2
$\begingroup$

Say we have an infrared camera which measures some amount of radiation, in a spectral bandwidth which is given, between wavelength $\lambda_1$ and $\lambda_2$ from a perfect black body. How is it possible to estimate its temperature? Planck's law and/or Stefan-Boltzmann's law? Wouldn't the amount of radiance that reaches the detector be strongly dependant on the distance to the object?

I am interested in the underlying physical concepts and a detailed description of how they are used to estimate the temperature of objects. A detailed dummy explanation would be appreciated.

Related: How does Infrared Relate to heat?

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
1
$\begingroup$

Emissivity of the material certainly does play a role. Also, if the material is particularly reflective, you might get stray energy bouncing in from the surrounding (i've seen this happen with aluminum).

If your sensor absorbed all radiation, you'd use the Stephan-Boltzman law, as you've mentioned. The trick that you need is to find the fraction of the energy that is stored in band of interest. For this you'd look at Planck's law, which lets you find the energy as a function of wavelength:

$$ B(T, \lambda) = \frac{2hc^2}{\lambda^5} \left[ e^{ \frac{hc}{\lambda k_B T} } -1 \right]^{-1} $$

so the fraction of the total energy emitted in your band becomes like an effective emissivity

$$ \epsilon_{eff} (T) = \frac{\int_{\lambda_1}^{\lambda_2}B(T, \lambda)d\lambda}{\int_{0}^{\infty}B(T, \lambda)d\lambda} $$

that you use with the S.B. law so that the signal from your sensor is:

$$V(T) = C_{cal} \,\epsilon_{eff}(T) \,T^4$$

I've avoided all of the solid angle inegrations involved with these intergral (as you'd see in the derivation of $\sigma$) under the assumption that you'll need to take at least one calibration measurement (with the target at known $T$) in order to account for optics, sensor sensitivity etc. anyway. The calibration constant $C_{cal}$ takes care of all of those shenanigans.

Distance will not change the result. Although energy reaching the lens goes as $d^{-2}$ (becoming smaller as you move away), that energy is concentrated over a smaller area on the imaging plane, also going as $d^{-2}$. So the flux incident on a given pixel remains constant (registering as the same temperature).

This will hold true until the image of the object becomes smaller than one pixel. At that point the average flux at the pixel will start to drop and temperature will appear to fall.

Note: because of the peak-shape of the black body curve, it is possible for some combinations of temperature ranges and sensor bandwidths that $e_{eff}$ could be non-monotonic, introducing ambiguity by allowing $V(T)$ to have the same value at different temperatures. A quick plot of the expected curve in advance will let you know if that'll be a problem.

Improve this answer
$\endgroup$
4
  • $\begingroup$ My question was a bit imprecisely formulated. Tried to make it a bit more clear now. $\endgroup$
    – Julian
    Commented Feb 14, 2015 at 17:07
  • $\begingroup$ Thanks. I realized yesterday that my answer wasn't very complete as it relates to the calculation/wavelength. Working on an edit now. $\endgroup$
    – user3823992
    Commented Feb 14, 2015 at 17:11
  • $\begingroup$ But doesn't the SB law originally give power/area, i.e. $watts/m^2$, so one would need knowledge of geometry? $\endgroup$
    – Julian
    Commented Feb 16, 2015 at 11:53
  • $\begingroup$ If you really wanted pre-compute the sensitivity of your system from first principles, yes, you'd need to calculate the actual energy transfer to your camera. But you'd also need to fully account for the imaging optics (including non-idealities) and the absolute response of your sensor (volts or amps per unit incident flux). In practice, all of that gets captured by that one calibration constant, which can be determined by viewing the object once at a known temperature. The only thing that you have to account mathematically is the spectral sensitivity. $\endgroup$
    – user3823992
    Commented Feb 16, 2015 at 15:20
1
$\begingroup$

An ideal blackbody has a unique emission spectrum for a given temperature: Blackbody spectrum

With a uniquely located peak, estimated by Wien's law: $$ \lambda_\text{max} = \frac{2.898\times 10^{-3}\text{m K}}{T} $$ A cheap estimate might interpolate the spectral maximum from several singular spectral measurements and infer the temperature from Wien's law, assuming the emitter is a blackbody. A professional IR camera will quite accurately determine the emission spectrum and use Planck's law to calculate the temperature.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Correct me if I'm wrong: An IR camera measures the spectral radiance throughout its spectral band, say for example from 4 to 12 $\mu m$, and calculates the peak of the resulting curve which gives the wavelength at which the curve peaks, $\lambda_{max}$. From this one can simply calculate the temperature from Wien's displacement law. Or can one use Planck's law directly somehow? Would it be possible to calculate temperature using the Stefan-Boltzmann law? If I'm completely off, would you care for a more detailed explanation? $\endgroup$
    – Julian
    Commented Feb 14, 2015 at 16:55
  • $\begingroup$ that would mean, however, that if the camera has a spectral range outside that of the peak of the spectral radiance, it would not be able to accurately measure temperature. I have edited the question a bit, as the inital post was a bit imprecise. $\endgroup$
    – Julian
    Commented Feb 16, 2015 at 8:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.