If a surface has zero transmissivity, what must be true about emissivity and reflectivity?

If a surface has zero transmissivity, what must be true about emissivity and reflectivity?

• A. They sum to 0
• B. They sum to 1
• C. They must both be 0
• D. They must both be 1

Answer: (B)

When a surface has zero transmissivity, it cannot let radiation pass through; all incident energy is either absorbed or reflected. For a given wavelength and direction, the portion absorbed is the same as the portion that is emissive (Kirchhoff’s law), so emissivity equals absorptivity. Since there’s no transmission, the sum of emissivity (absorptivity) and reflectivity must account for all incident energy, meaning they add up to 1. Put another way, a highly emissive surface reflects less, while a highly reflective surface emits less, with perfect black bodies having emissivity 1 and reflectivity 0, and perfect mirrors having emissivity 0 and reflectivity 1.

When a surface has zero transmissivity, it cannot let radiation pass through; all incident energy is either absorbed or reflected. For a given wavelength and direction, the portion absorbed is the same as the portion that is emissive (Kirchhoff’s law), so emissivity equals absorptivity. Since there’s no transmission, the sum of emissivity (absorptivity) and reflectivity must account for all incident energy, meaning they add up to 1. Put another way, a highly emissive surface reflects less, while a highly reflective surface emits less, with perfect black bodies having emissivity 1 and reflectivity 0, and perfect mirrors having emissivity 0 and reflectivity 1.