Find scattered flux density

[Earthland]

Homework Statement

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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 = F0/π sr), 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 σ_s = 1.6*10^-27 cm²

Answer given:

F≅F₀l*1.37*10^-11 (at 10 km)

Homework Equations

Scattered flux density is calculated as I*ΔΩ.

I = (I₀/r² )*σ_s*P(θ)/4π

The Attempt at a Solution

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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 )*σ_s*(3/16π)*πΔ²/4

where r = 10km, the answer would be something like ~10^-40

 

[Charles Link]

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.