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I Orthogonal Polarisation in EM waves and Interference

  1. Mar 26, 2017 #1
    I've been trying to get my head around Polarisation and how it achieves orthogonality. I'm not sure if this should be in Physics or Electrical Engineering Section. (Mods can move this where appropriate)

    I know that 2 EM wave with linear polarisations where one wave is shifted by π, they would interfere.
    But, if we now have those 2 same waves and assign opposing polarisations to each (1 wave with Horizontal Polarisation and the other with Vertical Polarisation), would this π shift still interfere, or they would be completely orthogonal to one another? --- Or basically give me an entirely new wave with the polarised vector additions?
     
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  3. Mar 26, 2017 #2

    blue_leaf77

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    You have two waves with perpendicular polarizations and are shifted by ##\pi## radians with each other. Mathematically, the sum of these two waves at a fixed position along its propagation direction (which is arbitrarily taken to be the z direction) will be
    $$
    \mathbf E(t) = \hat x E_{0x} \cos(\omega t +\phi) - \hat y E_{0y} \cos(\omega t +\phi)
    $$
    where ##E_{0x}## and ##E_{0y}## are the amplitudes of the perpendicularly interfering waves. Taking out the cosine factors gives you
    $$
    \mathbf E(t) = (\hat x E_{0x} - \hat y E_{0y}) \cos(\omega t +\phi)
    $$
    How would you describe the polarization state of the resultant wave, is it linear, circular, or elliptical?
     
  4. Mar 26, 2017 #3

    sophiecentaur

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    . . . . and what would happen if the two φ values were different in the first equation. In particular, if the difference between the two φs was π/2? But the E fields would always be orthogonal.

    Not sure what this wording means. Spatial orthogonality would mean that the E fields are at right angles to each other. Neither has a component in the direction of the other.
     
  5. Mar 27, 2017 #4
    It seems to be linear (although I read a few days ago, that Vertical+Horizontal results in circular polarisation, so I may/must be wrong), but I will need to brush up on the mathematical representation of the circular and elliptical waves. I haven't done physics for a long time, so I appreciate you teaching me the above. Thank you
     
    Last edited: Mar 27, 2017
  6. Mar 27, 2017 #5
    What could be orthogonal physically, is not necessarily orthogonal to EM systems. So, what I mean here is whether an EM system would differentiate between the 2 waves, although they interfere with one anther? Or will their polarisations prevent this interference, and have the system see them as independent waves, not a summation or difference from one another.

    Basically, looking at power intensity of the wave, and whether different polarisations could decrease power intensity due to interference like superimposed same linear polarisation waves do.

    And thank you very much for answering many of my questions. Your answers are really appreciated
     
  7. Mar 27, 2017 #6

    blue_leaf77

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    It is linear.
    If the phase shift is any integer multiple of ##\pi##, the resulting wave is linear. Other than those values, the polarization takes a different form. In order to convince yourself, I suggest that you try sophiecentaur's suggestion.
     
  8. Mar 27, 2017 #7

    Aha, I understand you. Thank you very much Blue Leaf
    Very helpful to always remember that a phase shift of any multiple of ##\pi## will result in a linear wave.
     
  9. Mar 27, 2017 #8

    sophiecentaur

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    With the (time) phase difference between the two orthogonal plane polarised waves is π/2 (or you could say one is a cos and the other is a sin) the result will be circular polarisation (ref Lissajous Figures) with the resultant vector turning through 2π every time it advances by one wavelength (aka after a period 1/f).
     
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