Question 12.2: A nearly transparent medium with extinction coefficient βλ i...

For the nearly transparent approximation, q_{rλ}=J_{λ,1}-J_{λ,2} Substituting J_{λ,1} and J_{λ,2} from Equation 11.27 yields

J_{\lambda ,1}=\epsilon _{\lambda ,1}\pi I_{\lambda b,1}+2(1-\epsilon _{\lambda ,1})\left[J_{\lambda ,2}E_3 (\tau_{\lambda ,D})+\pi \int_{\tau _\lambda ^*=0}^{\tau _{\lambda ,D}}{\widehat{I}_\lambda (\tau _\lambda ^*) E_2(\tau _\lambda ^*)}d\tau _\lambda ^* \right]                 (11.27a)

J_{\lambda ,2}=\epsilon _{\lambda ,2}\pi I_{\lambda b,2}+2(1-\epsilon _{\lambda ,2})\left[J_{\lambda ,1}E_3 (\tau_{\lambda ,D})+\pi \int_{\tau _\lambda ^*=0}^{\tau _{\lambda ,D}}{\widehat{I}_\lambda (\tau _\lambda ^*) E_2(\tau _{\lambda ,D}-\tau _\lambda ^*)}d\tau _\lambda ^* \right]                           (11.27b)

q_{r\lambda }d\lambda =\frac{\epsilon_{\lambda ,1}E_{\lambda b,1}+\epsilon_{\lambda ,2}E_{\lambda b,2}(1-\epsilon _{\lambda ,1})-\epsilon _{\lambda ,2}E_{\lambda b,2} +\epsilon_{\lambda ,1}E_{\lambda b,1}(1-\epsilon _{\lambda ,2})}{1-(1-\epsilon _{\lambda ,1})(1-\epsilon _{\lambda ,2})} d\lambda

Simplifying and integrating with respect to λ gives the required result,

q_{r }= \int_{\lambda =0}^{\infty }{\frac{E_{\lambda b,1}-E_{\lambda b,2}}{1/\epsilon _{\lambda ,1}+1/\epsilon _{\lambda ,2}-1} }d\lambda                       (12.13)

This is the same as Equation 6.6;

q_{1}=-q_2=\int_{\lambda =0}^{\infty }{q_{\lambda ,1}d\lambda } =\int_{\lambda =0}^{\infty }{\frac{E_{\lambda b,1}(T_1)-E_{\lambda b,2}(T_2)}{\frac{1}{\epsilon _{\lambda ,1}(T_1)}+\frac{1}{\epsilon _{\lambda ,2}(T_2)} -1 } } d\lambda          (6.6)

the radiant energy flux transferred is uninfluenced by the nearly transparent medium between the boundaries. The medium temperature is obtained in the next example.