Why is there emissivity + reflectivity + transmittance = 1? Foillowing this question, I understand that $emissivity + reflectivity + transmittance = 1$
Where does this relationship comes from? What are the physical phenomena explaining this relationship?
 A: So you can think about this as a statement of conservation of energy, with a little twist.  If you're shining light onto a material, you can generally state that light power in = light power out.  Call light power in $P_{\textrm{in}}$.  Light power out can take the form of transmitted, reflected, and absorbed power.  Normalized to $P_{\textrm{in}}$, those quantities are $T$, $R$, and $A$, respectively.  So $T$, $R$, and $A$ tell you the fraction of power that went into these different channels (assuming we aren't distinguishing between reflection and scattering).  If energy is conserved (and since power is proportional to energy), then $T+R+A=1$.  So far so good?
Okay, then we make a switch. According to Kirchhoff's Law of thermal radiation, then for any given wavelength, absorptivity = emissivity.  This general result is a consequence of thermal equilibrium.  So then, $E=A$, leading to the final relation, $T+R+E=1$. 
A: Light incident on a surface is energy, and energy is conserved. It can be split into three fractions: reflected, absorbed, and transmitted. 

Reflectivity,  Absorptivity, and Transmissivity are defined as being fractions of the original incident light. So you could have 30% reflected, 10% absorbed, and then you would know 60% was transmitted since the remaining light had to go somewhere (conservation of energy), for a total of 100% which is equal to 1.

Image source: Cengel, Yunus A., et al. Fundamentals of Thermal-Fluid Sciences. McGraw-Hill, 2008.
