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 qi = 40 W/cm2 in the air is incident on the glass surface.

Question

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 [latex]q_i = 40 W/cm^2[/latex] in the air is incident on the glass surface. The absorption coefficient of the glass is assumed constant at [latex]0.08 cm^{−1}[/latex] and the refractive index of the glass is n = 1.52. Scattering is neglected. Determine the energy absorption rate per unit volume in dV.

Accepted Answer

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 [latex]1 – ρ (θ_i)[/latex] , where [latex]ρ(θ_i)[/latex] is given for each component of polarization by Equations 3.6a and 3.6b.

[latex] ho _{\parallel } ( heta _i)=\left\{\frac{(n_2/n_1)^2\cos heta _i-[(n_2/n_1)^2-\sin ^2 heta _i]^{1/2}}{(n_2/n_1)^2\cos heta _i+[(n_2/n_1)^2-\sin ^2 heta _i]^{1/2}} ight\}^2 [/latex] (3.6a)

[latex] ho _{\bot } ( heta _i)=\left\{\frac{[(n_2/n_1)^2-\sin ^2 heta _i]^{1/2}-\cos heta _i}{[(n_2/n_1)^2-\sin^2 heta _i]^{1/2}+\cos heta _i} ight\}^2 [/latex] (3.6b)

From Equation 17.41

[latex]\frac{I_{\lambda ,1}[1- ho _\lambda ( heta_1)]d\lambda_1}{n_1^2} =\frac{I_{\lambda ,2}d\lambda _2}{n_2^2} [/latex] (17.41)

the intensity inside the medium for each component of polarization requires multiplying the incident intensity by [latex]n^2[/latex] in addition to a reflectivity factor. The intensity in the medium in direction χ then becomes [latex](G/2π)[1− ρ(θ(χ))]n^2,[/latex] 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 ,[latex]e^{-κx/cosχ}.[/latex] he fraction κdx/cos χ of this incident energy at [latex]dV, I_dV(χ) dA cosχ,[/latex] is then absorbed in the volume element dA dx. The energy absorption for all directions of the arriving energy is found by integrating over [latex]0 ≤ χ ≤ χ_{max},[/latex] where [latex]χ_{max}[/latex] is given by Snell’s law as [latex]sin^{–1}(I/n)[/latex]. 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

[latex]\frac{dQ}{dV} =kGn^2\int_{0}^{\chi _{max}}{[2- ho _{\parallel }(\chi )- ho _{\bot }]e^{-kX/\cos \chi }\sin \chi d\chi } [/latex]

From Equations 3.6a and 3.6b the reflectivities are given by

[latex] ho _{\parallel } (\chi )=\left[\frac{n^2\cos heta _i-(n^2-\sin ^2 heta _i)^{1/2}}{n^2\cos heta _i+(n^2-\sin ^2 heta _i)^{1/2}} ight]^2 [/latex]

[latex] ho _{\bot } (\chi )=\left[\frac{(n^2-\sin ^2 heta _i)^{1/2}-\cos heta _i}{(n^2-\sin ^2 heta _i)^{1/2}+\cos heta _i} ight]^2 [/latex]

where the [latex]θ_i = θ_i(χ)[/latex] is given by Snell’s law as [latex]θ_i(χ) = \sin^{-1}(n \sin χ).[/latex] Using the specified values, the integration is carried out numerically to give [latex]dQ/dV = 3.38 W/cm^3.[/latex]

Question 17.2: A volume element dV is located inside glass at x = 3 cm from...

Related Answered Questions

What is the fraction of externally incident unpolarized radiation that is transmitted through a glass window in air? The window is 0.75 cm thick, radiation is incident at θi = 50°, nglass = 1.53, and κglass = 0.1 cm–1. ...

Verified Answer: