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+ 2p(s)R(s) + q(s)R2(s)
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q(s) = (1 + rr2)b(s) - 2r12(h;e(S) - f(s)) (8.3.23) Similarly, differentiating (8.3.10) and using the (8.3.21) gives the differential equation for t( s) dt(s) (8.3.24) ~ = [P(s) + q(s)R(s)]t(s)
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Figure 8.3.2 Problem B: Thermal emission from an inhomogeneous slab.
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The initial conditions can be established by letting s = 0 in (8.3.9) and (8.3.10) and using (8.3.7) and (8.3.8)
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R(s = 0) = 1 t(s
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(8.3.26) 1 - TOlT12 The boundary value problem has been converted into an initial value problem which is readily solved by stepping forward in s. Equations (8.3.22) and (8.3.25) form an initial value problem for R(s). Equations (8.3.24) and (8.3.26) form an initial value problem for t(s) in terms of R(s). B. Thermal Emission from an Inhomogeneous Slab We next consider the problem of emission (Fig. 8.3.2). Medium 1 has a temperature distribution T(z), and medium 2 has temperature T 2 . The brightness temperature as a function of slab thickness is expressed in terms of the reflectivity and transmissivity function of Problem A. The radiative transfer equations for Problem Bare
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s) + j(z)1r(z, s) + b(z)Iz(z, s) + ~a(z)CT(z) -1l1(z, s) = - ~e(z)1z(z, s) + b(z)1r(z, s) + j(z)Iz(z, s) + ~a(z)CT(z)
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1rl(z, s) = ~e(z)1r(z,
(8.3.27) (8.3.28) (8.3.29) (8.3.30)
with boundary conditions
1r(O, s) =
1z(O, s)
11(s,s) = Tl21r (s,s) +tl2CT2
3.1 One-Dimensional Problem
We define the reflected temperature as 1 TR(S) = C Ir(s, s) From (8.3.30) and (8.3.31a) we obtain
(8.3.31b) Iz(s, s) = T1 2CTR(S) + t12CT2 The brightness temperature as observed by a radiometer in region 0 is TB(S)
= C t01 I z(0, s)
Differentiating (8.3.31) with respect to s gives dTR(s) 1 ds = C [Irl(s, s) + I r2 (s, s)]
The quantities Ir(s, s) and Iz(s, s) are expressed in terms of TR(S) through (8.3.30) and (8.3.31). Hence I r1(s, s) can be expressed in terms of TR(S) by letting z = s in (8.3.27):
Ir1(s, s)
= (-K;e(S) + f(s))CTR(S) + b(s)[r12CTR(S) + it2CT2] + K;a(s)CT(s)
(8.3.33b) To derive a relation for I r2 (s, s) in terms of TR (s), we differentiate radiative transfer equations (8.3.27) and (8.3.28) and boundary conditions (8.3.29) and (8.3.30) with respect to s. I r21 (z, s) = -K;e(z)Ir2 (z, s) + j(z)Ir2(z, s) + b(z)Iz 2(z, s) (8.3.34a) -Il2l(z, s) = -K;e(z)Idz, s) + b(z)Ir2 (z, s) + j(z)Idz, s) (8.3.34b) I r2 (0, s) = rOlIz 2(0, s) (8.3.35a) (8.3.35b) I z2 (s, s) = T12Ir2(S, s) + h2Ir1(s, s) - II1(s, s)] On comparison of (8.3.34) and (8.3.35) with Problem A of reflection and transmission, we note that they satisfy the same equations and boundary conditions with a change of source term from t12Q to r12Ir1(s, s) - I l1 (s, s) of Problem B. Hence, using the definitions of reflectivity and transmissivity functions of Problem A, we obtain I r2 (s, s) = R(s)[r12Irl(s, s) - II1(s, s)] IdO, s) = t(s)[r12Ir1(s, s) - Il1(s, s)] (8.3.36a) (8.3.36b)
The rest of the derivation is similar to the case of Problem A. We obtain dTR(s) ds = [P(s) + R(s)q(s)]TR(S) + [P(s)R(s) + b(S)]t12T2
+ K;a(s)T(s)[l + R(s)(l + T12)]
dTB(S) ds = tOlt(s)[TR(S)q(S)
+ p(S)t12 T2 + K;a(s)T(s)(l + r12)]
with the initial conditions
= 0) = = 0) =
rOliI2 T 2 rOlT12 tOlt12 T 2 rOlT12
(8.3.39) (8.3.40)
Hence, the initial value problem for TR(S) and TB(S) is in terms of R(s) and t(s) of Problem A and must be solved in conjunction with the latter two quantities, which are governed by (8.3.22) and (8.3.24) through (8.3.26). To illustrate the various effects due to nonuniform scattering, absorption, and temperature profiles, we use the laminar structure model of Section 2.1, which is now allowed to assume inhomogeneous profiles. We have the profiles ofT(z), 8(z), l(z), and It(z), where It(z) is defined as the loss tangent profile (8.3.41) We plot the numerical results corresponding to the case of a layer of ice over water. The parameters are taken to be E~m = 3.2E o , E2 = 77.2(1 + iO.17)E o , T2 = 273 K, and l(z) = 1 mm, with the following profiles, (a) Scattering profile:
<5(z) = 0.01 + 0.01 exp( -0.005z) T(z) = 273 K It(z) = 0.0009
(b) Scattering and temperature profiles: 0.01 + 0.01 exp( -0.005z) T(z) = 273 - 33exp(-,-0.01z)
<5 (z) =
It(z) = 0.0009
(c) Scattering and absorption profiles:
8(z) = 0.01 + 0.01 exp( -0.005z) T(z) = 273 K It(z) = 0.002 - 0.0011 exp( -0.005z)
(d) Uniform profile:
8(z) = 0.02 T(z) = 273 K It(z) = 0.0009