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B Wavelength, Path Difference, Phase Difference

  1. Mar 8, 2017 #1
    Hi, would it be possible to explain to me how does wave length, phase difference and path difference all link as I'm struggling with calculations involving these three things.
     
  2. jcsd
  3. Mar 8, 2017 #2

    Doc Al

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    What do you think they mean? Show us what you know already and we can add to that.
     
  4. Mar 8, 2017 #3
    When 2 or more waves meet they combine. To know the resultant wave you need to know whether the waves are 'in step'..in phase or 'out of step' ...out of phase, or something in between.
    Wave length is a fundamental property of the wave. Path difference between waves is one way (not the only way !!) that a phase difference can occur between waves. The 2 slits experiment is an elegant way to show how a path difference between 2 waves can produce an interference pattern.
     
  5. Mar 8, 2017 #4
    I understand wavelength is the distance between a point on a wave and the same point on the next cycle.

    Path Difference, how much a wave lags behind another usually measured in meters.

    Phase Difference how much a wave is behind or in front of another wave usually measured in radians or degrees.

    I clearly have a major misunderstanding as when faced with questions relating all three, I don't know how to begin to answer the question.
     
  6. Mar 8, 2017 #5
    Do you have a particular question that needs solving..... 2 slits, diffraction grating..... Etc
     
  7. Mar 8, 2017 #6

    Doc Al

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    When two waves take different paths, the length of those paths can be measured in wavelengths and the path difference would be the difference between those path lengths. Usually measured in wavelengths.

    Given the path difference of two waves, assuming they were in sync to start with, you can figure out their phase difference from the path difference.

    But it all depends on what specifically you are trying to do.
     
  8. Mar 8, 2017 #7

    Doc Al

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    Check this out:
     
  9. Mar 8, 2017 #8
    You seem to understand each concept separately. Do you know the equation that relates phase difference to path length difference? It is:

    ΔΦ = Δx (2π/λ)

    so if the path length difference between two waves that start out in phase is one wavelength, Δx = λ, the phase difference is ΔΦ = 2π, which means the waves are still in phase.

    If Δx = λ/2, then ΔΦ = π, so the wave are out of phase. So, for example, one wave can fit 10 wavelengths in its path, while the other can only fit 9.5 wavelengths.
     
  10. Mar 8, 2017 #9

    sophiecentaur

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    The geometrical lengths of the paths (in m) is not the only thing that counts. If the light travels through more than one medium (glass/air for instance) the number of wavelengths will be different, depending on the proportion of glass and air along the path. So it's the transit times that actually affect the relative phases where the waves interfere. That means the path difference should be calculated in terms of wavelengths of the wave (and it will be different for different wavelengths).
    The path difference can be many wavelengths in extent but the phase difference (what you use to calculate the resulting amplitude after the interference) will be between 0° and 360°.
     
  11. Mar 9, 2017 #10
    wbNTQmS.png
    Here is a slide from my lecturers power point, i'm having trouble understanding what he means by phase "different" in progressive transverse waves and "in phase between nodes, out of phase in adjacent nodes".
     
  12. Mar 9, 2017 #11

    Doc Al

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    A progressive wave (a traveling wave) will have a phase that varies with position and time, so I guess that's what he means by "different".

    The standing wave will have positions where the two waves destructively interfere (out of phase) at the nodes and constructively interfere (in phase) in between (at the anti-nodes).

    Does this help at all?
     
  13. Mar 9, 2017 #12
    Would you be able to further explain what you mean by the phase varies with position and time? What is the phase different relative to?
     
  14. Mar 9, 2017 #13

    sophiecentaur

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    A simple sinusoidal wave can be written
    A = A0 Cos(ωt-κx)
    Looking at a fixed point in time, the wave displacement A varies sinusoidally along the x axis. Looking at a fixed point in space, the wave displacement varies sinusoidally in time.
    What does he mean by this?:
    The two waves that go together to make the standing wave have a fixed phase relationship at any given point on the string (due to the fixed position of the reflecting point at the end) The diagram shows two phasors A and B, for the two waves, travelling on opposite directions (the sinusoids in the diagram show how the displacements of the waves vary in time and the angle between A and B corresponds to the phase relationship at a particular point on the string. R is the resultant. Over the cycles of waves A dan B, R stays in the same phase (points in the same direction on the phasor diagram), for all relative phases of A and B from 0 to 180° - it's just the resultant amplitude that varies from twice amplitude to zero at cancellation. BUT once the relative angle exceeds 180. R points in the opposite direction so there is a phase inversion as you cross the node. Not sure if this helps but drawing the diagram was worth while for me. :wink:
     

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  15. Mar 10, 2017 #14
    Hi this formula assumes that the wavelength of the two waves are the same, how would I deal with calculations where both wavelengths are different?
     
  16. Mar 10, 2017 #15

    sophiecentaur

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    If the wavelengths are different in the same medium then the frequencies are different. In that case, the interference pattern will not be stationary. Not what we are discussing here.
    Edit: You can produce a moving interference pattern when two waves on not-too different frequencies interfere. A good example of this is what you get in the area between two mf sound transmitters that are operating on the same nominal channel and with the same programme material. If their frequencies are just, say 1Hz apart, the maxes and mins in the reception areas where their received signals are of near equal amplitude will sweep across the land and give a 1Hz beat in the amplitude of the received signal. (Very annoying and it can be solved to some extent by locking the two carriers to an external reference or by using very high quality frequency references at each transmitter. But you also need to synchronise the audio feed arrival times at each transmitter.
     
    Last edited: Mar 11, 2017
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