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Homework Help: Find scattered flux density

  1. Dec 18, 2016 #1
    1. The problem statement, all variables and given/known data

    An unpolarized ruby laser operated at 0.7 μm is projected vertically into a clear sky to investigate the density of the atmosphere. A detector located 10 km from the base of the laser is used to receive the flux density scattered from the laser beam by air molecules. Assuming that the laser output has a uniform distribution of flux density F0 across the beam (i.e., I0 = F0sr), and neglecting the effects of multiple scattering, find the scattered flux density at 6 and 10 km received by a detector whose field of view in a plane is 0.05 rad. Use scattering cross section σs = 1.6*10-27 cm2

    Answer given:

    F≅F0l*1.37*10-11 (at 10 km)

    2. Relevant equations

    Scattered flux density is calculated as I*ΔΩ.

    I = (I0/r2 )*σs*P(θ)/4π

    3. The attempt at a solution

    Let the field of view in a plane be Δ=0.05 rad.

    Then ΔΩ should be ΔΩ=πΔ2/4

    P(θ) = 3/4 * (1+cos2(θ))=3/4

    However, if I try to calculate

    (F0/π*r2 )*σs*(3/16π)*πΔ2/4

    where r = 10km, the answer would be something like ~10-40
  2. jcsd
  3. Dec 18, 2016 #2

    Charles Link

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    Homework Helper
    Gold Member

    I'm having a little trouble understanding the geometry of this one. How is the scattered beam being observed? .05 rad in a plane=how high does this extend? Is it a square/circular .05 or does it follow the whole beam over an extended distance? I also see no info on the power output and beam divergence of the ruby laser.
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