# Measuring emissivity of material based on reflection

1. Feb 3, 2012

### lavalamp

1. The problem statement, all variables and given/known data

I have a university project that involves comparing various paints with regards to how they affect the insulating properties of houses.

As part of that I would like to measure the emissivity of the paint for a variety of wavelengths. Unfortunately I can not simply heat up a sample of the paints for this because I would like to measure emissivity in steps of 50 nm or so up to 300 or 250 nm, which is in the UV part of the spectrum.

So my question is, if I were to use a monochromator to select a narrow wavelength range of light, and then determine the fraction of that light that undergoes specular and diffusive reflection from the painted surface, would 1 minus that fraction be a good estimate of the emissivity of the paint for that wavelength range?

2. Feb 4, 2012

### marcusl

Puzzling post. Paint is too thin to provide significant insulation. Do you mean its effects on the energy balance for a house? Also measuring the UV emissivity makes no sense to me. I would pay attention to a) absorption or reflectance in the visible, where the sun's power spectral density is maximum, because this will determine the heat load added to the house, and b) the IR emissivity, to calculate radiative cooling.

3. Feb 5, 2012

### lavalamp

Indeed, I understand that a thin coating of paint, even if it had very low thermal conductivity, would be a decidedly sup-par insulator. That is why I decided to focus most of my attention on its emissivity.

I'm not going to measure emissivity too far into the UV, as I said perhaps to about 250 nm, which isn't that far outside visual range. Most of the suns power output is in the visual region, but does extend into the UV and back into the infra-red. This is why I would like to measure emissivity across this range, to more accurately determine the power absorbed by the paint when sunlight is shone upon it.

Of course the paint will only emit in the far infra-red due to its temperature, and I can similarly calculate the power loss by the paint due to this.

This is all just frosting though, my main question is; can emissivity be accurately determined solely by measuring the specular and diffuse reflection?

Intuitively I am inclined to say that yes it can since whatever is not absorbed must be reflected. However there could very well be other effects that I am not aware of that cause more light than would be expected to be reflected. Polarisation of light as it passes through the atmosphere for instance.