Options Futures and Other Derivatives Solutions , John C. Hull , 5ed , Prentice Hall 5판 솔루션

Problem Solutions for Chapter 2 2-1. 2-2. E = 100cos (2 π108 t + 30°) e x + 20 cos (2π10 8t ? 50°) e y + 40cos (2π10 8 t + 210°) e z The general form is: y = (amplitude) cos(ωt - kz) = A cos [2π(νt - z/λ)]. Therefore (a) amplitude = 8 μm (b) wavelength: 1/λ = 0.8 μm-1 so that λ = 1.25 μm (c) ω = 2πν = 2π(2) = 4π (d) At t = 0 and z = 4 μm we have y = 8 cos [2π(-0.8 μm-1)(4 μm)] = 8 cos [2π(-3.2)] = 2.472 2-3. For E in electron volts and λ in μm we have E = (a) At 0.82 μm, E = 1.240/0.82 = 1.512 eV At 1.32 μm, E = 1.240/1.32 = 0.939 eV At 1.55 μm, E = 1.240/1.55 = 0.800 eV (b) At 0.82 μm, k = 2π/λ = 7.662 μm-1 At 1.32 μm, k = 2π/λ = 4.760 μm-1 At 1.55 μm, k = 2π/λ = 4.054 μm-1 1.240 λ
2-4.

x1 = a1 cos (ωt - δ1) and x2 = a2 cos (ωt - δ2) Adding x1 and x2 yields x1 + x2 = a1 [cos ωt cos δ1 + sin ωt sin δ1] + a2 [cos ωt cos δ2 + sin ωt sin δ2] = [a1 cos δ1 + a2 cos δ2] cos ωt + [a1 sin δ1 + a2 sin δ2] sin ωt Since the a`s an