The rule breaker, emissivity + reflectivity = 1 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?
 A: 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.
A: 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.
A: I highly recommend to read "The Ultimate Infrared Handbook for R&D Professionals". There's a thing that many of you are not saying reflection, transmission and absoprtion depend on the wavelength. Your are probably measuring with normal spectrometers which measure up to 2000 nm. Must IR cameras for example InSb detecta between 3000 to 5000 nm in that region glass is opaque that is why is reported has a close to 1 value for example "glass has an emissivity of 0.85–0.90 in the 8–14 μm waveband" which is the region for MCT, QWIP and Microbolometer. The other part is that glass reflections are specular because of the polished. Also in the region below 8 μm there is a lot of objects reflecting as you may expect by Wiens law. Here you can see some considerations if you are trying to measure temperature in glass.
A: 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.


normally it works but not with glass! 


It is true for glass also, only transmissivity comes into picture, but still, reflectivity is reciprocal to emissivity.
A: 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.
A: 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... ;-)
A: One has to be careful with reflection.  If you take a car rear-view mirror, it might seem to be close to 100% reflectivity, but in practice may be around only 60% (this is helped partly because the inside of the car is usually darker than the outside).
If you couple this with the usual compressed, pseudo-color look up table that most thermal cameras use, the apparent reflectivity of objects can be 'accentuated'.
A: The law of energy conservation certainly also applies here. What is not reflected will be absorbed. R+A=1 has to hold. And according to Kirchoff's law E=A (absorptivity is emissivity) you could write R=1-E.
If you have to consider transmission through the glass as well, then R+A+T=1 would hold.
You do not specify what type of glass you are talking about, what wavelength (or wavelength band) and what incident angle you are talking.
E.g. for window glass you will have approx. 10% reflection and 90% emissivity at room-temperature radiation (brightness temperature) and normal incidence (=0° =perpend. to surface).
For incidence angles greater than 81° you will get 50% reflection and 50% emissivity.
If you state that glass has a high IR-reflectivity then you might think of lowE-coated glass as used in insulated glazing units. In this case you will have to consider all coated and uncoated layers of all windows pane. But again for each layer the law of energy conservation will apply.
You can find a lot read more about the emissivity and reflectivity of window glass and all relevant dependences in my open-access publication.
