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Problem Solutions for Chapter 2 2-1. 2-2. E = 100cos (2 ¥ð108 t + 30¡Æ) e x + 20 cos (2¥ð10 8t ? 50¡Æ) e y + 40cos (2¥ð10 8 t + 210¡Æ) e z The general form is: y = (amplitude) cos(¥øt - kz) = A cos [2¥ð(¥ít - z/¥ë)]. Therefore (a) amplitude = 8 ¥ìm (b) wavelength: 1/¥ë = 0.8 ¥ìm-1 so that ¥ë = 1.25 ¥ìm (c) ¥ø = 2¥ð¥í = 2¥ð(2) = 4¥ð (d) At t = 0 and z = 4 ¥ìm we have y = 8 cos [2¥ð(-0.8 ¥ìm-1)(4 ¥ìm)] = 8 cos [2¥ð(-3.2)] = 2.472 2-3. For E in electron volts and ¥ë in ¥ìm we have E = (a) At 0.82 ¥ìm, E = 1.240/0.82 = 1.512 eV At 1.32 ¥ìm, E = 1.240/1.32 = 0.939 eV At 1.55 ¥ìm, E = 1.240/1.55 = 0.800 eV (b) At 0.82 ¥ìm, k = 2¥ð/¥ë = 7.662 ¥ìm-1 At 1.32 ¥ìm, k = 2¥ð/¥ë = 4.760 ¥ìm-1 At 1.55 ¥ìm, k = 2¥ð/¥ë = 4.054 ¥ìm-1 1.240 ¥ë
2-4.
x1 = a1 cos (¥øt - ¥ä1) and x2 = a2 cos (¥øt - ¥ä2) Adding x1 and x2 yields x1 + x2 = a1 [cos ¥øt cos ¥ä1 + sin ¥øt sin ¥ä1] + a2 [cos ¥øt cos ¥ä2 + sin ¥øt sin ¥ä2] = [a1 cos ¥ä1 + a2 cos ¥ä2] cos ¥øt + [a1 sin ¥ä1 + a2 sin ¥ä2] sin ¥øt Since the a`s an
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