What would a piece of high emissivity polished metal inside a blackbody cavity look like? Darker than the walls of the cavity but the same color?

Question

I'm trying to work out the challenges to using an infrared pyrometer to measure the temperature of a piece of shiny metal. This device has no user-accessible emissivity setting, and my sample will have a very low emissivity in thermal IR.

I first want to confirm that my understanding of the underlying physics is correct, so I'll propose the following gedankenexperiment:

If I have a blackbody cavity with walls with unity emissivity at a uniform temperature say glowing orange-yellow when viewed through a small hole, and I put a sheet of clean, polished metal inside with a flat spectral emissivity of say 0.05 at this temperature and it's in an inert atmosphere or vacuum so it doesn't oxidize, my understanding is that it will equilibrate to the same temperature as the cavity walls.

The Stefan–Boltzmann law tells us that the radiated power of my sample will be proportional to $\sigma \epsilon T^4$ (as well as geometrical factors) so the total radiated power ("brightness") has to be 20x less if the temperature is the same as that of the cavity but the emissivity is flat at 0.05.

Though I've never seen Planck's law written with the emissivity (neither $\epsilon(\lambda)$ or just a constant $\epsilon$) in front, it seems to me that it needs to be there, and the implications are that for a flat emissivity my sample will be 20x darker but the same color as the cavity walls.

Am I correct to simply append $\epsilon(\lambda)$ to the front of Planck's law in this case? Is this a correct description of the visual appearance of the light coming from my gedankensample?

Answer 1

Though I've never seen Planck's law written with the emissivity (neither ϵ(λ) or just a constant ϵ) in front, it seems to me that it needs to be there

Yes, multiply by emissivity.

and the implications are that for a flat emissivity my sample will be 20x darker but the same color as the cavity walls.

A low emissivity piece of metal will be highly reflective. You are correct that it will emit only 5% of the energy as the blackbody. But you will measure quite a bit of reflected power as well unless you set up this experiment carefully.