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Effective index in bent fiber example
Posted Feb 17, 2021, 11:22 a.m. EST Wave Optics 0 Replies
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Hello,
I am struggling to understand what Mode analysis is actually calculating in the case of axial symmetry - such as in the Application Library example for optical fiber bend.
In particular, according to the description, the solution is sought in the form:
where lambda is the eigenvalue.
Also, according to the Wave Optics documentation, the effective index is defined as:
where k0 is the wave vector in vacuum. I am assuming this is also the case for the axisymmetrical example like the bent fiber.
Now, here is one thing which confuses me.
For the symmetry reasons, I expect any eigenvalue lambda to satisfy:
(in other words, I expect Im(lambda) to be an integer number).
In terms of n_eff, this should be:
However, this is not what I observe. Using global evaluation, if I calculate
it gives me something which is definitely not 1.
On the other hand, as an interesting observation, if I evaluate
it gives me 1 to a very high accuracy.
In other words, it looks like the stated value of n_eff is actually the imaginary part of lambda, and not lambda/k0?
On the other hand, when divided by the radius, it actually gives something very close to the effective index of the straight fiber mode. This is very confusing...
Am I missing something?
Hello Andriy Gorbach
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