For an opaque surface, what is the relationship between emissivity and reflectivity?

• A. Emissivity = 1 − Reflectivity
• B. Emissivity = Reflectivity
• C. Emissivity = 1 / Reflectivity
• D. Emissivity Independent of Reflectivity

Answer: (A)

On an opaque surface, no energy transmits through, so the radiative budget at a given wavelength is split between reflection and emission (with absorption filling the rest). Since there is no transmission, the fraction that is not reflected must be absorbed, and for a surface in thermal equilibrium the absorbed portion equals the emitted portion. That makes emissivity equal to the absorbed fraction, which is 1 minus reflectivity. In other words, emissivity = 1 − reflectivity. So a surface that reflects 20% would emit 80% at that wavelength.

The other options clash with energy balance or produce illogical results: emissivity equal to reflectivity would imply the emitted and reflected portions are always the same, which isn’t generally true; emissivity equal to 1 divided by reflectivity would give values outside the 0–1 range for many cases; and emissivity being independent of reflectivity ignores the fact that what isn’t reflected is what can be emitted (per Kirchhoff’s law).