When I look at pictures of the sun shield on the James Webb Space telescope (JWST), I see something that looks highly reflective (and hence must have a very low emissivity). My intuition tells me that this is no accident.
On the NASA website (https://webb.nasa.gov/content/about/innovations/coating.html) it is stated that the solar shield has a coating consisting of highly reflective (low emissivity) aluminum but also that the coating includes doped-silicon which has a high emissivity that "emits the most heat and light and acts to block the sun's heat from reaching the infrared instruments that will be located underneath it". My first issue is that if the emissivity of the solar shield is something that is needed to be optimized (that is, it is necessary for it to have either a very high value or a very low value) then why use a mixture of high and low emissivity materials? Surely this would lead to a middle-of-the-range emissivity?
My second issue is that when I apply the Stefan-Boltzmann law to determine the equilibrium temperature of the shield, I get a result that is independent of emissivity
$$P_{in} =P_{out}$$ $$\Rightarrow solarConstant * A*\epsilon =A*\epsilon*\sigma*T^4$$ $$\Rightarrow 1370 W/m^2 * 294m^2 = 294m^2* \sigma *T_{equilibrium}^4$$ $$T=121.26\deg C$$ Clearly the emissivity's, denoted $\epsilon$, in the above derivation cancel leading to an equilibrium temperature independent of emissive values. This leads me to conclude that the emissivity of the coating is completely inconsequential and that the solar shield could have been coated with a hypothetical black-body paint with perfect absorptivity and it would still function as well as it does now with aluminum. But if this is the case, then why does the NASA website even mention emissivity? Is it just an accident that the shield appears highly reflective but it doesn't actually need to be (since the final equilibrium temperature is the same regardless of the reflective/emissive properties of the shield)?