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Unformatted text preview: 2) + (0.01)(20) + (0.01)(40) = 28.9 so H is 8 km (b) [15 pt] Determine the decrease in intensity of visible (λ = 0.5 µm) and infrared (1 µm) light due to water vapor absorption in the atmosphere, as both a fraction and in magnitudes, over 3 scaleheights (95% of the atmosphere). Assume that the opacity of water vapor is similar to that of liquid water, and that the surface relative humidity is 50%, which translates into a surface partial pressure of water vapor of 0.3 kPa. To make this problem analytically solvable, you need to assume a constant temperature through the atmosphere, which is a terrible assumption. For +5 bonus points, use numerical integration to determine the intensity decline for both wavelengths assuming a temperature decrease with height of
6 K/km. Going back to our definition of light absorption: dI
(z ) = κλ ρ(z )dz
I
Note that the right term is positive because z is the height from the surface whereas light is absorbed going down from the top of the atmosphere. We can find ρ(z) using the ideal gas law and our pressure relation: ρ(z ) = P (z ) P (0) −z/H
µmH
=
e
kT...
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This document was uploaded on 02/26/2014 for the course PHYS 160 at UCSD.
 Fall '08
 Norman,M
 Physics, Work

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