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I'm cramming for my EMF exam, and a question came up:

Light in free space is an example for a TEM wave, isn't it? (Maybe thats already my mistake)

And we learned that in a hollow waveguide only TE or TM modes can exist, because a TEM wave cannot meet the boundary conditions.

So my question: How does it come that I can shine light through a hollow waveguide, for example a metal pipe or tube?

Does it get converted to a TE or TM wave at the entrance and leaves again as a TEM wave? Or is it just because the wave length of light is so small compared to the waveguide dimensions? But if so, where is the cutoff frequency between "TEM possible" and "TEM not possible"

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The attenuation of a waveguide is minimal when it is excited with the correct mode, but it isn't infinite for field configurations that aren't, so you can also send field configurations trough that don't fit the mode patterns, they will simply be attenuated very strongly for low frequencies. How strong that attenuation is depends on how much of the electrical field components gets "shorted out" by the wall.

Light in particular has a much smaller wavelength than the size of the dimension of a typical waveguide. Let's assume you are sending a gaussian wavefront trough a waveguide this way (many lasers produce nice gaussian wavefronts like that). The light will have initially vanishing electric field at the wall (if you inject right down the center of the waveguide) because the amplitude dependence as a function of beam radius is very strong. On the other hand the gaussian wavefront also has a non-zero divergence, so as the beam passes trough the waveguide it will eventually reach the walls and there will be attenuation and there will be grazing reflections, which will lead to destructive interference and eventually to complete attenuation.

What you know about the propagation modes of waveguides is therefor true, it's just a simplified go-no-go model that is sufficient for engineering purposes. The rule tells you not to try to use a waveguide any other ways than its eigenmodes dictate, but that rule is only simple because it avoids the complications of having to calculate the actual transfer function for em waves in the general case, which is an ugly business, especially if the wavelength is much smaller than the waveguide dimensions.

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