Maximum incidence angle in optic fiber. A circular glass cylinder is a simplified model for fiber optics. Calculate the maximum incidence angle $\theta$ at the entrance face at which the light will be guided into the fiber by successive total reflections as a function of the refractive index $n$ of the fiber.
  
  


I started studying optics a short time ago and I'm having a difficulty in some exercises in the book. I appreciate any hints.

I know that I just need to calculate the maximum incidence angle for total internal reflection, since for larger angles, the refracted ray will leak out of the fiber. Also, angle of incidence equals angle of reflection. But, I don't know how to use this informations to get an explicit equation.
There is an easy tutorial on Thorlabs web page explaining this.  It's just a single application of Snell's law assuming you know the index of the fiber and the cladding.  The numerical aperture is defined by the the maximum half-acceptance angle $\theta_a$ associated with total internal reflection.


Let $n_i$ be the index of the outside material (usually 1) and $\theta_c$ be the critical angle for total internal reflection.  Then

$$\sin \theta_c = n_c/n_f = \cos \theta_t = \sqrt{1- \sin^2 \theta _t}$$
So the numerical aperture $NA$, is 
$$ NA = n_i \sin \theta_a = \sqrt{n_f ^2 - n_c ^2}$$

If $n_i$ is 1 and $\theta_a$ is small, then you just the NA is approximately $\theta_a$.

Here is the link to Thorlabs:  https://www.thorlabs.com/tutorials.cfm?tabID=17485628-68dd-4d22-ad17-6dd4520974c7
You know that the maximum angle of incidence that can be used in an optical fiber is called the critical angle (and this angle produces and angle of refraction equal to $90$ degrees). 
To solve for maximum angle of incidence $i$, use the formula
$\sin (i)/\sin (r) =n_2/n_1=C_1/C_2$

Where $i$ is the maximum incidence angle $u$ want to find,
$r$ is $90$ degrees giving $\sin(90)=1$,
$n_2$ is the index of the fiber core and $n_1$ is the index of air which is equal to $1$. Then $C_1$ is the speed of light in air which is equal to $3 \cdot 10^8$ m/s, then $C_2$ is the speed of light in the glass core.