- #1

EmilyRuck

- 136

- 6

In Optical fibers, let ##k_1## and ##k_2## be respectively the propagation constants in core and cladding, ##\beta## the propagation costant of a mode along the direction ##z##, ##a## the radius of the fiber. Using the normalized quantities ##u=a \sqrt{k_1^2 − \beta^2}## and ##w=a \sqrt{\beta^2 − k_2^2}##, the characteristic equation for modes is:

$$\left[ \displaystyle \frac{J'_{\nu} (u)}{u J_{\nu}(u)} + \frac{K'_{\nu} (w)}{w K_{\nu}(w)} \right] \left[ \displaystyle \frac{k_1^2 J'_{\nu} (u)}{u J_{\nu}(u)} + \frac{k_2^2 K'_{\nu} (w)}{w K_{\nu}(w)} \right] = \nu^2 \beta^2 \left( \displaystyle \frac{1}{u^2} + \frac{1}{w^2} \right)^2$$

where ##n_1## and ##n_2## are the refractive indices of core and cladding, ##\nu## is the order of Bessel functions.

As specified in this document, page 15, the cutoff condition for modes is obtained taking the limit of the above expression for ##w \to 0##. But, before this,

**I can't obtain the same expression as in the document**. Consider:

$$\frac{J'_{\nu} (u)}{u J_{\nu}(u)} = \frac{J_{\nu - 1} (u)}{u J_{\nu}(u)} - \frac{\nu}{u^2} = \xi_1(u) - \frac{\nu}{u^2}\\

\frac{K'_{\nu} (w)}{w K_{\nu}(u)} = \frac{K_{\nu - 1} (w)}{w K_{\nu}(w)} + \frac{\nu}{w^2} = \xi_2(w) + \frac{\nu}{w^2}$$

Substituting in the characteristic equation, dividing both sides by ##k_1^2## and rearranging the terms, I obtain:

$$\xi_1^2(u) + \xi_1(u) \left[ \frac{k_1^2 + k_2^2}{k_1^2} \cdot \xi_2(w) - \nu \left( \frac{2}{u^2} - \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] + \left[ \frac{k_2^2}{k_1^2} \cdot \xi_2^2(w) - \xi_2(w) \nu \left( \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{u^2} - 2 \frac{k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] + \left( \frac{\nu}{u^2} \right)^2 - \frac{\nu^2}{u^2 w^2} - \frac{k_2^2}{k_1^2} \cdot \frac{\nu^2}{u^2 w^2} + \frac{k_2^2}{k_1^2} \left( \frac{\nu}{w^2} \right)^2 - \frac{1}{k_1^2} \left( \frac{\nu^2 \beta^2}{u^4} + 2 \frac{\nu^2 \beta^2}{u^2 w^2} + \frac{\nu^2 \beta^2}{w^4} \right) = 0$$

I double checked this result, but the document shows instead:

$$\xi_1^2(u) - \xi_1(u) \left[ \frac{k_1^2 + k_2^2}{k_1^2} \cdot \xi_2(w) + \nu \left( \frac{2}{u^2} + \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] + \left[ \frac{k_2^2}{k_1^2} \cdot \xi_2^2(w) + \xi_2(w) \nu \left( \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{u^2} + 2 \frac{k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] = 0$$

First, terms with ##\frac{k_1^2 + k_2^2}{k_1^2}## have swapped signs; moreover, I can't figure out how it could be

$$\left( \frac{\nu}{u^2} \right)^2 - \frac{\nu^2}{u^2 w^2} - \frac{k_2^2}{k_1^2} \cdot \frac{\nu^2}{u^2 w^2} + \frac{k_2^2}{k_1^2} \left( \frac{\nu}{w^2} \right)^2 - \frac{1}{k_1^2} \left( \frac{\nu^2 \beta^2}{u^4} + 2 \frac{\nu^2 \beta^2}{u^2 w^2} + \frac{\nu^2 \beta^2}{w^4} \right) = 0$$

Is the result in the document correct? Or am I doing something wrong? As an alternative, I'm also looking for a textbook which deals with the procedure to obtain the cutoff condition.EmilyP. S.

The linked document should be similar, or equal, to this article:

*Cylindrical Dielectric Waveguide Modes*, E. Snitzer,

*Journal of the Optical Society of America*Vol. 51, Issue 5, pp. 491-498 (1961)

which I can't find for free.