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B Superimposing 2 sources of light slightly out of phase

  1. Oct 19, 2016 #1
    Assuming for now a simple photon particle model for light, where you have a stream of photons traveling with a certain frequency.

    Picturing this in my head on a number line, with just 3 photons at a time, and lets just say the wavelength is 1 and speed is 1, and at time t=0 we have a photon at x=0, x=1, x=2, then at time t=1 we have a photon at x=1, x=2, x=3, etc

    So let's say I shine a flashlight with the above setup on a spot. Then I shine another flashlight, which has the same wavelength and speed (and thus the same color), on the same spot. However this 2nd flash light is slightly out of phase from the first one. For this second flash light, at time t=0 the 3 photons are at x=0.5, x=1.5, x=2.5, and at t=1 they are at x=1.5, x=2.5, x=3.5

    If I shine both these flash lights at the same spot, and the light reflects back to my eye, both streams of photons overlap, and when they combine there are now more photons per unit of length, so the wavelength has changed, so I would assume that I should see a different color. However, both flashlights I shined on the spot have the same color, so it seems odd.

    Hopefully I am misunderstanding something here that someone can clarify for me, thanks!
  2. jcsd
  3. Oct 19, 2016 #2

    Jonathan Scott

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    Gold Member

    Frequency of photon arrival (which is a questionable concept anyway) only affects intensity of light. Each photon has its own frequency. When multiple photons of the same frequency arrive around the same time, that increases the intensity but does not change the frequency. (There can of course also be interference effects which increase the intensity in some directions and decrease it in others, but the general effect is that more photons means more intensity).
  4. Oct 19, 2016 #3
    Well it would be nonsense to talk about a light as a particle when you talk about light as a wave. If we wanted the microscopic scale of things, sure we would need some quantum mechanics to see what would happen.
    Since you are talking about flashlights and seeing what would they make when out of phase, the light as a particle would be at best unnecessary, except for one thing: The frequency (and thus both wavelength and speed) are not dependent on the frequency of how many photons arrive, instead on the energy and thus the frequency, and other properties of the particle itself as was described ~110 years ago, E=hf, where h is planck's constant. It would be better to just describe light as waves for a practical solution. (the speed of the particle is also not determined by what light is, but where it propagates)

    The description itself is a little complicated, and going to avoid it for now, but basically the amplitude and phase are going to change, but the frequency would remain as is. You can actually do this graphically, or mathematically using super position of the two waves.(i.e. just add them together, but don't forget their not in phase!) Needless to say, a middle school student can do it graphically, it is basically adding together two functions at each point on the graph.\

    Experimentally, however, you'd need a monochromatic source, flashlights produce a lot of noise if not filtered properly. The easiest way would be performing the double slit experiment. For that you'd need a light source, preferably a monochromatic laser and two tiny slits. The latter can be bought online on various sites such as Ali-express, amazon or ebay, just need to look for one. I do not recommend making your own slits as you'd need extremely tiny ones that for ones made by hand you would need to hire Willard Wigan.
    The double slit ensures that not at all times would the two light waves be in the same phase.
    Last edited: Oct 19, 2016
  5. Oct 19, 2016 #4


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    Staff: Mentor

    When you hear that a photon is a "particle of light" it's natural to think that way... But this model turns out to be hopelessly wrong and misleading. Feynman's book "QED: the strange theory of light and matter" is a good layman-friendly description of what's really going on, and it's nothing like what you're thinking.

    For a question like this one, you don't think in terms of photons at all. Light is an electromagnetic wave, and you calculate superposition and interference effects by adding and subtracting the strength of these waves at each point in space. Photons only come in when you make detailed calculations of exactly how these waves interact, and we don't need that for this problem.
  6. Oct 20, 2016 #5
    Thank you all for the replies, it cleared up some misunderstandings I had about what frequency means wrt light.
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