Question 17.2: A volume element dV is located inside glass at x = 3 cm from...
A volume element dV is located inside glass at x = 3 cm from an optically smooth planar interface in air, as in Figure 17.20. A diffuse-gray radiation flux q_i = 40 W/cm^2 in the air is incident on the glass surface. The absorption coefficient of the glass is assumed constant at 0.08 cm^{−1} and the refractive index of the glass is n = 1.52. Scattering is neglected. Determine the energy absorption rate per unit volume in dV.
Since the incident energy is diffuse and unpolarized, the incident intensity in each component of polarization is G/2π. The fraction of the intensity transmitted through the interface depends on the angle of incidence and is 1 – ρ (θ_i) , where ρ(θ_i) is given for each component of polarization by Equations 3.6a and 3.6b.
\rho _{\parallel } (\theta _i)=\left\{\frac{(n_2/n_1)^2\cos \theta _i-[(n_2/n_1)^2-\sin ^2\theta _i]^{1/2}}{(n_2/n_1)^2\cos \theta _i+[(n_2/n_1)^2-\sin ^2\theta _i]^{1/2}} \right\}^2 (3.6a)
\rho _{\bot } (\theta _i)=\left\{\frac{[(n_2/n_1)^2-\sin ^2\theta _i]^{1/2}-\cos \theta _i}{[(n_2/n_1)^2-\sin^2 \theta _i]^{1/2}+\cos \theta _i} \right\}^2 (3.6b)
From Equation 17.41
\frac{I_{\lambda ,1}[1-\rho _\lambda (\theta_1)]d\lambda_1}{n_1^2} =\frac{I_{\lambda ,2}d\lambda _2}{n_2^2} (17.41)
the intensity inside the medium for each component of polarization requires multiplying the incident intensity by n^2 in addition to a reflectivity factor. The intensity in the medium in direction χ then becomes (G/2π)[1− ρ(θ(χ))]n^2, where χ is the angle of refraction. The path length traveled from the interface to the volume element is x/ cosχ. Using Bouguer’s law, the fraction reaching dV is ,e^{-κx/cosχ}. he fraction κdx/cos χ of this incident energy at dV, I_dV(χ) dA cosχ, is then absorbed in the volume element dA dx. The energy absorption for all directions of the arriving energy is found by integrating over 0 ≤ χ ≤ χ_{max}, where χ_{max} is given by Snell’s law as sin^{–1}(I/n). The integration over all incident solid angles introduces the factor for solid angle, 2π sin χ dχ. The energy absorbed per unit volume in dV is obtained by doing the integration for each component of polarization and summing the results. This yields
\frac{dQ}{dV} =kGn^2\int_{0}^{\chi _{max}}{[2-\rho _{\parallel }(\chi )-\rho _{\bot }]e^{-kX/\cos \chi }\sin \chi d\chi }
From Equations 3.6a and 3.6b the reflectivities are given by
\rho _{\parallel } (\chi )=\left[\frac{n^2\cos \theta _i-(n^2-\sin ^2\theta _i)^{1/2}}{n^2\cos \theta _i+(n^2-\sin ^2\theta _i)^{1/2}} \right]^2
\rho _{\bot } (\chi )=\left[\frac{(n^2-\sin ^2\theta _i)^{1/2}-\cos \theta _i}{(n^2-\sin ^2\theta _i)^{1/2}+\cos \theta _i} \right]^2
where the θ_i = θ_i(χ) is given by Snell’s law as θ_i(χ) = \sin^{-1}(n \sin χ). Using the specified values, the integration is carried out numerically to give dQ/dV = 3.38 W/cm^3.