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According to Wikipedia:

Long-wavelength infrared (8–15 µm, 20–37 THz, 83–155 meV): The "thermal imaging" region, in which sensors can obtain a completely passive image of objects only slightly higher in temperature than room temperature - for example, the human body - based on thermal emissions only and requiring no illumination such as the sun, moon, or infrared illuminator. This region is also called the "thermal infrared".

However, using $\frac{hc}{\lambda}=k_\mathrm B T$, the temperature range 288–308 K (15–35 °C) is equivalent to 50–46.7 µm, while 8–15 µm is equivalent to 1800–960 K (using the same equation).

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up vote 12 down vote accepted

Your formula is wrong, is the short answer. It works fine as an order of mangnitude approximation, but it's only an approximation. What you need is Wien's displacement law:


where Wien's displacement constant $b=2.8977729(17)×10^{−3} m K$. Put $T=288K,308K$ into that and you get:

$\lambda=9.4-10.0\mu m$, which as you'd expect is at the lower range of the thermal infra-red sensors.

Note that $b=\frac{hc}{xk_B}$ where $x$ can be determined from the finding the peak of the black body spectrum from Planck's law - which has to be done numerically. It turns out that $x=4.96$, as opposed the value of 1 you in effect used in your original estimate. Hence your values are around a factor of 5 too high.

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The dark noise in the semiconductor is dependent on kT so if the SNR is ~5, why thermographic cameras need cooling? – Sparkler Mar 26 at 16:13
You can't work out the signal to noise ratio that way: that factor of 5 or so is nothing to do with the signal to noise, it's about where the peak of the black body spectrum is at a given temperature. The dark noise depends on the details of the electronics and the temperature. You lower the noise by reducing the temperature of your camera, while the signal depends on the temperature of what you are looking at. So you always improve the signal to noise by cooling the camera. – PhillS Mar 26 at 16:25

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