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Energy of Infrared Light

  1. Nov 2, 2011 #1
    I thought IR light has the least energy since UV light has the most (shorter wavelengths more energy). The article states that the IR light makes the air temperature at the highest, what is happening here?
     
  2. jcsd
  3. Nov 2, 2011 #2

    Drakkith

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    The IR radiation is directly heating the thermometer, not the air. UV light is almost entirely blocked by the atmosphere before ever reaching the surface.
     
  4. Nov 2, 2011 #3
    Ah, then why is the IR heat the thermometer more than the visible light? E=p/lambda and thus shorter wavelength means more energy? IR has the greatest wavelength out of all of them.
     
  5. Nov 2, 2011 #4
    I suspect that mercury (assuming a mercury thermometer) might absorb IR radiation more efficiently than visible.
     
  6. Nov 2, 2011 #5

    Drakkith

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    I don't know how effectively visible light heats the thermometer compared to IR, but the point of the article was that the thermometer wasn't in the visible light at all, but in the IR light. Since IR light is invisible Herschel didn't know it was there and the increasing temperature of the thermometer enabled him to discover it.
     
  7. Nov 2, 2011 #6

    DaveC426913

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    You're missing the point of the OP's question though:
     
  8. Nov 2, 2011 #7

    Drakkith

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    Whoops! Guess I didn't read the quote well enough.
     
  9. Nov 3, 2011 #8
  10. Nov 3, 2011 #9
    But Herschel held the thermometer just beyond the red end of the spectrum, and those wavelengths are mostly not absorbed.
     
  11. Nov 3, 2011 #10
    Wait what? The thermometer just beyond the red end must be in the Infrared region correct? I hope I'm not misinterpreting your statement.
     
  12. Nov 3, 2011 #11
    Yes, and if you look at the link you gave, you'll see the atmosphere won't stop much of this near-infrared radiation
     
  13. Nov 3, 2011 #12
    You mean this link?
    http://upload.wikimedia.org/wikipedi...ic_opacity.svg [Broken]

    It says "Most of the infrared spectrum absorbed by atmospheric gases (best observed from space)". I think it is pretty clear in saying that the infrared spectrum is absorbed.
     
    Last edited by a moderator: May 5, 2017
  14. Nov 3, 2011 #13
    Note the valley just to the right of red light at 1 μm. That valley is infrared and is absorbed less than red light (although more than green light).

    Remember that visible light is a small section of the spectrum. Infrared is a huge range in comparison. While much of that range is blocked by the atmosphere, there are specific frequencies that are not. These appear as valleys on that graphic.

    Edit to add:
    The link you gave only considers how much of each frequency is absorbed. It is also important how much is emitted. This link shows how much actually reaches sea level:
    http://upload.wikimedia.org/wikipedia/commons/4/4c/Solar_Spectrum.png

    There you see there is still a problem. Since the highest point in infrared is still lower than almost all of visible light. I can only guess at an explanation:
    1. the thermometer absorbed infrared more than visible.
    2. the prism absorbed visible more than infrared.
    3. the thermometer happened to be near the ends of visible where the intensity falls lower than near infrared.

    As I said, these are only guesses. I'll be interested to see if anyone else can provide a more certain answer.
     
    Last edited by a moderator: May 5, 2017
  15. Nov 3, 2011 #14
    Oh, so the small valley accounts for it having a higher temperature. Also, maybe the fact that it is more concentrated than the other light rays that are scattered makes it heat up the thermometer more than the green - violet light.
     
  16. Nov 3, 2011 #15

    fluidistic

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    My thoughts:
    Visible light passes through the glass material containing the mercury (thermometer). It gets mostly reflected from the mercury so overall it doesn't heat that much the thermometer. However the infrared light is more absorbed than the visible light by the glass (I have no idea about the mercury though). The glass then heats up and since it's in direct contact with the mercury, the mercury heats up and the thermometer shows an increase of temperature greater than the one with visible light.
     
  17. Nov 4, 2011 #16

    Redbelly98

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    IR has less energy per photon than visible or UV. So if we had the same number of photons of IR , visible, and UV, then yes the IR would have the least amount of energy.

    But that doesn't apply here -- there are unequal numbers of photons of the different wavelengths, and so if there are enough photons of IR then it can have more energy than is in the visible portion of the spectrum.
     
  18. Nov 4, 2011 #17

    A.T.

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    The irradiance in the visible portion of the spectrum is greater than for for IR:

    Solar_Spectrum.png

    From: http://en.wikipedia.org/wiki/Sunlight

    But there are some holes due to filtering by the atmosphere. The prism might also have filtered more visible than IR light.

    It should also be noted that the dispersion of light is not proportional to wavelength:

    Dispersion-curve.png

    Visible range is orange. From: http://en.wikipedia.org/wiki/Dispersion_(optics)

    So holding the thermometer into the IR area might capture a wider range of wavelengths, than within the visible and UV area (more surface under the first graph due to a wider wavelength range, despite lower irradiance).
     
    Last edited: Nov 4, 2011
  19. Nov 4, 2011 #18

    fluidistic

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    I don't think the holes really matter. He placed the thermometer right after the end of the visible spectrum. So just before the O2 hole.
    According to wikipedia,
    but I don't find anything related to the near visible infrared.
     
  20. Nov 4, 2011 #19

    Redbelly98

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    Yup, I was aware of the spectrum peaking in the visible, but as it seemed the OP's main confusion was about the energy-per-photon, I just wanted to clear that up. It seemed that nobody else was addressing that.

    I thought about the lesser dispersion of n in the IR, and you bring up a good point that the transmission of the prism could have played a role. Another possibility is that the thermometer had a greater absorptivity for near-IR wavelengths -- effectively acting as a filter.
     
    Last edited by a moderator: May 5, 2017
  21. Nov 6, 2011 #20
    Wiki is correct, but "mostly" doesn't mean all. It is this infrared that escapes absorption by the atmosphere that warms you when you step from the shade into the direct sunlight.

    Almost all surfaces absorb different proportions of infrared radiation and visible radiation. This is usually measured by their visible light albedos and their infrared albedos.

    Glass is largely transparent to visible light and largely opaque to infrared. Thus it transmits the one and absorbs the other. Mercury absorbs both. This is why the thermometer heated up more in the infrared than it did in the visible. It was absorbing more radiant energy per unit of surface area in the infrared zone.

    Edit: Idle thought. Was Herschel observing indoors (through a glass window) or outdoors?
     
    Last edited: Nov 6, 2011
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