Since this layer is isothermal and since its absorptivity a equals its emis

Since this layer is isothermal and since its

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Since this layer is isothermal and since its absorptivity a equals its emis- sivity ε , it follows that it must emit fl ux density F 1 in both the upward and downward directions. Hence, it loses energy at a rate 2 F 1 . The loss due to emission is balanced by the absorption of upwelling radiation from below. The upwelling radiation from below is ( F + F 1 ) and the fraction of it that is absorbed in the topmost layer is a . Hence 2 F 1 = a ( F + F 1 ) (1) Solving, we obtain F 1 = αF (2 α ) (2) But from the Stefan Boltzmann law and Kirchho ff ’s law F 1 = aσT 4 1 (3) where T 1 is the equivalent blackbody temperature of the topmost layer. Combining Eqs. (2) and (3) yields aσT 4 1 = αF (2 α ) In the limit, as a 0 , σT 4 1 = F 2 = σT 4 E 2 Solving, we obtain T 1 = (0 . 5) 1 / 4 T E For the Earth;s atmosphere, T E = 255 K and T 1 , the so-called skin tem- perature, is 214 K, which is quite comparable to the mean temperature of the lower stratosphere. 63
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4.40 Consider an idealized aerosol consisting of spherical particles of radius r with a refractive index of 1.5. Using Fig. 4.13, estimate the smallest radius for which the particles would impart a bluish cast to transmitted white light, as in the rarely observed "blue moon". To impart a bluish (as opposed to a reddish) cast to transmitted light, the particles would need to exhibit a range of sizes for which the scattering e ciency K λ decreases with increasing values of the size parameter, the reverse of the situation for Rayleigh scattering. It is evident from Fig. 4.13, that the smallest range of values of x for which dK λ /dx < 0 occurs around x = 2 πr λ 5 For visible light with λ 0 . 5 µ m 2 πr 5 × 0 . 5 µ m = 2 . 5 µ m r 0 . 4 µ m 4.41 Consider an idealized cloud consisting of spherical droplets with a uniform radius of 20 µ m and concentrations of 1 cm 3 . How long a path through such a cloud would be required to deplete a beam of visible radiation by a factor of e due to scattering alone? (Assume that none of the scattered radiation is subsequently scattered back into the path of the beam.) From Eq. (4.16), dI λ = I λ K λ Nσds The size parameter x = 2 πr/λ = 2 π × 20 µ m ÷ 0 . 5 µ m > 50 . Hence, if absorption is assumed to be negligible, Fig. 4.13 yields a value of K λ = 2 . The number density N = 1 cm 3 = 10 6 m 3 and the cross-sectional area σ = π × ¡ 20 × 10 6 ¢ 2 = 1 . 2 × 10 9 . Hence, the volume scattering coe cient K λ = 2 . 4 × 10 3 m 1 . The incident radiation is depleted by a factor of e over a path in which the optical depth τ λ = R K λ Nσds is equal to unity. If we assume that conditions are uniform along the ray path, the path length is given by s = ( K λ ) 1 = ¡ 2 . 4 × 10 3 m 1 ¢ 1 400 m 4.43 Consider radiation with wavelength λ and zero zenith angle passing through a gas with an volume absorption coe cient k λ of 0.01 m 2 kg 1 . What frac- tion of the beam is absorbed in passing through a layer containing 1 kg m 2 of the gas? What mass of gas would the layer have to contain on order to absorb half the incident radiation?
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