Optics Solution Manual

OPTICS\nFourth Edition\n\nINSTRUCTOR'S SOLUTIONS\nMANUAL\n\nEugene Hecht\nAdelphi University\n\nMark Coffey\nUniversity of Colorado\n\nPaul Dolan\nNortheastern Illinois University\n\nAddison\nWesley\n\nSan Francisco Boston New York\nCape Town Hong Kong London \nMadrid Mexico City Montreal Munich Paris Singapore Sydney Tokyo Toronto ISBN 0-8033-8379-0\n\nCopyright © 2002 Pearson Education, Inc., publishing as Addison Wesley, 1301 Sansome St., San Francisco, CA 94111. All rights reserved. Manufactured in the United States of America. This publication is protected by copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form by any means, electronic, mechanical, photocopying, recording, or otherwise. To obtain permission to use material from this work, please submit a written request to Pearson Education, Inc., Permissions Department, 1900 E. Lake Ave., Glenview, IL 60025. For information regarding permissions, call 847/486-2653.\n\nMany of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks. Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed in initial caps or all.\n\n45678910—BTP—06 04 03 02 01\n\nwww.aw.com/physics Contents\n\nChapter 2 Solutions 1\nChapter 3 Solutions 7\nChapter 4 Solutions 15\nChapter 5 Solutions 30\nChapter 6 Solutions 45\nChapter 7 Solutions 52\nChapter 8 Solutions 61\nChapter 9 Solutions 72\nChapter 10 Solutions 80\nChapter 11 Solutions 87\nChapter 12 Solutions 93\nChapter 13 Solutions 97\n\niii (0.003)(2.54 x 10^-2/580 x 10^-9) = number of waves = 131, c = v/\\lambda, \\lambda = c/v = 3 x 10^8/10^10, \\lambda = 3 cm. Waves extend 3.9 m.\n\n2.2 \\lambda = c/v = 3 x 10^8/5 x 10^14 = 6 x 10^-7 m = 0.6\\mu m.\n\\lambda = 3 x 10^8/60 = 5 x 10^6 m = 5 x 10^6 km.\n\n2.3 v = v/\\lambda = 5 x 10^-7 x 6 x 10^6 = 300 m/s.\n\n2.4 The time between the crests is the period, so \\tau = 1/f; hence \\nu = 1/\\tau = 2.0 Hz. As for the speed v = L/t = 4.5 m/1.5 s = 3.0 m/s. We now know \\tau, v, and u and must determine \\lambda. Thus, \\lambda = v/\\nu = 3.0 m/s/2.0 Hz = 1.5 m.\n\n2.5 v = v/\\lambda = 3.5 x 10^3 m/s = \\sqrt{(4.3 m)}; \\nu = 0.81 kHz.\n\n2.6 v = \\lambda = 1498 m/s = (440 Hz)\\lambda; \\lambda = 3.40 m.\n\n2.7 v = (10 m)/2.0 s = 5.0 m/s; \\nu = v/\\lambda = (5.0 m/s)/(0.50 m) = 10 Hz.\n\n2.8 v = \\lambda = (\\omega/2\\pi)\\lambda and so \\omega = (2\\pi/\\lambda)v.\n\n2.9 \\begin{tabular}{|c|c|c|c|c|c|}\n\\hline\nθ & -\\frac{\\pi}{2} & -\\frac{\\pi}{4} & 0 & \\frac{\\pi}{4} & \\frac{\\pi}{2} & 3\\frac{\\pi}{4} \\\\\n\\hline\n\\sin \\theta & -1 & -\\frac{\\sqrt{2}}{2} & 0 & \\frac{\\sqrt{2}}{2} & 1 & \\frac{\\sqrt{2}}{2} \\\\\n\\cos \\theta & 0 & \\frac{\\sqrt{2}}{2} & 1 & \\frac{\\sqrt{2}}{2} & 0 & -\\frac{\\sqrt{2}}{2} \\\\\n\\sin(θ - \\frac{\\pi}{4}) & -\\frac{\\sqrt{2}}{2} & -1 & -\\frac{\\sqrt{2}}{2} & 0 & \\frac{\\sqrt{2}}{2} & \\frac{\\sqrt{2}}{2} \\\\\n\\sin(θ - \\frac{3}{4}) & 0 & -\\frac{\\sqrt{2}}{2} & 1 & -1 & -\\frac{\\sqrt{2}}{2} & 0 \\\\\n\\sin(θ + \\frac{\\pi}{2}) & 0 & \\frac{\\sqrt{2}}{2} & 1 & \\frac{\\sqrt{2}}{2} & 0 & -\\frac{\\sqrt{2}}{2} \\\\\n\\hline\n\\end{tabular} 2.10 x = -\\frac{\\lambda}{2} -\\frac{\\lambda}{4} + 0 + \\frac{\\lambda}{4} + \\frac{3\\lambda}{4} \\n\n\\kappa = 2\\pi/\\lambda\\n\\cos(kx - \\frac{\\pi}{2}) = -\\sin(kx - \\frac{\\pi}{2})\\n\\sin(kx + \\frac{3\\pi}{4})\\n2.11\\nu = \\frac{2\\pi}{\\tau} = -\\frac{3\\pi}{4}\\n\\sin(\\frac{kx - \\pi}{4})\\n\\sin(\\frac{\\pi}{4} - t)\\n2.12 Comparing \\nu with Eq. (2.13) tells us that A = 0.02 m. Moreover, 2\\pi/\\tau = 157 m^-1 and so \\lambda = 2\\pi/(157 m^-1) = 0.0400 m. The relationship between frequency and wavelength is \\nu = v/\\lambda, and so \\nu = \\frac{v}{\\lambda} = 1.2 m/s/0.0400 m = 30 Hz. The period is the inverse of the frequency, and therefore \\tau = 1/\\nu = 0.033 s.\n\n2.13 (a) \\lambda = (4.0 - 0.0) m = 4.0 m. (b) \\nu = v/\\lambda, so \\nu = \\frac{v}{\\lambda} = (20.0 m/s)/(4.0 m) = 5.0 Hz. (c) \\nu = v/\\lambda = (40.0 cm/s)/(5.0 cm) = 8.00 cm.\n\n2.14 (a) \\lambda = (30.0 - 0.0) cm = 30.0 cm. (c) \\nu = v/\\lambda, so \\nu = \\frac{v}{\\lambda} = (100 cm/s)/(30.0 cm) = 3.33 Hz. 2.15 (a) \\tau = (0.20 - 0.00) s = 0.20 s. (b) \\nu = 1/\\tau = 1/(0.20s) = 5.00 Hz. (c) v = \\lambda, so A = v = (440 cm/s)/(5.00 s) = 8.00 cm.\n\n2.16 \\psi = A \\sin(2\\pi(x/\\lambda - \\nu t)) = v = 4 sin(2\\pi(0.2x - 3t). (a) x = 1/0.2, (b) A = 1/3, (d) A = 15, (f) positive x \\n\\psi = A \\sin(kx - \\omega t + e). From the figure, A = 0.020 m; k = 2\\pi/\\lambda = 2\\pi/(4.0 m) = 0.5\\pi m^{-1}; \\omega = 2\\pi\\times(5.0 Hz) = 10.0\\pi rad/s.\\n\\nAt t = 0, x = 0, \\psi(0,0) = -0.020 m; \\psi(0, t) = (0.020 m) \\sin(0.5\\pi(0) - 10.0t(0) + e) = (0.020 m) \\sin(e); \\sin(e) = -1; c = -\\pi/2. \\psi(x, t) = (0.020 m) \\sin(0.5\\pi x - 10.0t - \\frac{\\pi}{2})\\n2.21 \\nu = -\\omega A \\cos(kx - \\omega t + e), a_y = -\\omega^2y. Simple harmonic motion since a_y \\propto y. \\n2.22 \\tau = 2.2 x 10^{-15} s; therefore \\nu = 1/\\tau = 4.5 x 10^{14} Hz; \\nu = v/\\lambda, \\lambda = v/\\nu = 6.6 x 10^{-7} m and k = 2\\pi/\\lambda = 9.5 x 10^6 m^{-1}.\\n\\psi(x,t) = (10^6 V/m) \\cos(9.5 x 10^6 m^{-1}(x + 3 x 10^8 (m/s)t)). \\text{It's cosine because cos 0 = 1.}\\n2.23 y(x,t) = C/2 + (x + vt)^{2}. [(4 - 0)z + (2 - 0)j + (1 - 0)i]e^{√(4 + 2² + 1²)} = (4i + 2j + k)/√21 and k = k(4i + 2j + k)/√21. r = xi + yj + zk, hence ψ(r, x, y, z, t) = A sin(4k/√21)x + (2k/√21)y + (k/√21)z - ωt.\n2.43 k = (1 + 0j + 0k), r = xi + yj + zk, so, ψ = A sin(k·r - ωt + e) = A sin(kz - ωt + e) where k = 2π/λ (could use cos instead of sin).\n2.44 ψ(r₁, t) = ψ(r₁ - r₂( r₂ - r₁), t) = ψ(k·r, t) = ψ(k·(r₂ - r₁), t) = ψ(k·r - (r₂ - r₁), t) = ψ(r₂ - r₁) = 0.\nψ(r₂ - r₁) = ψ(r₁, t) since k·(r₂ - r₁) = 0.\n2.45 θ = -π/2 - π/4 = 3π/4 = 5π/4, 3π/4 = 3/4, 5π/4 = 3π/4, sin(θ) = -1/√2, sin(2θ) = -1/√2, sin(3θ) = -√3/√2, sin(3θ) = 0, sin(3θ) = 1. \n2.46 sin(e - θ = -π/2) = sin(-π/2) = -1, (2θ + sin(θ - π/2) = sin(e - θ = -π/2) = 0, hence sin(e + θ) = 1.\n2.47 Note that the amplitude of {sin(θ) + sin(θ - π/2)} is greater than 1, the amplitude of {sin(θ) + sin(θ - 4π/3)} is less than 1. The phase difference is π/8.\n2.48 kx = -π/2 - π/2, π/2 π/2 2π.\ncos kx -1 0 1 0 -1 0 1\ncos(kx + π) 1 0 -1 0 1 0 -1 Chapter 3 Solutions\n\n3.1 Compare E_y = 2 cos(2π × 10^14(t - z/c) + π/2) to\nE_y = A cos[2πvt(t - z/v) + π/2]. (a) v = 10^14 Hz, v = c, and\nλ = c/v = 3 × 10^8/10^14 = 3 × 10^-6 m, moves in positive x-direction,\nA = 2 √2 V/m, ε = r/2 linearly polarized in the y-direction. (b) B_z = 0,\nB_y = 0, B_x = E_y/c.\n\n3.2 E_x = 0, E_y = E_0 sin(kz - ωt) or cosine; B_z = 0, B_y = -B_x = E_y/c,\nor if you like,\nE = E_0/√2 (i + j) sin(kz - ωt), B = E_0/c√2 (j - i) sin(kz - ωt).\n\n3.3 First, by the right-hand rule, the directions of the vectors are right. Then\nkE = ωB and so (2π/λ)E = ωB = 2πνB, hence E = λB = c.\n\n3.4 ∂E/∂x = -kE sin(kz - ωt); -∂B/∂t = -ωB sin(kz - ωt);\n-kE = -ωB; E_0 = -ωB_0 and Eq. (2.33) μ = κ = c.\n\n3.5 (a) The electric field oscillates along the line specified by the vector\n-i + √3; (b) To find E_0, dot E_0 with itself and take the square root, thus\nE_0 = √9 + 27 × 10^4 V/m = 6 × 10^4 V/m. (c) From the exponential\nk·r = (√5x + 2y)(3/7) × 10^7, hence k = (√5 + 2j) × 10^7 and\nbecause the phase is k·r - ωt rather than k·r + ωt the wave moves in the\ndirection of k. (d) k·k = (x × 10^7)², k = x × 10^7 m⁻¹ and\nλ = 2π/k = 200 nm. (e) ω = 9.42 × 10^{15} rad/s and\nν = ω/2π = 1.5 × 10^{15} Hz. (f) ν = νλ = 3.00 × 10^8 m/s.