If emissivity and reflectivity are inversely proportionate, why does glass have a high emissivity of around 0.95-0.97 as well as being very reflective for IR Radiation?
normally it works but not with glass!
Can anyone explain this?
According to this article the lenses on thermal cameras are not made from glass, but rether from Germanium, CZinc Selenide or Zinc Sulfide. These materials are not transparent to light so it's quite reasonable for them to have a high reflectivity.
Response to comment:
The emissivity and reflectivity only have to add up to one at the same wavelength. So if the emissivity is high for infra-red that doesn't clash with the reflectivity being high for visible light. This (or rather it's converse) is exactly why greenhouses heat up in visible light. They have a high emissivity and low reflectivity at visible wavelengths but a low emissivity and high reflectivity at IR wavelengths.
I experienced the same problem: too often, the emission value of glass is misquoted in different sources on the internet, as I found to my own dismay. My own calculations found about an average reflectivity R of 0.2, transmittance T of 0.4 and absorption of 0.4 for glass of about 2mm thick. (I calculated those values using absorption spectrum graphs of wikipedia for soda lime glass, http://en.wikipedia.org/wiki/Soda-lime_glass, using black body radiation curve to get a weighed average, and hemispherical averaging to account for the different reflectivity, transmittance and absorbance for different angles at the same wavelength). The emission value for glass quoted is often 0.82, however: this cannot be right. The error is in the fact that only the 0.2 reflectivity of glass is deducted from 1, instead of both transmittance and reflectivity. The glass absorption coefficient for glass is (kirchhoff law) equal to the emissivity of glass, and thus equal to approximately 0.4, or 40 percent. When I used this emission value in the heat transport model that I created, the outcome was a calculated U value of about 6 W/m2*Kelvin for the glass plate, in good agreement with the value to be expected for a glass plate.
In an isothermal steady state condition, meaning when the temperature is uniform and not changing with time,
%Reflected + %Transmitted + % Absorbed = 100%
For opaque system,
%Reflected + % Absorbed = 100% .........(1)
Now, if the object absorbs infrared radiation, its energy (and thus temperature) will increase, but as the object is in steady state, to offset that increase in temperature, the rate of emission must be equal to the rate of absorption. So,
% Absorbed = % Emitted.
Substituting in equation (1),
%Reflected + % Emitted = 100% .........(2).
So reflectivity is reciprocal to emissivity.
For translucent system,
%Reflected + %Transmitted + % Emitted = 100%.........(3).
This is the associated physics.
It is true for glass also, only transmissivity comes into picture, but still, reflectivity is reciprocal to emissivity.
Based on the experiments you describe in the comments, it seems like you might very well have a reflectivity of 20-30% in your window, for the spectral region where your camera measure. The question is where you got the high emissivity numebers from. It seems likely that the problem is that you're assuming the emissivity and reflectivity is the same throughout the infrared region. The high emissivity might be for another part of the spectrum than where your camera measure.
To me the problem is not linked to the reflectance, but to the surface finish.
Picking up an image of a reflection in the visible part of the spectrum is possible even on a black carbon surface if that surface is polished well enough. To convince yourself of this, simply pick a dark material with a well polished surface (or a not so well polished surface but placed at an highly inclined angle). Then place the surface to reflect the sky day light. And you will see the surface reflecting the color of the sky even if the surface color is blue, red or any other dark color/tint.
Thus, since the glass window is a well polished surface, it will reflect any image of the surrounding in the thermal infrared. This is also true for the sky in the visible part of the spectrum as would tell you any bird that survives hitting a cleaned glass window at spring time... ;-)