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What Exactly is an Optical Mode in a Waveguide?

  1. May 23, 2014 #1
    I have been working with equations throughout the semester and using them to find cutoff frequencies and such, but when you say something like TE10, TE20, TE21, etc., I am realizing that I am having trouble getting a physical idea of what exactly these things are. Of course if you transmit at a frequency below the cutoff frequency of TE21, it will not propagate, but maybe TE10 modes can.

    But this is what confuses me. Let's say you are transmitting at a frequency greater than the TE21 cutoff frequency. That means you can have TE10 & TE21, among other possibilities. It seems to be in this case that both of these modes are transmitting a beam of light with the same wavelength. If that is correct, then what exactly is different about these modes? The QM I remember makes me want to say that one is in an excited state, but I wouldn't know what that really means, even if true. So what different properties would two waves (of the same wavelength, if that is true) from different modes have exactly?

    Finally, I am struggling to see how this applies to multimode fibers too. I heard that you can separate different modes at the end of a fiber with a grating, which makes sense if the wavelengths are indeed different. So if different modes = different wavelengths, well I am not exactly sure how these different modes run in different wavelengths anyhow...

    Thanks for any of these concepts you might be able to elucidate.
     
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  3. May 23, 2014 #2

    Simon Bridge

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    Optical mode: A transverse mode of a beam of electromagnetic radiation is a particular electromagnetic field pattern of radiation measured in a plane perpendicular to the propagation direction of the beam. ... it's like the modes of vibration on a string only for the electromagnetic field.
     
  4. May 23, 2014 #3
    With the string, it is a physical object already, so it's wavelength is twice as long in the n=2 vs n=1 mode. , and if it's a guitar, the pitch will be different. With an electromagnetic radiation, I understand wavelength in the direction of propagation of course, but it is a little tricky for me to think of what it means for the electromagnetic field to have a different "wavelength," sort of. And it leaves me with three questions:

    1) If the field has these different patterns based on what mode it is in, what physical difference will there be between these propagating waves? For example, the energy should be identical, but what physical aspect isn't?

    2) Why would a higher mode (in the transverse direction) not be able to propagate in the z-direction at certain (lower) frequencies?

    3) What causes a wave to be in one mode or another?

    Thanks!
     
  5. May 23, 2014 #4

    Simon Bridge

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    It may be easier for you to think of it like pressure waves in a tube.

    Solve the wave equations for an EM field in an idealized 3D box, dimensions LxWxD, with perfectly reflecting sides. You are going to get standing wave solutions in each dimension - which will have different wavelengths (there have to be nodes at each wall).

    A waveguide is like one of these boxes, with one dimension very much longer than the others.
    say: W=D=2R and L>>R.

    IRL. the walls are not perfectly reflecting either - which gives you all the wrinkles you've probably been learning about.

    Aside:
    For a pressure wave in a tube, you will have seen the approximation for a long thin tube so only the "length" is considered. But the width of the tube also affects things. You can understand that the air pressure by the walls of the tube will be different from in the middle - any perpendicular motion in the molecules must be a minimum there. You can see the effect if you upend the tube, replace the air with water - and tap the "top" ... see the waves ripple out to the edges, as well as travel down the length of the tube?
     
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