Hi so I have this senior design project involving cooling human skin as it is irradiated by an infrared LED. Human skin not only is super absorbent to IR, but that IR goes THROUGH the thickness of it really well. There are some very long path lengths of IR photons within the human body.

In my heat transfer class, we only saw radiation as dumping all its energy right at the very surface of some block of mystery meat or whatever. In that class we only cared about absorption in terms of how much radiation isn't reflected. We then treated all the radiation that got absorbed as being "stopped" by that surface essentially. After that the heat from this interaction would be distributed by conduction within the receiving block thing.

Million dollar question: What is happening to that topmost surface temperature if absorbed radiation is going THROUGH the surface? So now my deeper layers are being heated by both conduction and direct radiation?

The end goal is to figure out the surface temperature of this irradiated spot of skin, then cool it by cold air convection until a goal surface temperature is reached. Any diagrams in answers would be super helpful. I am a visual creature. Part of my trouble is photons are hard to visualize and I keep picturing bullets stopped by walls when I know photons don't have mass and it's a poor analogy. It's really tripping me up.

  • $\begingroup$ How can your skin model both be "superabsorbent" and "IR goes through... really well" ? THose are contradictory. $\endgroup$ Mar 8, 2021 at 15:30
  • $\begingroup$ This is nontrivial. You need to read up on Blackbody Radiation as well, as that is one of several mechanisms by which the skin dumps energy(heat). Then you'll need the heat transfer rate for the different skin layers, and so on. $\endgroup$ Mar 8, 2021 at 15:52

1 Answer 1


Photons either pass through a given material or they don't.

If they pass through, it's as if the material isn't there as far as they're concerned (ignoring refraction).

If they get absorbed, then they dump all their energy into the material. This might excite the incident atom such that it then re-emits the photon with the same or different wavelength, such that the effective "heating" of the material is just the difference between the energy given by the incoming and outgoing photons.

What you need to remember is that a beam of light is composed of an almost infinite number of photons. So whenever you talk about light going through a material and heating it, you're talking about a few different groups: photons that went through the material completely undisturbed (i.e. with the same energy as before passing through) and which don't contribute at all to heating the material, photons which are entirely absorbed by the material without any "output" photons and therefore heat the material by their energy, and intermediary photons which are absorbed but also emit photons themselves and therefore heat the material less.

And if it is possible for light to pass through a material, you'll get an iterative process: some fraction of the photons (let's say 70%) will be absorbed by the material's surface, and the rest (30%) will pass through. On the second layer, 70% of the photons which managed to pass through will be absorbed and 30% will pass through, rinse and repeat until you get entirely past the material. (Real life obviously doesn't really have "layers", but it's a useful mental image). Only some fraction of the absorbed photons will emit photons of their own in random directions; these outgoing photons will then also have to gamble to see how many "layers" they can pass through before being absorbed.

So, to answer your million dollar question: the surface temperature will be increased by the fraction of the photons which don't pass through. It will also have some "inner" heating from "outgoing" photons emitted by inner layers which have absorbed some of the photons which managed to pass through the surface, but that will likely be a very minor complement.

  • $\begingroup$ Thanks! Does the path length matter? If I have two of the same intensity lights (with same power) but one penetrates deeper in one block...is the temperature of the surface of that block any lower than the surface of a block where the light didn't penetrate as deep? $\endgroup$
    – MeowMix
    Mar 6, 2021 at 1:16
  • $\begingroup$ I have had severe skin burns skiing on very cold cloudy days, like 8 Fahrenheit degrees. just saying keeping the surface cold won't help much. $\endgroup$
    – kamran
    Mar 6, 2021 at 5:29
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    $\begingroup$ @MeowMix: Most likely yes; the surface temperature will be most impacted by the fraction of photons which are immediately absorbed. If more photons manage to make it deeper into the material, then they'll have almost no impact on the surface. $\endgroup$
    – Wasabi
    Mar 6, 2021 at 13:13
  • $\begingroup$ @kamran I had that exact same thought but I learned that sunburns are not thermal burns due to infrared. Sunburns are UV damage to your DNA and so are totally different. You are not sunburned because your skin had too much heat energy. $\endgroup$
    – MeowMix
    Mar 6, 2021 at 19:55

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