Solution Manual for Radiative Heat Transfer, 3rd Edition

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 θ1 ≤ (n2 /n1 ), or θ1 ≤ θc = sin−1 (n2 /n1 ). cos θ2 = 0 (θ2 = π/2): Substituting the first relation into the second gives w00 t = η0 q n21 sin2 θ1 − n22 , i.e., a legitimate nonzero solution for n21 sin2 θ1 −n22 ≥ 0 or θ1 ≥ θc . Inspection of the reflection coefficients, equations (2.109), shows that e rk = 2 0 in21 w00 t + n2 wi cos θ1 2 0 in21 w00 t − n2 wi cos θ1 , e r⊥ = w0i cos θ1 + iw00 t w0i cos θ1 − iw00 t Since, in both reflection coefficients, there are no sign changes within the real and imaginary parts, it follows readily that ρk = e rke r∗k = ρ⊥ = e r⊥e r∗⊥ = 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 − iw00 = w0 ŝ − iw00 n̂; this implies that E0 is not a vector normal to ŝ. It is best to assume E0 = Ek êk + E⊥ ê⊥ + Es ŝ. Solution 0 00 Inside the absorbing medium wt = w0t − iw00 t = wt ŝt − iwt n̂, and the electric field vector does not lie in a plane normal to ŝ. Thus, we assume a general three-dimensional representation, or E0 = Ek êk + E⊥ ê⊥ + Es ŝ. Following the development for nonabsorbing media, equations (2.77) through (2.88), then leads to νµH0 = w × E0 = (w0 ŝ − iw00 n̂) × (Ek êk + E⊥ ê⊥ + Es ŝ). = 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 = −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 νµH0 = w0 (Ek ê⊥ − E⊥ êk ) − iw00 (Ek ê⊥ cos θ − E⊥ t̂ + Es ê⊥ sin θ). For the interface condition (with νµ the same everywhere) νµH0 × n̂ = w0 (Ek t̂ + E⊥ ê⊥ cos θ) − iw00 (Ek cos θt̂ + E⊥ ê⊥ + Es sin θt̂). Thus, from equation (2.78) w0i (Eik t̂ + Ei⊥ ê⊥ cos θ1 ) − w0i (Erk t̂ + Er⊥ ê⊥ cos θ1 ) = w0t (Etk t̂ + Et⊥ ê⊥ cos θ2 ) − iw00 t (Etk cos θ2 t̂ + Et⊥ ê⊥ + Ets sin θ2 t̂) or 00 w0i (Eik − Erk ) = (w0t − iw00 t cos θ2 ) Etk − iwt sin θ2 Ets w0i (Ei⊥ − Er⊥ ) cos θ1 = (w0t cos θ2 − iw00 t ) Et⊥ (2.2-A) (2.2-B) Similarly, from equation (2.77), E0 × n̂ = (Ek êk + E⊥ ê⊥ + Es ŝ) × n̂ = −Ek ê⊥ cos θ + E⊥ t̂ − Es ê⊥ sin θ and (Eik + Erk ) cos θ1 = Etk cos θ2 + Ets sin θ2 Ei⊥ + Er⊥ = Et⊥ (2.2-C) (2.2-D) These four equations have 5 unknowns (Erk , Ers , Etk , Et⊥ , 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 · E0 = 0 = (w0t ŝ − iw00 t n̂) · (Etk êtk + Et⊥ ê⊥ + Ets ŝ) = w0t Ets + iw00 t (Etk sin θ2 − Ets cos θ2 ). Eliminating Et⊥ from equations (2.2-B) and (2.2-D), with e r⊥ = Er⊥ /Ei⊥ , gives r⊥ ), w0i (1 − e r⊥ ) cos θ1 = (w0t cos θ2 − iw00 t )(1 + e or e r⊥ = which is identical to equation (2.109). w0i cos θ1 − (w0t cos θ2 − iw00 t ) w0i cos θ1 + (w0t cos θ2 − 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 − Erk ) + iw00 t cos θ1 (Eik + Erk ) = wt Etk . (2.2-F) Eliminating Ets from equations (2.2-C) and (2.2-E) leads to Ets = −iw00 t Etk sin θ2 0 wt − iw00 t cos θ2   2 iw00 w0 cos θ2 − iw00   t sin θ2 t   = Etk t (Eik + Erk ) cos θ1 = Etk cos θ2 − 0 . w − iw00 cos θ  w0 − iw00 cos θ t t 2 t t 2 Using this to eliminate Etk from equation (2.2-F), with e rk = Erk /Eik , gives w0i (1 − e rk ) + iw00 rk ) = w0t cos θ1 (1 + e rk ) t cos θ1 (1 + e w0t − iw00 t cos θ2 w0t cos θ2 − iw00 t w0i (w0t cos θ2 − iw00 rk ) t )(1 − e 0 0 00 0 00 = wt (wt − iwt cos θ2 ) − iw00 rk ) t (wt cos θ2 − iwt ) cos θ1 (1 + e = η20 m22 cos θ1 (1 + e rk ), e rk = 2 2 w0i (w0t cos θ2 − iw00 t ) − η0 m2 cos θ1 2 2 w0i (w0t cos θ2 − iw00 t ) + η0 m2 cos θ1 , 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 êik · n̂ + Erk êrk · n̂) = m22 (Etk êtk · n̂ + Ets ŝ · n̂) or n21 (Eik − Erk ) sin θ1 = m22 (Etk sin θ2 − Ets cos θ2 ), 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 − iw00 = (w0 −iw00 ) n̂ for all three cases. From equation (2.31) w·w = (w0 −iw00 )2 n̂·n̂ = η20 m2 it follows that w0 −iw00 = ±η0 m. Thus w0i − iw00 i = η0 m1 , w0r − iw00 r = −η0 m1 (reflected wave is moving in a direction of − n̂), w0t − iw00 t = η0 m2 . From equations (2.23) and (2.24), it follows that the electric and magnetic field vectors are normal to n̂, i.e., tangential to the surface, say E0 = E0 t̂. Then, from equation (2.77) (Ei + Er ) t̂ × n̂ = Et t̂ × n̂, or Ei + Er = Et From equation (2.25) νµH0 = w × E0 = (w0 − iw00 ) E n̂ × t̂, and from equation (2.78) n1 (Ei − Er ) = m2 Et . Substituting for Et and dividing by Ei , with e r = Er /Ei : m1 (1 − e r) = m2 (1 + e r) or e r = m1 − m2 m1 + m2 and ρn = e re r∗ = (m1 − m2 )(m1 − m2 )∗ (n1 − n2 ) + i(k1 − k2 ) 2 = (m1 + m2 )(m1 + m2 )∗ (n1 + n2 ) + i(k1 + k2 ) ρn = (n1 − n2 )2 + (k1 − 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 45◦ off normal. What are the normalized Stokes’ parameters before and after the reflection, and what are the degrees of polarization? Solution From the definition of Stokes’ parameters the incident wave has degrees of polarization Qi Ui = = 0, Ii Ii Vi = ±1, Ii the sign giving the handedness of the circular polarization. With Erk = Eik rk and Er⊥ = Ei⊥ r⊥ , from equations (2.50) through (2.53): Ir = Eik E∗ik r2k + Ei⊥ E∗i⊥ r2⊥ = Eik E∗ik (ρk + ρ⊥ ) = 1 (ρk + ρ⊥ ) Ii 2 Since Eik E∗ik = Ei⊥ E∗i⊥ [from equation (2.51)] and ρ = r2 . Similarly, Qr = Eik E∗ik r2k − Ei⊥ E∗i⊥ r2⊥ = Eik E∗ik (ρk − ρ⊥ ) Ur = Eik E∗i⊥ rk r⊥ + Ei⊥ E∗ik r⊥ rk = Ui rk r⊥ = 0 Vr = i(Eik E∗i⊥ − Ei⊥ E∗ik ) rk r⊥ = Vi rk r⊥ ρk − ρ⊥ Qr = , Ir ρk + ρ⊥ 2rk r⊥ Vi Vr = . Ir ρk + ρ⊥ Ii From Snell’s law sin θ1 ; cos θ2 = sin θ2 = n2 s sin2 θ1 1− = n22 r 0.5 = 1− 1.52 r 7 , 9 and from equations (2.89) and (2.90) √ √ n1 cos θ2 − n2 cos θ1 7/9 − 1.5 1/2 rk = = −0.0920, ρk = 0.0085 = √ √ n1 cos θ2 + n2 cos θ1 7/9 + 1.5 1/2 √ √ n1 cos θ1 − n2 cos θ2 1/2 − 1.5 7/9 r⊥ = = √ = −0.3033, ρ⊥ = 0.0920 √ n1 cos θ1 + n2 cos θ2 1/2 + 1.5 7/9 Qr 0.0085 − 0.0920 Ur = = −0.8315, =0 Ir 0.0085 + 0.0920 Ir Vr 2×0.0920×0.0085 =± = ±0.5556 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 Er⊥ vanish. From Fig. 2-9 it follows that r⊥ , 0 and, therefore Er⊥ , 0 for all incidence directions, while rk = 0 for θ = θp (Brewster’s angle), or n2 θp = tan−1 = tan−1 1.5 = 56.31◦ . n1 The resulting wave is purely perpendicularpolarized, i.e., ê⊥ must lie in the y − z plane, or êk must be in the x−z plane. Therefore, the surface may be expressed in terms of its surface normal as .√ n̂ = ı̂ sin θp − k̂ cos θp = (ı̂ − 1.5k̂) 3.25. x Ei z n p p Er CHAPTER 2 25 2.6 A polished platinum surface is coated with a 1 µm thick layer of MgO. (a) Determine the material’s reflectivity in the vicinity of λ = 2 µm (for platinum at 2 µm mPt = 5.29 − 6.71 i, for MgO mMgO = 1.65 − 0.0001 i). (b) Estimate the thickness of MgO required to reduce the average reflectivity in the vicinity of 2 µm 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 − m2 1 − n2 1 − 1.65 = −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 − 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) δ23 lies in the second quadrant, π/2 < δ23 < π, or δ23 = 2.8364. Also ζ12 = 4π × 1.65 × 1 µm/2 µm = 10.3673, and cos [δ12 ± (δ23 − ζ12 )] = cos [π ± (2.8364 − 10.3673)] = −0.3175. Also κ2 d = 4π × 10−4 × 1 µm/2 µm = 2π × 10−4 and τ = e−κ2 d = 0.9994 ' 1. Thus R= 0.24532 + 2×0.2453×0.7908×(−0.3175) + 0.79082 1 + 2×0.2453×0.7908×(−0.3175) + 0.24532 ×0.79082 R = 0.6149. (b) The cos in the numerator fluctuates between −1 < cos < +1. The average value for R is obtained by dropping the cos-term. Then Rav = or τ2 = r212 + r223 τ2 1 + r212 r223 τ2 Rav − r212 r223 (1 − r212 ) d = − , = 0.4 − 0.24532 = 0.5782, 0.79082 (1 − 0.24532 ) − ln 0.5782 1 1 ln τ = − ln τ2 = = 43.6 µm. κ2 2κ2 4π × 10−4 µm−1 More accurate is the averaged expression, equation (2.129) Rav = ρ12 + ρ23 (1 − ρ12 )2 τ2 1 − ρ12 ρ23 τ2 or τ2 = = Rav − ρ12 Rav − ρ12 = 2 ρ23 (Rav − ρ12 )ρ12 + (1 − ρ12 ) ρ23 1 − (2 − Rav )ρ12 0.4 − 0.24532 = 0.6013 0.79082 [1 − 1.6 × 0.24532 ] 26 RADIATIVE HEAT TRANSFER and d= − ln 0.6013 = 40.47 µm 4π × 10−4 µm−1 For such a large d, it follows that ζ2 ' 40 × 10.3673 ' 450. A full interference period is traversed if ζ2 ' 450 ± π. Around λ = 2 µm this implies a full period is traversed between 2 µm ± 0.014 µm. 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 λ0 = 2 µm has a wavelength of λ = λ0 /n1 = 2/1.65 = 1.2. µm, mPt should really be evaluated at 1.21 µm.