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I Infrared Light passing though Atmpshere

  1. Jul 26, 2016 #1
    Question, exactly what frequencies in the infrared spectrum can partialy penertarate our atmosphere and under what circumstances

    transmission.jpg
     
  2. jcsd
  3. Jul 26, 2016 #2

    Drakkith

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    Staff: Mentor

    It looks to me like your picture answers your question. Everywhere there's a dip in the graph, that part of the spectrum can get through at least partially.
     
  4. Jul 26, 2016 #3
    Well, if you want to be more exact, you'll have to decide what you mean by partially penetrate and by infrared. Since spectroscopic intstruments have incredible sensitivity these days, you might still be able to see some infra-red even if the atmosphere was 99.999% opaque at that particular wavelength/frequency. People often think about opacity in terms of "optical density" which is essentially a log scale for opacity. OD 5 means only 1 photon in 10^5 gets through, for example. Laser goggles are usually at least OD 5 for the wavelength of the laser.

    But let's say you're happy with knowing penetration on the % level as shown in this chart you've included. I've found a similar one that roughly "assigns" (as a spectroscopists would say) the various IR features to the responsible molecules.

    ?u=http%3A%2F%2Fwww.helpsavetheclimate.com%2Fatmoswindows1.gif

    There's a few fun things to understand here. The first is that the wavelengths that are absorbed are absorbed by very specific, simple but abundant atmospheric molecules- the ones shown on this chart. Wavelengths that penetrate do so because they are not absorbed by any of these most abundant atmospheric molecules. That is why the molecules only label peaks, not troughs. The second is why these molecules absorb this infrared light. For atoms, absorbed light can only go into a change in electron motion of the atom, which usually involves large changes of energy corresponding to visible light. For molecules however, the absorbed light can go into a change in rotation or vibration of the constituent atoms of the molecule relative to one another. These motions can have much lower energies than electrons, and so molecules can absorb in the IR. It is also possible though for molecules to absorb light that changes their electron motion also. Because covalently bonded molecules tend to have a very tight hold on their electrons compared with atoms, this takes even more energy, in the UV where O2 and O3 are labeled.

    Now, even though molecules can absorb light to change their rotation and vibration, from quantum mechanics you might expect that only very specific frequencies can be absorbed, and you'd be correct. However, because the molecules are moving around so fast, they experience the doppler effect. IR light that isn't quite right for a certain rotational or vibrational transition can still be absorbed by an H2O molecule that happens to be flying toward or away from the light at the right speed for the light to hit the transition exactly. Since at typical temperatures these molecules are moving at 100s of miles per hour, the narrow lines are broadened into the wide features seen on these graphs.

    So finally, I can answer your question about the conditions under which the absorption happens. There are no conditions. These molecules are always there, and they're always pretty hot on the kelvin scale, so this chart will hardly change with day of the year or location on the earth. The only thing that can really impact this chart is cloud formation. Clouds can absorb much more broadly and completely than individual molecules, because liquid and solid objects have many more ways to absorb energy than the simple rotations and vibrations allowed for gaseous simple molecules.

    Finally, I'm assuming you know how to convert wavelength to frequency, but I suppose maybe that was your actual question, since you asked about frequency but the charts are wavelength. You can do this by taking the speed of light and dividing by the wavelength. For example, the penetration window around 10um wavelength corresponds to the frequency:
    (3x10^8 m/s )/ (10um=10^-5 m) = 3x10^13 Hz = 30 THz.
     
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