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I am slightly confused about infrared radiation and the equations related to it.

$P = A \epsilon \sigma T^4$ (1)

and

$B_{\lambda}(\lambda,T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k_B T}}-1}$ (2)

I understand that equation one is just integral form of equation two. Are P and B just the power output of the black body radiation?

Perhaps I should explain. I am working with an infrared thermometer (MLX90614) and I am wondering exactly what this picks up. Does it pick up the power of the infrared radiation through a lens and convert it to an electric signal so we can get the temperature of the object? Or does it measure the wavelength of the infrared radiation that is coming in and get temperature that way? THANKS!

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  • $\begingroup$ I'm not sure abt the equation but the intensity of infrared received by yr "thermometer" will affect its resistance and your sys will then measure this varying current or voltage. I'm sure there is a preset values of range of wavelength on your sys. $\endgroup$
    – user6760
    Commented Apr 9, 2015 at 7:29

2 Answers 2

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This thermometer is sensitive in a band 5.5um-14um. The equation (2) (I think) shows your black-body spectrum and the IR-th integrates the intensity in the above range.

Normally, such a probe should also measure the ambient temperature and it should know the emissivity $\epsilon$ value of the material you point at, to properly calculate the temperature. E.g. shiny metals are more difficult to measure, because the large part is reflected from other objects. Also consult the manual, page 45.

edit: In summary, IR thermometer can see only one 'color' and the only thing it can do is to measure the intensity of this single color. Imagine a heated iron rod. Forget about what you see in visible spectrum. If you select only this 'IR deep-red color', it's intensity will grow up with higher temperature.

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  • $\begingroup$ Ya still wondering how the ir-sensor actually gets temperature. Those two equations have T as the independent variable - I don't see a T= anywhere, so how can we determine the temperature or something from the wavelength it is emitting? $\endgroup$
    – sci-guy
    Commented Apr 9, 2015 at 11:52
  • $\begingroup$ Yes, it looks strange, but it is not determined from the wavelength(!). It is determined form the intensity in a narrow IR band. We are very biased because we think in terms of visible light and colors. Just imagine the thermal camera - the hotter = the brighter. No colors, no frequencies. $\endgroup$
    – jaromrax
    Commented Apr 9, 2015 at 12:41
  • $\begingroup$ @renegade - you look at (almost) one frequency only, so what remains to measure is only the intensity. And at this frequency - black body radiation (usually) prevails over the reflection (those we are used to from visible region) $\endgroup$
    – jaromrax
    Commented Apr 10, 2015 at 6:57
  • $\begingroup$ I don't think you can use the absolute intensity of a specific wavelength, because then how far away you are standing would have a huge impact on the measured temperature. $\endgroup$
    – pentane
    Commented Apr 15, 2015 at 19:12
  • $\begingroup$ The distance problem fortunately compensates: the farther you are, the lower intensity per area you see with $\frac{1}{R^2}$. However, the larger area you see, because your optics has a defined viewing angle - it goes with $R^2$. $\endgroup$
    – jaromrax
    Commented Apr 15, 2015 at 20:23
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I think, if you plot $B_{\lambda}$ in equation (2) with respect to $\lambda$ you get a distribution. As in

https://en.wikipedia.org/wiki/Black-body_radiation#/media/File:Black_body.svg

So, if you want to find the temperature from wavelength you need the find the point where the intensity peaks. That is to say you need to find the maximum point of intensity. Therefore you need to solve $$ \frac{\partial B_{\lambda}}{\partial \lambda}=0 $$

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