# A Erroneous results when solving fiber mode eigen equation

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1. Aug 4, 2016

### p_rob

Hi,

I'm new to this forum and I couldn't find any specific sub-forum for fiber optics/waveguide theory, which my problem is regarding. Please do let me know if I should post this question some where else (and if so, where) on this forum. Anyways, here's my problem:

I want to find the effective index for the fundamental mode in a circularly symmetric fiber for different wavelengths. To do this, I used the standard eigenvalue equation for the propagation constant, see for instance Nonlinear fiberoptics by Agrawal:
$$\left(\frac{J'_0(pa)}{pJ_0(pa)}+\frac{K'_0(qa)}{qK_0(qa)}\right)\left(\frac{J'_0(pa)}{pJ_0(pa)}+\frac{n^2_2}{n^2_1}\frac{K'_0(qa)}{qK_0(qa)}\right)=0$$
where $p=\sqrt{n^2_1k^2_0-\beta^2}, q=\sqrt{\beta^2-n^2_2k^2_0}$, $J_0$ denotes the zeroth order Bessel function of the first kind, $K_0$ denotes the the zeroth order modified Bessel function and the primed versions denote their derivatives. $a$ is the fiber's core radius, $n_1$ is the refractive index in the core and $n_2$ is the refractive index in the cladding, $k_0=\frac{2\pi}{\lambda}$ where $\lambda$ is the wavelength and $\beta$ is the propagation constant.

My problem is that using this equation, I don't get the same solutions (and sometimes I don't get any solutions) in cases where I can see that there are solutions by simulating the electric field distribution in COMSOL for a straight circularly symmetric fiber (using the application provided in the wave optics library). So for instance, if I'm using the following parameters:
$$n_1=1.4632, n_2=1.4322, \lambda=2.3865 \mu m, a=2.75 \mu m$$
COMSOL gives me an effective refractive index, $n_{eff}$, (which is related to the propagation constant by $\beta=n_{eff}k_0$) of 1.4466. But if I plot the absolute value of the left hand side of the eigenvalue equation ranging from 1.4322 to 1.4632, there is no dip in the curve around 1.4466 as one would expect. It doesn't seem to be a discretization problem either as I've tried to use more points within the interval without being able to find any dip around the the value of 1.4466.

So my question is why I can't find this dip when I'm using the eigenvalue equation?

Rob

2. Aug 9, 2016