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Thermal cameras designed to image wavelengths between 7-14µm (which is the peak for blackbody radiation around "room" temperatures) use lenses made of germanium, zinc sulfide, zinc selenide, or various chalcogenide glasses. All of these lens materials are solid substances, but they are practically transparent to thermal radiation at the blackbody temperatures where they operate. How does this work?

E.g., a thermal camera consisting of a microbolometer sensor array behind a germanium lens operating at a stable temperature of 20°C is able to resolve the radiant temperature of external objects also around 20°C. Why doesn't the lens itself blind the sensor with its own 20°C radiation?

I.e., do these materials deviate markedly in terms of their thermal radiation from a blackbody model? If so, how and why? Or is this transparency the result of some other physics?

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    $\begingroup$ Those materials all have bandgaps wider then the IR wavelengths, so the IR will pass through with minimal absorption. And, as real materials, they are not ideal blackbodies. $\endgroup$ – Jon Custer Jan 9 '18 at 19:58
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    $\begingroup$ It is mainly that there is no absorption by phonons. And these substances do not radiate because absorption is so low. $\endgroup$ – Pieter Jan 9 '18 at 20:04
  • $\begingroup$ @JonCuster – so due to some known physics (but not familiar to me), photons of the desired spectrum are minimally absorbed. And of course they are not ideal blackbodies. My uninformed assumption was that real materials in typical conditions radiate something like an ideal blackbody. Does practice diverge so much from the blackbody model that these substances do not radiate significantly in that wavelength range? For analogy: I'm trying to imagine a solid, transparent to light, that when heated to ~5000K does not emit so much light itself as to overwhelm transmitted light. $\endgroup$ – feetwet Jan 9 '18 at 22:34
  • $\begingroup$ The physics is Kirchoff's law. Metals like aluminium hardly radiate in the IR. Same for substances that are IR transparent. $\endgroup$ – Pieter Jan 9 '18 at 22:39
  • $\begingroup$ @Pieter: Kirchoff's law looks sort of applicable: I can see with the thermal camera I have on loan that metals have high thermal reflectivity. But they are not thermally transparent. And they do radiate: For example, I put a sheet of aluminum foil on a burner, and its heating and cooling was plainly visible to the microbolometers. I.e., once it was heated above the temperature of any other surrounding heat source, its thermal reflectivity was overwhelmed by its thermal radiation. $\endgroup$ – feetwet Jan 9 '18 at 23:26
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How can a solid be thermally transparent?

Thermal cameras designed to image wavelengths between 7-14µm (which is the peak for blackbody radiation around "room" temperatures) use lenses made of germanium, zinc sulfide, zinc selenide, or various chalcogenide glasses. All of these lens materials are solid substances, but they are practically transparent to thermal radiation at the blackbody temperatures where they operate. How does this work?

Light Absorption

Atoms and molecules contain electrons. Think of these electrons as being attached to the atoms by springs. The electrons have a natural frequency at which they tend to vibrate. When a light wave with that same natural frequency impinges upon an atom, then the electrons of that atom will be set into vibrational motion. If a light wave of a given frequency strikes a material with electrons having the same vibrational frequencies then those electrons will absorb the energy. Different atoms and molecules have different natural frequencies of vibration, they will selectively absorb different frequencies of visible light.

Light Reflection and Transmission

Reflection and transmission of light waves occur because the frequencies of the light waves do not match the natural frequencies of vibration of the objects, instead of vibrating in resonance at a large amplitude the electrons vibrate for brief periods of time with small amplitudes of vibration and the energy is reemitted as a light wave.

If the object is transparent then the vibrations of the electrons are passed on to neighboring atoms through the bulk of the material and reemitted on the opposite side of the object.

If the object is opaque then the vibrations of the electrons on the material's surface vibrate for short periods of time and then reemit the energy as a reflected light wave.

An ideal lens, for whatever frequency, transmits far more energy than it absorbs or reflects.

Absorption can especially be a problem with infrared lenses where the Coefficient of Thermal Expansion (CTE) of most IR materials is orders of magnitude higher than those of visible glasses, creating large changes in the refractive index.

Much as with an achromatic doublet, which uses a positive and negative element of different materials with equal and opposite amounts of chromatic aberration to correct for color, you would solve an athermal doublet equation to ensure that temperature changes are compensated for and focus is maintained. Calculations must also compensate for expansion and contraction of the lens body.

The use of materials more expensive than glass, difficult to work with (both grinding/polishing and the additional calculations) and the small market for such lenses tend to make them extremely expensive.

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  • $\begingroup$ This is a good foundation for the answer – these materials don't strongly absorb in the desired bands – but now: Aren't the atoms in the lenses (at whatever ambient temperature) themselves emitting photons concentrated in the wavelength band around their temperature? Is it obvious that/why the IR radiated by the atoms in these lenses would be orders of magnitude smaller than the transmitted IR radiation they are lensing, and that's why they don't wash out the IR images they are focusing? $\endgroup$ – feetwet Jul 25 '18 at 15:02

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