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Unformatted text preview: Exam-type Problems #1 Solutions (Revised)
2004/09/14 Problem 1
E(x, t) = (5 V/m) cos(ky + ωt)ˆ
cos(ky + ωt)ˆ
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 = 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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- Spring '08
- Quantum Physics