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How do we see things?

  1. Sep 15, 2007 #1
    When all the light waves (or, any kind of EM waves) mix around, shouldn't there be just one resultant wave which has a particular frequency and wavelength (at a given time)? But why is it that the waves do not mix and loose their property.

    Similarly, in the visible light spectra, all the 7 color frequencies are allowed to mix and form white color. But, why can't, say radio waves and micro wave, cannot mix and form visible light?
  2. jcsd
  3. Sep 15, 2007 #2


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    The ampliltudes (strength) of waves add, not their frequencies. If you combine two waves of different frequencies the result is not a wave with the "average" frequency.
  4. Sep 15, 2007 #3


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    And there are not 7 visible frequencies. It's an essentially seamless progression of frequencies from near-infrared (for some) to upper violet. That's for humans; other species have sensitivity into deeper infrared and ultraviolet.
    Eyes are only one component of vision, as well. While they're extremely competent photon detectors, it's ultimately up to the brain to assemble and interpret the data that they supply.
  5. Sep 15, 2007 #4
    If you drop two pebbles in a pond, the receding waves travel through each other and reach the other sides of the pond. If 6 pebbles are dropped each individual resultant wave are still there and are just travelling through each other, they dont add up together and become one wave. Now instead of pebbles we have little vibrator/oscillators each vibrating at different frequencies. Same thing happens and the frequency of one wave does not speed up or slow down the frequency of another, they still just travel through each other no matter how many pebbles or oscillators, frequencies unchanged.
    Last edited: Sep 15, 2007
  6. Sep 15, 2007 #5
    As Danger pointed out, the visible spectrum is continuous. Any visual separation due to seemingly different colors is fundamentally psychological.

    Folks, can't one interfere two invisible E-M waves whose difference (or sum) in frequency is of a wave possibly visible?
  7. Sep 16, 2007 #6
    Few questions here:

    1. Is photon a name used only for the light context, or is it a general term for any EM waves?

    2. Assume I've 2 parallel mirrors and there is one stream of photons going back and forth between these mirrors. If the light waves add together and if the total sum of the wave aplitude is zero, where did the energy go?

    3. If the amplitudes can be doubled by two photons in the same phase, can I add 100 photons together and get an amplitude 100 times more than the original photon? What is the limit to it?

    4. If I sum up 2 photons perfectly out of phase, can I say that pair has only particle nature and no wave nature (and that way, learning physics would be much easier :))
  8. Sep 16, 2007 #7
    1. A photon is the particle (carrier) for electromagnetism in general.

    2. Where did what energy go? What initial state is this final state relative to?

    3. You seem to be describing a laser being pumped up.

    "The efficiency of a CO2 laser is over 10%."


    4. Great question. I guess there would be no photon at all, of either particle or wave nature.
  9. Sep 16, 2007 #8
    I once asked myself the same question.

    It turns that out that you can't have a perfect stream of photons, there will always be some diffraction.

    When you talk about the waves adding up to zero, this is only happening at one particular point in space. You'll find that in order for your system to be physically possible, there will have to be points nearby where the interference will be constructive. This is where the energy goes.

    If you want to visualize my explanation, draw two point sources on a sheet of paper. Then draw circles around them representing wave crests at regular intervals. Look at their intersections and you'll see what I mean.
  10. Sep 16, 2007 #9

    Claude Bile

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    1. Any EM wave.

    2. High-reflection dielectric stacks (which essentially behave as a whole bunch of parallel mirrors stacked on top of one another) are often called 1-dimensional photonic crystals because they exhibit the photonic band-gap effect in a single direction.


    To answer you question - Photon's cannot be generated within the "cavity" as there is technically no optical mode available to radiate into, and photons cannot be injected into the "cavity" as they would be reflected, again, due to no optical mode being available within the "cavity" - this is the essence of the so-called photonic band-gap effect. Essentially, there is no way the energy could have gotten where it is in the 1st place - your initial proposition is flawed.

    For questions 3 & 4, it is more correct in these instances to refer to the phase of a photons wavefunction rather than the phase of a photon.

    3. Here you want to speak in terms of photons/sec, to establish and maintain an EM wave of any amplitude, you need a constant stream of photons, since the wave is constantly radiating energy. Photon flux (photons/sec/m^2) is proportional to irradiance (in W/m^2). For a coherent wave, the irradiance is proportional to the amplitude squared, hence increasing the photon flux by a factor of 100 results in a 10x increase in amplitude.

    4. No, you can't. Again, it makes no sense to "add photons in perfectly out of phase" it only makes sense to add two photon wavefunctions that are perfectly out of phase. When you do this, the result is obvious, there is no EM wave, no photons, and thus your philosophical proposition becomes redundant. Note though there is no implication that there are photons zipping in and out of existence, as is the implication if one were to refer to the phase of a photon rather than a photons wavefunction.

    Last edited: Sep 16, 2007
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