism below is designed such that light enters at Brewster’s angle, and once inside, is turned around using total reflection. Using the result of Example 13.9, we find that with glass, θB = 55.4◦, which, by the geometry, is also the incident angle for total reflection at the back of the prism. For this to work, the Brewster angle must be greater than or equal to the critical angle. This is in fact the case, since θc = sin−1(n2/n1) = sin−1(1/1.45) = 43.6◦. 13.27. Using Eq. (79) in Chapter 12 as a starting point, determine the ratio of the group and phase velocities of an electromagnetic wave in a good conductor. Assume conductivity does not vary with frequency: In a good conductor: β = √ πfµσ = √ ωµσ 2 → dβ dω = 1 2 [ωµσ 2 ]−1/2 µσ 2 Thus dω dβ = ( dβ dω )−1 = 2 √ 2ω µσ = vg and vp = ω β = ω√ ωµσ/2 = √ 2ω µσ Therefore vg/vp = 2. 13 13.28. Over a small wavelength range, the refractive index of a certain material varies approximately linearly with wavelength as n(λ) .= na + nb(λ− λa), where na, nb, and λa are constants, and where λ is the free space wavelength. a) Show that d/dω = −(2πc/ω2)d/dλ: With λ as the free space wavelength, we use λ = 2πc/ω, from which dλ/dω = −2πc/ω2. Then d/dω = (dλ/dω) d/dλ = −(2πc/ω2) d/dλ. b) Using β(λ) = 2πn/λ, determine the wavelength-dependent (or independent) group delay over a unit distance: This will be tg = 1 vg = dβ dω = d dω [ 2πn(λ) λ ] = −2πc ω2 d dλ [ 2π λ [na + nb(λ− λa)] ] = −2πc ω2 [ −2π λ2 [na + nb(λ− λa)] + 2π λ nb ] = − λ 2 2πc [ −2πna λ2 + 2πnbλa λ2 ] = 1 c (na − nbλa) s/m c) Determine β2 from your result of part b: β2 = d2β/dω2|ω0 . Since the part b result is independent of wavelength (and of frequency), it follows that β2 = 0. d) Discuss the implications of these results, if any, on pulse broadening: A wavelength- independent group delay (leading to zero β2) means that there will simply be no pulse broadening at all. All frequency components arrive simultaneously. This sort of thing happens in most transparent materials – that is, there will be a certain wavelength, known as the zero dispersion wavelength, around which the variation of n with λ is locally linear. Transmitting pulses at this wavelength will result in no pulse broadening (to first order). 13.29. A T = 5 ps transform-limited pulse propagates in a dispersive channel for which β2 = 10 ps2/km. Over what distance will the pulse spread to twice its initial width? After prop- agation, the width is T ′ = √ T 2 + (∆τ)2 = 2T . Thus ∆τ = √ 3T , where ∆τ = β2z/T . Therefore β2z T = √ 3T or z = √ 3T 2 β2 = √ 3(5 ps)2 10 ps2/km = 4.3 km 13.30. A T = 20 ps transform-limited pulse propagates through 10 km of a dispersive channel for which β2 = 12 ps2/km. The pulse then propagates through a second 10 km channel for which β2 = −12 ps2/km. Describe the pulse at the output of the second channel and give a physical explanation for what happened. Our theory of pulse spreading will allow for changes in β2 down the length of the channel. In fact, we may write in general: ∆τ = 1 T ∫ L 0 β2(z) dz Having β2 change sign at the midpoint, yields a zero ∆τ , and so the pulse emerges from the output unchanged! Physically, the pulse acquires a positive linear chirp (frequency increases with time over the pulse envelope) during the first half of the channel. When β2 switches sign, the pulse begins to acquire a negative chirp in the second half, which, over an equal distance, will completely eliminate the chirp acquired during the first half. The pulse, if originally transform-limited at input, will emerge, again transform-limited, at its original width. More generally, complete dispersion compensation is achieved using a two-segment channel when β2L = −β′2L′, assuming dispersion terms of higher order than β2 do not exist. 14