Thermal cameras are often calibrated and display the temperature color-coded in $^\circ$C and I wonder how that works...

A black body emits photons because of it's thermal energy. The hotter it is, the smaller the wavelength $\lambda$ of the emitted photon (e.g., see https://en.wikipedia.org/wiki/Black-body_radiation#/media/File:Black_body.svg). The photon has thus larger energy $E_{Photon} = \frac{h \cdot c}{\lambda}$. That means, a pixel of a thermal camera will receive more energy (and therefore outputs a larger gray value), when the observed object is hotter. In addition, the hotter the object, the more photons are emitted (also obvious from the graph of black-body radiation). Thus, both effects go into the same direction: Hotter object $\rightarrow$ more energy received (more photons + photons have shorter wavelengths).

I guess manufacturers now simply do a calibration of a reference panel of different temperature (at the same time accounting for quantum efficiency of the pixel and other instrumental effects).

However, reason $1$ (large $T \rightarrow $ large $E_{Photon}$) is independent of the distance to an object (even if in 500 m the photon energy will be directly related to the temperature $T$). But reason 2 is not: The amount of photons reaching the sensor strongly depends on the distance as they are emitted in every direction...

So every temperature calibration will fail. But how can a thermal camera then detect heat?

  • $\begingroup$ en.wikipedia.org/wiki/Thermographic_camera $\endgroup$
    – anna v
    May 19, 2021 at 13:04
  • 1
    $\begingroup$ Sorry, but found nothing in that link that answers my quesion... $\endgroup$ May 19, 2021 at 13:14
  • $\begingroup$ Do you know what a bolometer is and how it works? $\endgroup$
    – Jon Custer
    May 19, 2021 at 13:26
  • $\begingroup$ @Jon Custer: Not really. Is it sensitive to photon energy only (independent on photon number)? $\endgroup$ May 19, 2021 at 13:30
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    $\begingroup$ You need to ponder more on how black bodies come into equilibrium. $\endgroup$
    – Jon Custer
    May 19, 2021 at 14:30

1 Answer 1


The amount of photons reaching the sensor strongly depends on the distance as they are emitted in every direction

This is true, and it'll give smaller brightness of the pixel—if all the light received from this object goes to this single pixel. But thermal cameras are mainly used to image extended objects.

Consider what happens when you image an extended object, i.e. one large enough to have its image across multiple pixels. As this object comes closer to the camera, the light from object's surface that gets to the image plane grows according to the inverse-squares law. But the image of the object also becomes larger, covering more pixels. Thus the area of the image also grows as per inverse square law. This means that the intensity at each pixel remains the same: the increase of light flux is "used" to cover increasing area of the image.

In fact, this is exactly the same as you can see in visible-light photography: brightness of an extended object doesn't depend on the distance to it.

  • $\begingroup$ Of course, good explanation! So the number of photons per pixel stays constant, and as thermal camera pixels seem to work with metal (free electrons) the quantum efficiency is not very wavelength-dependent. So you can calibrate for heat... Got it, thanks. $\endgroup$ May 19, 2021 at 15:15

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