Optical fiber cable

If a light beam in the glass fiber would be partially transmitted at the interface between glass and air, it would lose intensity and disappears virtually completely after a short distance. Therefore, all guided beams must meet the condition of total reflection.

The refraction is descriped by Snell's law $n_1 \sin \alpha_1 = n_2 \sin \alpha_2$ with the angles of incidence $\alpha_1$ and refraction $\alpha_2$ (transmitted light). But, for $n_1 > n_2$ this equation has no solution for $\alpha_2$ , if $\frac{n_1}{n_2} \sin \alpha_1 > 1$ since the sine function $\sin \alpha_2 \in [0,1]$ takes only values on the unit interval. Therefore, no transmission occurs and the light is totally reflected. The limiting case defines a minimal total reflection angle $n \sin \alpha^\ast = 1 \quad \Rightarrow \quad \alpha^\ast = \arcsin \frac{1}{n}$ so that the corresponding light beam emerges at the widest possible opening angle. For the refraction at the end of the glass fiber, the beam has an angle $\beta^\ast = 90^\circ - \alpha^\ast$ of incidence. Snell's law yields $\begin{aligned} \sin \gamma^\ast &= n \sin \beta^\ast = n \sin(90^\circ - \alpha^\ast) = n \cos \alpha^\ast \\ &= n \sqrt{1 - \sin^2 \alpha^\ast} = n \sqrt{1 - \frac{1}{n^2}} = \sqrt{n^2 - 1 } \\ \Rightarrow \gamma^\ast &= \arcsin \sqrt{n^2 - 1 } \approx 72^\circ \end{aligned}$

The relations are

• from geometry, $\beta = 90^\circ - \alpha\ \ \ \ \ \therefore\ \ \ \ \ \sin\beta = \sqrt{1 - \sin^2\alpha};$

• from Snell's law, $\sin\gamma = n\sin\beta;$

• from Snell's law in the limiting case of total internal reflection, $\sin\alpha_{\text{min}} = \frac 1 n.$

Combining these, $\sin\gamma_{\text{max}} = n\sin\beta_{\text{max}} = n\sqrt{1 - \sin^2\alpha_{\text{min}}} = n\sqrt{1 - \left(\frac 1 n\right)^2} = \sqrt{n^2 - 1}.$ With $n = 1.38$ we find $\gamma_{\text{max}} = \text{inv sin}\ \sqrt{1.38^2 - 1} \approx \text{inv sin}\ 0.951 \approx 72^\circ.$

I'll assume a working knowledge of Snell's law and total internal reflection. They are nicely summarised in the first two sections of this article .

Looking at the ray coming out of the end of the fibre we see that

$\sin\gamma=1.38 \sin \beta$

Easy geometry from the diagram shows that $\beta = 90 - \alpha$ so we may rewrite this as

$\sin\gamma=1.38 \cos \alpha$

We wish to make this as large as possible, which means that $\alpha$ must be as small as possible. However for the internal reflections to be possible at all we know that $\alpha$ cannot be smaller than the critical angle, $\sin^{-1}{\frac{1}{1.38}}$ , and so the largest possible value of $\sin{\gamma}$ is

$\sin\gamma_{max}=1.38 \cos\left(\sin^{-1}\left(\frac{1}{1.38}\right)\right)$

and so finally

$\gamma_{max} = \sin^{-1}\left(1.38 \cos\left(\sin^{-1}\left(\frac{1}{1.38}\right)\right)\right)= 71.99^{\circ}\approx\boxed{72^{\circ}}$