# Reflectivity, emissivity, and absorptivity of opaque materials

 P: 30 First, for any material, $$\alpha ( \lambda, \theta, \phi ) = \epsilon ( \lambda, \theta, \phi )$$ Where $$\lambda, \theta, and \phi$$ represent wavelength, incident angle, and radial angle respectively (basically, $$\theta and \phi$$ determine what direction the incoming or outgoing ray is). If we have a diffuse surface, angle is no longer part of the equation, and thus: $$\alpha ( \lambda ) = \epsilon ( \lambda )$$ If we have a grey, non-diffuse surface, then: $$\alpha ( \theta, \phi ) = \epsilon ( \theta, \phi )$$ If we have a diffuse, grey surface (a favorite assumption in radiative heat transfer): $$\alpha = \epsilon$$ Now, if diffuse, grey, and opaque (transmissivity is zero), we can say $$\rho + \epsilon = \rho + \alpha = 1$$ If not, we can at least say: $$\rho ( \lambda, \theta, \phi ) + \alpha ( \lambda, \theta, \phi ) = \rho ( \lambda, \theta, \phi ) + \epsilon ( \lambda, \theta, \phi )$$ So, the exact assumption we make depends on whether we can assume diffuse, grey, etc. And emissiviy and absorptivity will be functions of different variables depending one which ones we can assume. Hope this helps a bit!