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[Phys-L] Re: fiber optic cables and waveguides



On 02/26/05 02:56, James McLean wrote:

How can it be that a fiber optic cable "works like a wave guide"?

It's a waveguide in the sense that it guides the wave.

I've seen several kilometers of fiber wound on a spool.
The light goes in one end and comes out the other.
(This was in a lab where they were busy inventing
new types of amplifier etc. for undersea communication
cables. I walked past it almost every day of my life
for several years.)

My
impression was that a defining characteristic of waveguides is that
their transverse dimensions are comparable to the wavelength.

I don't see why that would be required. For over a
hundred years people have known how to make waveguides
for visible light using fountains and jets of water.

OK, in the *microwave* region it is common to use
waveguides that are so small (relative to the wavelength)
that only the 0,0 mode can propagate -- but that's
merely a matter of convenience, because the waves
are so darn big that you don't want to mess with a
waveguide that's bigger than the minimum.


The apparatus that we have has a single-fiber cable, the fiber being
maybe 0.75mm in diameter (I don't have it handy to measure exactly).

That's core+cladding+protective jacket. The physically-
relevant dimension is the core diameter, which is a
whooole lot smaller.

I'm sure that the light source is probably not visible but it probably
is in the IR. So if the wavelength is 1 micron, we're in mode 1500 of
the waveguide? That doesn't sound very "scrunched" to me!

Assuming what you've got is communication fiber, it's
probably 1.06 micron light in a 50-micron core. OK,
that's not _very_ scrunched, but even for the 0,0 mode
there would be an easily-observable effect on the speed
of propagation (which is where this thread started).

Also, there's no law that says everything winds up
in the 0,0 mode. In fact, given the setup as
described, it would be a miracle if you could reliably
inject into the 0,0 mode.

I'm also a bit confused by the fact that microwave waveguides are made
of conducting materials.

1) The key idea in both cases is to have _reflecting_
walls.

2a) In one case, we have external reflection from a
conductive mirror wall.
2b) In the other case, we have total internal reflection
inside a non-conductive dielectric.

For defining what's a waveguide, item (1) is the
important similarity, and item (2) is a detail of
secondary importance.

I gather that the wall conductivity

(or lack thereof :-) (in waveguides)

is what
determines the transmission power loss. Fiber optics obviously don't
involve conductors. Of course, both involve solutions of Maxwell's
equations under zero-field boundary conditions. But is the difference
in the reason for those boundary conditions really totally unimportant?

Certainly it's _technically_ important, in the sense
that total internal reflection is very very efficient,
and therefore makes possible stunningly efficient
fiber networks.

But it's not crucial to the definition of what's a
waveguide and what's not.