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In this previous post, most replies point out that it is not possible to predict the angle of refraction (and so the frequency) at a certain interface, given the wavelength of the original signal.

In particular,

But when dealing with optical waveguides, it seems to be different. I am referring for example to this link, slide 23: the lower the mode in the fiber, the lower the angle of incidence (and the angle of refraction will vary as a consequence, knowing that [itex]n_2/n_1[/itex] is a given value).The way that index of refraction depends on frequency is very complicated and material dependent. There is not a single simple equation to describe it.

It seems to be the same for the dielectric slabs and for the graded index optical fibers like this (Graded Index Fibers paragraph).

I know that in a certain waveguide, if [itex]\beta[/itex] is the propagation constant along [itex]z[/itex] (the direction of propagation), the following relations exist:

[itex]\omega^2 \mu \epsilon = k_c^2 + \beta^2[/itex]

[itex]\beta = \sqrt{\omega^2 \mu \epsilon - k_c^2}[/itex]

where [itex]k_c[/itex] is the transverse wavenumber. So, depending on the frequency [itex]\omega[/itex], there will be waves with a higher [itex]\beta[/itex] and a lower [itex]k_c[/itex] and vice-versa: this will vary the angle of incidence according to the mode. But what exactly is the relation between the wave frequency and this angle? And why the lowest modes always have an almost straight path and a high [itex]\beta[/itex], while the highest have a low [itex]\beta[/itex], so bouncing more frequently between the core-cladding interface?