Can Two EM Waves, Same Frequency,Out of Phase, Affect Wavelength ?

Discussion in 'Physics' started by morrobay, Nov 24, 2013.

  1. morrobay

    Thread Starter New Member

    Nov 7, 2013
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    Huygens' Construction for the refractive index (geometric optics ) is: sin theta 1/sin theta 2 = λ1/λ2 = v1/v2. The wavelength of light and velocity decreases in a medium with greater refractive index. (air to water ,λ1 to λ2 ) λ2 = λ v2/v1
    The explanation for the shorter wavelength is that when an EM wave in the optical range enters a medium the time varying electric field causes the electrons in the medium to oscillate proportional to the permittivity. This oscillation of charges causes the radiation of an EM wave of the same frequency and slightly out of phase with the original. The sum of these waves creates a wave with same frequency and shorter wavelength, leading to a slowing of the waves travel.
    ( since the frequency of blue light is closer to the natural resonate frequency of the electrons, violet - uv, the refractive index is greater for blue light).
    The question here is : Can the summation of two EM waves of same frequency and slightly out of phase produce a wave with shorter wavelength ? My understanding is that only the amplitude is affected ?
     
  2. Wendy

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    Mar 24, 2008
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    I believe your understanding is correct. You can not change the base frequency of a EM wave except with conversion or Doppler effects. Conversion generally involves a photon being absorbed, then re-emitted at a different energy level from an electron.
     
  3. morrobay

    Thread Starter New Member

    Nov 7, 2013
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    Re general info above : This is a question and it is starting a thread ?
     
  4. Wendy

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    You started this thread, I answered it. You will likely get other answers from other people, but this is not chat, it is a forum. It can take days or weeks for people to respond. These kind of sites are run by volunteers.
     
  5. morrobay

    Thread Starter New Member

    Nov 7, 2013
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    Well I got some answers here, if interested : www.physicshelpforum.com
    (light and optics) & www.physicsforums.com (general physics) Refractive Phase Difference topic
     
  6. alfacliff

    Well-Known Member

    Dec 13, 2013
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    switching the phase or phase modulation causes an instantaneous frequency change, but thats another story.
    cliff
     
  7. morrobay

    Thread Starter New Member

    Nov 7, 2013
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    In refraction the mainstream view is the the slowing of lights travel in the medium is that the frequency (color ) stays the same but the wavelength is shorter. Analogy : Consider a wave at boundary of a medium with wave length of 4 meters and frequency 10 cycles/sec. So velocity = 40m/sec.
    In the medium the wave length is shortened to 2 meters with same frequency,10 cycles/ sec and velocity 20m/sec,the refracted wave.
    So what we have here are two waves : One is the incident radiation and the second is the wave produced in the dielectric
    medium as a result of the electric field vector of the EM wave causing the electrons in medium to oscillate at same frequency.
    This new wave with same frequency and wave length is slightly out of phase, lags behind, original wave. So my question is on how the summation of these two wave produces a wave of same frequency and shorter wave length leading to a slowing of the waves travel in medium, refraction.
     
    Last edited: Dec 14, 2013
  8. Wendy

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    Mar 24, 2008
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    Once the wave leaves the other medium it resumes its original speed, so while you get a phase shift the speed has not really changed, and the frequency is still constant.
     
  9. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Are you actually asking for a wave theory explanation of how refraction works?
     
  10. morrobay

    Thread Starter New Member

    Nov 7, 2013
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    In this paper: http://www.en.wikipedia.org/wiki/Refractive_index
    See microscale section. The question is on how the wavelength is shortened
    with the summation of the incident wave + the radiated wave in the medium.
     
  11. studiot

    AAC Fanatic!

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    OK so I understand you question to be

    Given
    The velocity of a wave is given by v = frequency times wavelength

    When we develop the theory of refraction we say that refraction occurs because the velocity varies by change in transmission medium.

    Since the velocity varies either the frequency or wavelength or both must change for their product to equal the new velocity.

    Why do we take all the change to be in the wavelength and keep the frequency constant?

    The full answer to this involves some quite advanced mathematics to prove the underlying theorem, which is known as Fermat's Principle (FP).

    The advance maths involves applying the calculus of variations to the differential equation known as the wave equation.

    Here is a simplified argument, if you are willing to accept this statement of FP.

    Light (and other waves) always take the path of least transit time between two points.

    Note that this time will obviously depend upon the wave velocity.

    Now the line joining all points at a particlar transit time away from a source is called a wavefront. All points on a wavefront take the same time to reach from a source.

    Consider particularly the wavefront that is exactly one cycle away from the source. All points are in phase.
    The transit time to this first wavefront is the period of the wave.
    So all points on this wavefront have the same period.
    That is the period is constant.
    But period is 1/frequency by definition.
    So the frequency is constant.

    Note that I have not said that all points on the wavefront are the same distance from the source. That too depends upon the velocity along the path.

    If the velocity varies (because the refractive index, n, varies) then the distance travelled in the given fixed time must vary.

    Since we are considering one complete cycle this distance is the wavelength.

    So the wavelength must vary with velocity, whilst the frequency does not.

    We can extend this argument to any distance by observing that the same number of cycles must fit into the distance travelled in a given fixed transit time from source to any wavefront.

    Conventionally this is often treated by creating an auxiliary variable, known as the optical distance which is constant for a given wavfront. That is all points on a wavefront are the same optical distance from the source.
    If the physical path length is measured as x metres the optical path length is nx.

    How are we getting on?

    Incidentally this is in contrast to the Doppler effect, where the transit time is not constant, allowing frequency variation.
     
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