hw4_solution

# Hw4_solution - Power output =[Power input x[efficiency Power input =[Power output[efficiency = 2.6e6W/0.3 = 8.67e6 W A = Power/I =(8.67e6W(1100

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Physics 3C Webassign Homework 4 Solutions: a. t = D/v = (9.8e18m/3.e8m/s) = 3.26e10 s (1yr / 3.155 e7 s) = 1040 yr b. t = (7.78e11m)/(3.e8m/s) = 2593s = 43.2 minutes c. t = 2(3.84e8m)/(3.e8 m/s) = 2.56s d. t = 2pi(6.37e6m)/(3.e8 m/2) = 0.133s e. t = 1.01e4m/3.e8m/s = 3.37e-5 s E/B = c or 210/B = 3.e8 => B = 7e-7 T = 700nT a. B = E/c = (116 V/m)/(3.e8m/s) = 0.377e-6 T b. λ = 2 π /k = 2 π /(1.8e7 m-1) = 0.349e-6m c. f = c/ λ = 8.59e14 Hz

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The distance between antinodes is λ /2 = 0.04m or λ = 0.08m. V = λ f = (0.08 m)(3.56e9 Hz) = 2.85e8 m/s.
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Unformatted text preview: Power output = [Power input] x [efficiency] Power input = [Power output]/[efficiency] = 2.6e6W/0.3 = 8.67e6 W A = Power/I = (8.67e6W)/(1100 W/m^2) = 7880 m^2 I = I max cos 2 θ => θ = acos(I/I max ) 1/2 a. (I/Imax) = (1/2.4), θ = acos(0.645) = 49.8 degrees etc. The average value of the cosine squared function is one-half, so the first polarizer transmits ½ the light. The second transmits cos 2 25 o = 0.822. So we have (0.822/2) = 0.411...
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## This note was uploaded on 09/08/2008 for the course PHYS 3B taught by Professor Wu during the Spring '08 term at UC Irvine.

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Hw4_solution - Power output =[Power input x[efficiency Power input =[Power output[efficiency = 2.6e6W/0.3 = 8.67e6 W A = Power/I =(8.67e6W(1100

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