# Solution Manual for Radiative Heat Transfer, 3rd Edition

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CHAPTER 2 2.1 Show that for an electromagnetic wave traveling through a dielectric (m1 = n1 ), impinging on the interface with another, optically less dense dielectric (n2 n2 this is a legitimate solution only for sin ฮธ1 โค (n2 /n1 ), or ฮธ1 โค ฮธc = sinโ1 (n2 /n1 ). cos ฮธ2 = 0 (ฮธ2 = ฯ/2): Substituting the first relation into the second gives w00 t = ฮท0 q n21 sin2 ฮธ1 โ n22 , i.e., a legitimate nonzero solution for n21 sin2 ฮธ1 โn22 โฅ 0 or ฮธ1 โฅ ฮธc . Inspection of the reflection coefficients, equations (2.109), shows that e rk = 2 0 in21 w00 t + n2 wi cos ฮธ1 2 0 in21 w00 t โ n2 wi cos ฮธ1 , e rโฅ = w0i cos ฮธ1 + iw00 t w0i cos ฮธ1 โ iw00 t Since, in both reflection coefficients, there are no sign changes within the real and imaginary parts, it follows readily that ฯk = e rke rโk = ฯโฅ = e rโฅe rโโฅ = 1. 19 20 RADIATIVE HEAT TRANSFER 2.2 Derive equations (2.109) using the same approach as in the development of equations (2.89) through (2.92). Hint: Remember that within the absorbing medium, w = w0 โ iw00 = w0 sฬ โ iw00 nฬ; this implies that E0 is not a vector normal to sฬ. It is best to assume E0 = Ek eฬk + Eโฅ eฬโฅ + Es sฬ. Solution 0 00 Inside the absorbing medium wt = w0t โ iw00 t = wt sฬt โ iwt nฬ, and the electric field vector does not lie in a plane normal to sฬ. Thus, we assume a general three-dimensional representation, or E0 = Ek eฬk + Eโฅ eฬโฅ + Es sฬ. Following the development for nonabsorbing media, equations (2.77) through (2.88), then leads to ฮฝยตH0 = w ร E0 = (w0 sฬ โ iw00 nฬ) ร (Ek eฬk + Eโฅ eฬโฅ + Es sฬ). = 0) and reflected wave This formulation is valid for the transmitted wave, but also for the incident wave (w00 i 0 0 (w00 r = 0, wr = โwi ). The contribution from Es vanishes for incident and reflected wave. Using the same vector relations as given for the nonabsorbing media interface, one obtains ฮฝยตH0 = w0 (Ek eฬโฅ โ Eโฅ eฬk ) โ iw00 (Ek eฬโฅ cos ฮธ โ Eโฅ tฬ + Es eฬโฅ sin ฮธ). For the interface condition (with ฮฝยต the same everywhere) ฮฝยตH0 ร nฬ = w0 (Ek tฬ + Eโฅ eฬโฅ cos ฮธ) โ iw00 (Ek cos ฮธtฬ + Eโฅ eฬโฅ + Es sin ฮธtฬ). Thus, from equation (2.78) w0i (Eik tฬ + Eiโฅ eฬโฅ cos ฮธ1 ) โ w0i (Erk tฬ + Erโฅ eฬโฅ cos ฮธ1 ) = w0t (Etk tฬ + Etโฅ eฬโฅ cos ฮธ2 ) โ iw00 t (Etk cos ฮธ2 tฬ + Etโฅ eฬโฅ + Ets sin ฮธ2 tฬ) or 00 w0i (Eik โ Erk ) = (w0t โ iw00 t cos ฮธ2 ) Etk โ iwt sin ฮธ2 Ets w0i (Eiโฅ โ Erโฅ ) cos ฮธ1 = (w0t cos ฮธ2 โ iw00 t ) Etโฅ (2.2-A) (2.2-B) Similarly, from equation (2.77), E0 ร nฬ = (Ek eฬk + Eโฅ eฬโฅ + Es sฬ) ร nฬ = โEk eฬโฅ cos ฮธ + Eโฅ tฬ โ Es eฬโฅ sin ฮธ and (Eik + Erk ) cos ฮธ1 = Etk cos ฮธ2 + Ets sin ฮธ2 Eiโฅ + Erโฅ = Etโฅ (2.2-C) (2.2-D) These four equations have 5 unknowns (Erk , Ers , Etk , Etโฅ , and Ets ), and an additional condition is needed, e.g., equation (2.23) or equation (2.64). Choosing equation (2.23) we obtain, inside the absorbing medium, w ยท E0 = 0 = (w0t sฬ โ iw00 t nฬ) ยท (Etk eฬtk + Etโฅ eฬโฅ + Ets sฬ) = w0t Ets + iw00 t (Etk sin ฮธ2 โ Ets cos ฮธ2 ). Eliminating Etโฅ from equations (2.2-B) and (2.2-D), with e rโฅ = Erโฅ /Eiโฅ , gives rโฅ ), w0i (1 โ e rโฅ ) cos ฮธ1 = (w0t cos ฮธ2 โ iw00 t )(1 + e or e rโฅ = which is identical to equation (2.109). w0i cos ฮธ1 โ (w0t cos ฮธ2 โ iw00 t ) w0i cos ฮธ1 + (w0t cos ฮธ2 โ iw00 t ) , (2.2-E) CHAPTER 2 21 Now, eliminating Ets from equations (2.2-A) and (2.2-C) [multiplying equation (2.2-C) by iw00 t and adding]: 0 w0i (Eik โ Erk ) + iw00 t cos ฮธ1 (Eik + Erk ) = wt Etk . (2.2-F) Eliminating Ets from equations (2.2-C) and (2.2-E) leads to Ets = โiw00 t Etk sin ฮธ2 0 wt โ iw00 t cos ฮธ2 ๏ฃฎ ๏ฃน 2 iw00 w0 cos ฮธ2 โ iw00 ๏ฃฏ๏ฃฏ ๏ฃบ๏ฃบ t sin ฮธ2 t ๏ฃฏ ๏ฃบ๏ฃบ = Etk t (Eik + Erk ) cos ฮธ1 = Etk ๏ฃฏ๏ฃฐcos ฮธ2 โ 0 . w โ iw00 cos ฮธ ๏ฃป w0 โ iw00 cos ฮธ t t 2 t t 2 Using this to eliminate Etk from equation (2.2-F), with e rk = Erk /Eik , gives w0i (1 โ e rk ) + iw00 rk ) = w0t cos ฮธ1 (1 + e rk ) t cos ฮธ1 (1 + e w0t โ iw00 t cos ฮธ2 w0t cos ฮธ2 โ iw00 t w0i (w0t cos ฮธ2 โ iw00 rk ) t )(1 โ e 0 0 00 0 00 = wt (wt โ iwt cos ฮธ2 ) โ iw00 rk ) t (wt cos ฮธ2 โ iwt ) cos ฮธ1 (1 + e = ฮท20 m22 cos ฮธ1 (1 + e rk ), e rk = 2 2 w0i (w0t cos ฮธ2 โ iw00 t ) โ ฮท0 m2 cos ฮธ1 2 2 w0i (w0t cos ฮธ2 โ iw00 t ) + ฮท0 m2 cos ฮธ1 , which is the same as equation (2.109). It is a simple matter to show that other conditions give the same result. For example, from equation (2.64) n21 (Eik eฬik ยท nฬ + Erk eฬrk ยท nฬ) = m22 (Etk eฬtk ยท nฬ + Ets sฬ ยท nฬ) or n21 (Eik โ Erk ) sin ฮธ1 = m22 (Etk sin ฮธ2 โ Ets cos ฮธ2 ), etc. 22 RADIATIVE HEAT TRANSFER 2.3 Find the normal spectral reflectivity at the interface between two absorbing media. [Hint: Use an approach similar to the one that led to equations (2.89) and (2.90), keeping in mind that all wave vectors will be complex, but that the wave will be homogeneous in both media, i.e., all components of the wave vectors are colinear with the surface normal]. Solution Equations (2.19) and (2.20) remain valid for incident, reflected and transmitted waves, with w = w0 โ iw00 = (w0 โiw00 ) nฬ for all three cases. From equation (2.31) wยทw = (w0 โiw00 )2 nฬยทnฬ = ฮท20 m2 it follows that w0 โiw00 = ยฑฮท0 m. Thus w0i โ iw00 i = ฮท0 m1 , w0r โ iw00 r = โฮท0 m1 (reflected wave is moving in a direction of โ nฬ), w0t โ iw00 t = ฮท0 m2 . From equations (2.23) and (2.24), it follows that the electric and magnetic field vectors are normal to nฬ, i.e., tangential to the surface, say E0 = E0 tฬ. Then, from equation (2.77) (Ei + Er ) tฬ ร nฬ = Et tฬ ร nฬ, or Ei + Er = Et From equation (2.25) ฮฝยตH0 = w ร E0 = (w0 โ iw00 ) E nฬ ร tฬ, and from equation (2.78) n1 (Ei โ Er ) = m2 Et . Substituting for Et and dividing by Ei , with e r = Er /Ei : m1 (1 โ e r) = m2 (1 + e r) or e r = m1 โ m2 m1 + m2 and ฯn = e re rโ = (m1 โ m2 )(m1 โ m2 )โ (n1 โ n2 ) + i(k1 โ k2 ) 2 = (m1 + m2 )(m1 + m2 )โ (n1 + n2 ) + i(k1 + k2 ) ฯn = (n1 โ n2 )2 + (k1 โ k2 )2 (n1 + n2 )2 + (k1 + k2 )2 CHAPTER 2 23 2.4 A circularly polarized wave in air is incident upon a smooth dielectric surface (n = 1.5) with a direction of 45โฆ off normal. What are the normalized Stokesโ parameters before and after the reflection, and what are the degrees of polarization? Solution From the definition of Stokesโ parameters the incident wave has degrees of polarization Qi Ui = = 0, Ii Ii Vi = ยฑ1, Ii the sign giving the handedness of the circular polarization. With Erk = Eik rk and Erโฅ = Eiโฅ rโฅ , from equations (2.50) through (2.53): Ir = Eik Eโik r2k + Eiโฅ Eโiโฅ r2โฅ = Eik Eโik (ฯk + ฯโฅ ) = 1 (ฯk + ฯโฅ ) Ii 2 Since Eik Eโik = Eiโฅ Eโiโฅ [from equation (2.51)] and ฯ = r2 . Similarly, Qr = Eik Eโik r2k โ Eiโฅ Eโiโฅ r2โฅ = Eik Eโik (ฯk โ ฯโฅ ) Ur = Eik Eโiโฅ rk rโฅ + Eiโฅ Eโik rโฅ rk = Ui rk rโฅ = 0 Vr = i(Eik Eโiโฅ โ Eiโฅ Eโik ) rk rโฅ = Vi rk rโฅ ฯk โ ฯโฅ Qr = , Ir ฯk + ฯโฅ 2rk rโฅ Vi Vr = . Ir ฯk + ฯโฅ Ii From Snellโs law sin ฮธ1 ; cos ฮธ2 = sin ฮธ2 = n2 s sin2 ฮธ1 1โ = n22 r 0.5 = 1โ 1.52 r 7 , 9 and from equations (2.89) and (2.90) โ โ n1 cos ฮธ2 โ n2 cos ฮธ1 7/9 โ 1.5 1/2 rk = = โ0.0920, ฯk = 0.0085 = โ โ n1 cos ฮธ2 + n2 cos ฮธ1 7/9 + 1.5 1/2 โ โ n1 cos ฮธ1 โ n2 cos ฮธ2 1/2 โ 1.5 7/9 rโฅ = = โ = โ0.3033, ฯโฅ = 0.0920 โ n1 cos ฮธ1 + n2 cos ฮธ2 1/2 + 1.5 7/9 Qr 0.0085 โ 0.0920 Ur = = โ0.8315, =0 Ir 0.0085 + 0.0920 Ir Vr 2ร0.0920ร0.0085 =ยฑ = ยฑ0.5556 Ir 0.0085 + 0.0920 Since the perpendicular polarization is much more strongly reflected, the resulting wave is no longer circularly polarized, but to a large degree linearly polarized (in the perpendicular direction). 24 RADIATIVE HEAT TRANSFER 2.5 A circularly polarized wave in air traveling along the z-axis is incident upon a dielectric surface (n = 1.5). How must the dielectric-air interface be oriented so that the reflected wave is a linearly polarized wave in the y-z-plane? Solution From equations (2.50) through (2.53) it follows that Qr /Ir = 1, Ur = Vr = 0 (i.e., linear polarization), if either Erk or Erโฅ vanish. From Fig. 2-9 it follows that rโฅ , 0 and, therefore Erโฅ , 0 for all incidence directions, while rk = 0 for ฮธ = ฮธp (Brewsterโs angle), or n2 ฮธp = tanโ1 = tanโ1 1.5 = 56.31โฆ . n1 The resulting wave is purely perpendicularpolarized, i.e., eฬโฅ must lie in the y โ z plane, or eฬk must be in the xโz plane. Therefore, the surface may be expressed in terms of its surface normal as .โ nฬ = ฤฑฬ sin ฮธp โ kฬ cos ฮธp = (ฤฑฬ โ 1.5kฬ) 3.25. x Ei z n p p Er CHAPTER 2 25 2.6 A polished platinum surface is coated with a 1 ยตm thick layer of MgO. (a) Determine the materialโs reflectivity in the vicinity of ฮป = 2 ยตm (for platinum at 2 ยตm mPt = 5.29 โ 6.71 i, for MgO mMgO = 1.65 โ 0.0001 i). (b) Estimate the thickness of MgO required to reduce the average reflectivity in the vicinity of 2 ยตm to 0.4. What happens to the interference effects for this case? Solution (a) The desired overall reflectivity must be calculated from equation (2.124) after determining the relevant reflection coefficient. From equation (2.122) e r12 = 1 โ m2 1 โ n2 1 โ 1.65 = โ0.2453 ‘ = 1 + m2 1 + n2 1 + 1.65 since k2 1, and r12 = 0.2453. e r23 may also be calculated from equation (2.122) or, more conveniently, from equation (2.126): r223 = (1.65 โ 5.29)2 + 6.712 = 0.6253 or r23 = 0.7908. (1.65 + 5.29)2 + 6.712 Since the real part of e r12 0 (numerator) and <(r23 ) < 0 (denominator) ฮด23 lies in the second quadrant, ฯ/2 < ฮด23 < ฯ, or ฮด23 = 2.8364. Also ฮถ12 = 4ฯ ร 1.65 ร 1 ยตm/2 ยตm = 10.3673, and cos [ฮด12 ยฑ (ฮด23 โ ฮถ12 )] = cos [ฯ ยฑ (2.8364 โ 10.3673)] = โ0.3175. Also ฮบ2 d = 4ฯ ร 10โ4 ร 1 ยตm/2 ยตm = 2ฯ ร 10โ4 and ฯ = eโฮบ2 d = 0.9994 ' 1. Thus R= 0.24532 + 2ร0.2453ร0.7908ร(โ0.3175) + 0.79082 1 + 2ร0.2453ร0.7908ร(โ0.3175) + 0.24532 ร0.79082 R = 0.6149. (b) The cos in the numerator fluctuates between โ1 < cos < +1. The average value for R is obtained by dropping the cos-term. Then Rav = or ฯ2 = r212 + r223 ฯ2 1 + r212 r223 ฯ2 Rav โ r212 r223 (1 โ r212 ) d = โ , = 0.4 โ 0.24532 = 0.5782, 0.79082 (1 โ 0.24532 ) โ ln 0.5782 1 1 ln ฯ = โ ln ฯ2 = = 43.6 ยตm. ฮบ2 2ฮบ2 4ฯ ร 10โ4 ยตmโ1 More accurate is the averaged expression, equation (2.129) Rav = ฯ12 + ฯ23 (1 โ ฯ12 )2 ฯ2 1 โ ฯ12 ฯ23 ฯ2 or ฯ2 = = Rav โ ฯ12 Rav โ ฯ12 = 2 ฯ23 (Rav โ ฯ12 )ฯ12 + (1 โ ฯ12 ) ฯ23 1 โ (2 โ Rav )ฯ12 0.4 โ 0.24532 = 0.6013 0.79082 [1 โ 1.6 ร 0.24532 ] 26 RADIATIVE HEAT TRANSFER and d= โ ln 0.6013 = 40.47 ยตm 4ฯ ร 10โ4 ยตmโ1 For such a large d, it follows that ฮถ2 ' 40 ร 10.3673 ' 450. A full interference period is traversed if ฮถ2 ' 450 ยฑ ฯ. Around ฮป = 2 ยตm this implies a full period is traversed between 2 ยตm ยฑ 0.014 ยตm. Such interference effects will rarely be observed because (i) the detector will not respond to such small wavelength changes, and (ii) the slightest inaccuracies in layer thickness will eliminate the interference effects. Note: since incoming radiation at ฮป0 = 2 ยตm has a wavelength of ฮป = ฮป0 /n1 = 2/1.65 = 1.2. ยตm, mPt should really be evaluated at 1.21 ยตm.

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### Solution Manual for Radiative Heat Transfer, 3rd Edition

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