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WeekProb_Jan_26Sol - Exam-type Problems#1 Solutions(Revised...

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Unformatted text preview: Exam-type Problems #1 Solutions (Revised) 7C Staff 2004/09/14 Problem 1 E(x, t) = (5 V/m) cos(ky + ωt)ˆ x (5 V/m) cos(ky + ωt)ˆ z c ω = 2π · (500 Hz) B(x, t) = k = ω/c Problem 2 In the case of finite reflectivity Q (Q = 0 is perfect absorption, and Q = 1 is perfect reflectivity), the value of the effective cross-section of a sphere is not immediately obvious because the impulse is only ∼ 1 + Q when a photon hits the surface at normal incidence. If the beads were perfectly absorptive, the effective cross-section would be πR2 by simple opacity. We begin by noting that a photon traveling straight upward will reflect through an angle of π − 2θ when it strikes a lower hemisphere at polar angle θ. The upward component of the reflected photon’s momentum changes from p to p cos(π − 2θ), so the upward impulse on the hemisphere is p (1 − cos(π − 2θ)) = p (1 + cos(2θ)). Therefore, the total upward force on the hemisphere for photon flux N is π/2 F = pN · dA(θ) (1 + Q cos(2θ)) 0 π/2 = 0 = I πR2 c I c cos θ [(2πR)(R sin θdθ)] (1 + Q cos(2θ)) π/2 sin 2θ (1 + Q cos(2θ)) 0 1 PROBLEM 2 ALTERNATIVE INITIAL DERIVATION. After this page everything continues the same way. ...
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