Absorption, emission and scattering from a medium

From Thermal-FluidsPedia

Propagation of intensity

The intensity $I$ was defined and discussed in planck distribution. A property of the intensity that was not discussed is found by looking at the diagram in figure on the right, where we can calculate the energy from the two equal area elements d $A 1$ and d $A 2$ that is incident on surface d $A s$ .

The intensity as defined in the following equation: ,

${I_{\lambda b}} = \frac{{d{e_\lambda }}}{{dA\cos \theta d\omega d\lambda }}$

is used to determine the energy striking d $A s$ from the two equal area elements d $A 1$ and d $A 2$ , resulting in

$d{e_{\lambda ,1}} = {I_{\lambda ,1}}d{A_1}d{\Omega _1}d\lambda = {I_{\lambda ,1}}d{A_1}\frac{{d{A_s}\cos \theta }}{{S_1^2}}d\lambda \qquad \qquad(1)$ $d{e_{\lambda ,2}} = {I_{\lambda ,2}}d{A_2}d{\Omega _2}d\lambda = {I_{\lambda ,2}}d{A_2}\frac{{d{A_s}\cos \theta }}{{S_2^2}}d\lambda \qquad \qquad(2)$

Let d $A 1 = d A 2 = d A$ , and impose the inverse square law for energy attenuation with distance,

$d{e_{\lambda ,2}} = \frac{{S_1^2}}{{S_2^2}}d{e_{\lambda ,1}}$

Making these substitutions results in the intensity from the two surfaces that reach d $A s$ being

${I_{\lambda ,1}} = {I_{\lambda ,2}}\qquad \qquad(3)$

This shows that, in the absence of attenuation or emission by a medium along the path $S$ , the intensity is constant with distance. The intensity thus serves as a metric for determining the effect of attenuation.

References

Faghri, A., Zhang, Y., and Howell, J. R., 2010, Advanced Heat and Mass Transfer, Global Digital Press, Columbia, MO.

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Retrieved from "http://thermalfluidscentral.org/encyclopedia/index.php/Absorption,_emission_and_scattering_from_a_medium"

Absorption, emission and scattering from a medium

From Thermal-FluidsPedia

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