Optical Waveguide's "Base Bandwidth" 
Consider a dielectric slab waveguide (lossless, isotropic) illuminated transversally from the vacuum (with coherent, monochromatic light).
We define the base bandwidth of a waveguide (or optical fiber), $AB$, to be the inverse of the time retardation, $\Delta t$, at 1 km of the waveguide between the energy of a guided mode (transmitted following the zig-zag model) with a $\theta_{c}$ critical angle and the energy transmitted without total internal reflections. Let $n_{f}$ and $n_{s}$ be the refractive indexes of the film and the substrate respectively. Prove that if $\Delta n=n_{f}-n_{s}<<1$, then:
$$
{\rm{AB}} = {\left( {\Delta t} \right)^{ - 1}} \simeq \frac{{2c{n_s}}}{{{{\left( {{\rm{AN}}} \right)}^2}}} = \frac{c}{{{n_f} - {n_s}}}
$$
where $AN$ is the numerical aperture of the guide.

This problem has been on my mind for 2 days now and it seems I can't find a method to calculate that time difference... Any ideas?
My thinking so far:
We have to look at the Ray Optics picture of Dielectric Waveguide Theory (see e.g. Tamir et al: Integrated Optics (chapter 2)). A guided mode is propagated through the waveguide following a series of total internal reflections at an angle $\theta_{c}$ with the normal to the film-substrate or film-cover surfaces, and therefore its energy is "trapped" in the film. A mode which is not a guided mode will travel through the waveguide suffering reflections and refractions and therefore radiating some energy to the cover and substrate.

The rays travel through the film at the same speed $c/n_{f}$ but follow different paths, so it will take different times for them to advance 1 km in the waveguide. We could then try to find the components of these velocities in the direction of propagation ($z$ axis), $v_{i}$, and use the simple relation $t_{i}=d/v_{i}$. The trouble with this, is that the problem doesn't specify the angle at which the non-guided mode incides in the surfaces of the film. Am I missing something here?

On the other hand, the numerical aperture of the waveguide is:
$$
n\sin \left( {{\theta _{\max }}} \right) = {n_f}\sin \left( {90 - {\theta _{\rm{c}}}} \right) = {n_f}\cos \left( {{\theta _{\rm{c}}}} \right)
$$
where n=1, and $\theta_{\max}$ is the angle of incidence of the illumination beam with the normal to the $x-y$ plane so that doing some work we find:
$$
{\rm{AN}} = \sin \left( {{\theta _{\max }}} \right) = \sqrt {{n_f}^2 - {n_s}^2}
$$

UPDATE: Some calculations:

The effective refractive indexes of the 3 rays (ray 1: guided mode, ray 2: radiation mode, ray 3: no reflection at all) are:
$$
{N_1} = \frac{{{\beta _1}}}{{{k_1}}} = \frac{c}{{{V_1}}} = {n_f}\sin \left( {{\theta _{\rm{c}}}} \right)
$$
$$
{N_2} = \frac{{{\beta _2}}}{{{k_2}}} = \frac{c}{{{V_2}}} = {n_f}\sin \left( \theta  \right)
$$
$$
{N_3} = {n_f} = \frac{c}{{{v_3}}}
$$
So that the retardations would be: ($d=1km$)
Between rays 1 and 2:
$$
\Delta {t_{1 - 2}} = {t_2} - {t_1} = d\left( {\frac{1}{{{V_2}}} - \frac{1}{{{V_1}}}} \right) = \left[ {...} \right] = \left( {d\frac{{{n_f}}}{c}} \right)\left( {\sin {\theta _c} - \sin \theta } \right)
$$
Between rays 1 and 3:
$$
\Delta {t_{1 - 3}} = {t_3} - {t_1} = d\left( {\frac{1}{{{v_3}}} - \frac{1}{{{V_1}}}} \right) = \left[ {...} \right] = \left( {d\frac{{{n_f}}}{c}} \right)\left( {\sin {\theta _c} - 1} \right)
$$
And the respective base bandwidths:
$$
{\rm{A}}{{\rm{B}}_{1 - 2}} = \frac{{\frac{c}{{d{n_f}}}}}{{{\rm{AN}} - \sin \theta }}
$$
$$
{\rm{A}}{{\rm{B}}_{1 - 3}} = \frac{{\frac{c}{{d{n_f}}}}}{{{\rm{AN}} - 1}}
$$
Are any of these results equal to the equation given at the beginning for $AB$? How would one use the approximation $n_{f}-n_{s}<<1$?
 A: What you should be comparing is the time it takes for direct propagation (which I would guess is the "energy transmitted without total internal reflection") versus the time it takes for guided propagation at the critical angle, which is the longest delay/broadening you will get out of the fibre at the other end. Modes at angles higher than $\theta_c$ will leak energy into the substrate and will not make it to the other end, so you don't need to consider them.
Your error is in the calculation of the times each beam travels. For each length $l$ that the direct beam travels, the critical-angle beam travels a length $l'$ given by
$$
\frac l{l'}=\sin(\theta_c).
$$

Thus, if the direct beam travels a total length $d$, the critical-angle beam will travel a length $d'=\frac{d}{\sin(\theta_c)}>d$. Since they are both travelling in the same medium, the real index of refraction is the same, and hence their travel times are
$$
t_\text{direct}=\frac{d}{v}=\frac {dn_f}{ c}\text{ and } t_\text{c.a.}=\frac{d'}{v}=\frac{dn_f}{c}\frac1{\sin(\theta_c)}. \tag0
$$
The critical angle will be given by the total internal reflection limit at the boundary with either the substrate or the cover, whichever has a larger index of refraction. Assuming wlog that $n_s>n_f>n_c$, the critical angle is given by $\sin(\theta_c)=n_s/n_f$. This means that the time delay is
$$
\Delta t=t_\text{c.a.}-t_\text{direct}=\frac{dn_f}{c}\left(\frac{n_f}{n_s}-1\right)=d\frac{n_f}{n_s}\frac{n_f-n_s}{c}. \tag1
$$
The inverse of this is the bandwidth of the fibre, given by
$$
\frac{d}{\Delta t}=\frac{n_s}{n_f}\frac{c}{n_f-n_s}. \tag2
$$
This is pretty close to the result you were asked for, $\frac{c}{n_f-n_s}$. For one, it has a factor of $d$, which is eliminated in your result, effectively, by calculating the 'bandwidth per unit length' of the fibre, $1\text{ km}/\Delta t$, in the understanding that the actual bandwidth will vary inversely with the actual length. This makes a lot of sense: longer fibres make for longer distances travelled by the different beams and therefore longer delays. This is to be expected and should be factored out.
Other than that, some of the prefactors don't quite match up. For one, I must note that one of the equalities that you write as exact isn't really so:
$$
\frac{{2c{n_s}}}{{{{\left( {{\rm{AN}}} \right)}^2}}}=\frac{2cn_s}{n_f^2-n_s^2}=\frac{2n_s}{n_f+n_s}\frac{c}{n_f-n_s},
$$
and this only equals $\cfrac{c}{n_f-n_s}$ in the limit where $n_f$ and $n_s$ are really quite close together. Similarly, in that limit, $\cfrac{n_s}{n_f}\approx 1$, so in that sense all three answers match.
A bit further along those lines, the factor of $\cfrac{2n_s}{n_f+n_s}=\cfrac{2 n_s/n_f}{1+\frac{n_s}{n_f}}$ from the $1/\rm{AN}^2$ answer sits kind of "in between" the exact answer, $n_s/n_f$, so it is not so bad an approximation.


I would therefore sum up the situation as saying that 
$$
\frac{d}{\Delta t}
=\frac{n_s}{n_f}\frac{c}{n_f-n_s}
\approx
\frac{{2c{n_s}}}{{{{\left( {{\rm{AN}}} \right)}^2}}}
=\frac{2n_s}{n_f+n_s}\frac{c}{n_f-n_s}
\approx
\frac{c}{n_f-n_s},
$$
where each approximation accumulates a slight loss of accuracy, from left to right, though of course everything tends to equality as $n_s/n_f\to1^-$. Thus, if it is convenient for some reason to include the numerical aperture in the formula for the bandwidth, then it makes some sense to put it in the picture.
