# Does a thermal camera detect distant objects colder?

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?

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