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Refractive Phase Difference

  1. Sep 23, 2013 #1
    I'm wondering about small changes of phase due to a refractive medium. For example, suppose there is an emitter of radio or light waves, and two detectors equidistant from the emitter. One of the detectors is behind a medium, while the other is a straight shot through a vacuum.

    How would the difference in the detectors evolve over time? Would it just be a constant phase shift, or would the angular frequency of the refracted wave change?
    How can I think about this properly?

    Thanks in advance.
     
  2. jcsd
  3. Sep 24, 2013 #2

    UltrafastPED

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    Draw a diagram ... if one wave is subject to a refractive medium it will have a change of wavelength in that medium because v=c/n, but the frequency will not change.

    The one through vacuum has n=1, so v=c and there is no refraction.
     
  4. Sep 24, 2013 #3
    Hmm, perhaps my question was not explained well. The detector behind the medium is not in the medium. I know that the wavelength and the velocity will change, but I'm not sure whether the detected phase change will by proportional to time or constant. My confusion is coming from two perspectives:

    1. As an analogy, the speed of the convoy is the slowest ship. Since the speed of light is slowed through the medium, all of the light going through the medium should travel at a slower rate. Let's say that that in the vacuum, n = 2. Then c / 2 is proportional to the number of photons (regardless of frequency in the spectrum?) hitting the detector. It seems like the detector behind the medium will detect photons half as as fast, which in turn seems to imply that somehow the frequency has changed?

    Number of photons = c * k1 * t
    Number of photons = c / 2 * k1 * t
    where k1 is some proportion relating speed of light and number of photons. As time goes on, the difference in photons detected at each receiver will increase.

    2. Since the frequency is constant and determined by the emitter, the frequency should not change. Thus, there should be a constant difference of photons detected.

    Number of photons = c * k1 * t
    Number of photons = c * k1 * (t - delta)
    where delta is the constant shift.

    Any advice is appreciated.

    EDIT: Additionally, I'm not sure how to think about it because light can be thought of as a particle or a wave. Does the particle model apply to radio waves as well?
     
  5. Sep 24, 2013 #4

    UltrafastPED

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    The frequency never changes while passing through a medium - it is always the wavelength.

    Radio and light are both electromagnetic waves, so the theory is the same, but have many practical differences.
    One is that the index of refraction barely exists for radio waves because almost everything is through air ... hence n=1.

    But using light you can use the "optical path length" technique: http://en.wikipedia.org/wiki/Optical_path_length

    This should provide your answer.
     
  6. Nov 27, 2013 #5

    morrobay

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    What is the description of the mechanism of how the wavelength changes in the medium ?
    The geometry was first shown by Huygen : λ2 = λ v/c
    I understand that the frequency has to be the same because of the boundary conditions.
    So the apparent slowing of light transmission in refraction is because of a shorter wavelength .Im not looking for a geometric optics explanation. Rather the exact physical mechanism at the micro scale in terms of the interaction of the time varying electric field of the EM wave with the electrons in the medium.
     
    Last edited: Nov 27, 2013
  7. Nov 28, 2013 #6

    sophiecentaur

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    Would as 'classical' explanation satisfy you? You could think in terms of a distribution of charges with mass, loading the wave in its progress and delaying the perturbations as the wave progresses through the medium as they are forced to oscillate by the impressed varying EM fields and then re radiate a bit later.
     
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