Question

Figure 35-57 shows an optical fiber in which a central platic core of index of refraction${\mathbf{n}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{58}$-is surrounded by a plastic sheath of index of refraction${\mathbf{n}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{53}$. Light can travel along different paths within the central core, leading to different travel times through the fiber, resulting in information loss. Consider light that travels directly along the central axis of the fiber and light that is repeatedly reflected at the critical angle along the core-sheath interface, reflecting from side to side as it travels down the central core. If the fiber length is 300 m, what is the difference in the travel times along these two routes?

Short answer

Thus, the difference in the travel times along two routes is $51.6ns$.

Step-by-step solution

Light travel along different paths within the central core.

A light ray travelling directly along the central axis reaches the end in time

$$\begin{gathered} {\mathbf{t}}_{\mathbf{direct}}\mathbf{=}\frac{\mathbf{L}}{{\mathbf{v}}_{\mathbf{1}}} \\ \mathbf{=}\frac{{\mathbf{n}}_{\mathbf{1}}\mathbf{L}}{\mathbf{c}} \end{gathered}$$

For the ray taking the critical zig –zag path, only its velocity component along the core axis direction contributes to reach the other end of the fiber. That component is${\mathbf{v}}_{\mathbf{1}}\mathbf{cos\theta }$ , so the time of travel for this ray is:

$$\begin{gathered} {\mathbf{t}}_{\mathbf{zig}\mathbf{-}\mathbf{zag}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{v}}_{\mathbf{1}}\mathbf{cos\theta }} \\ \mathbf{=}\frac{{\mathbf{n}}_{\mathbf{1}}\mathbf{L}}{\mathbf{c}\sqrt{\mathbf{1}\mathbf{-}{\left(\frac{\mathrm{sin\theta }}{n_1}\right)}^{\mathbf{2}}}} \end{gathered}$$

Using results from the previous solution. Plugging in $\mathbf{sin\theta }\mathbf{=}\sqrt{{\mathbf{n}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{-}{\mathbf{n}}_{\mathbf{2}}^{\mathbf{2}}}$and simplifying gives:

$$\begin{gathered} {\mathbf{t}}_{\mathbf{zig}\mathbf{-}\mathbf{zag}}\mathbf{=}\frac{{\mathbf{n}}_{\mathbf{1}}\mathbf{L}}{\mathbf{c}\left(\frac{n_2}{n_1}\right)} \\ \mathbf{=}\frac{{\mathbf{n}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{L}}{{\mathbf{cn}}_{\mathbf{3}}} \end{gathered}$$

The difference in the travel times along two routes.

The difference is written as follows:

$$\begin{gathered} \Delta t=t_{zig-zag}-t_{direct} \\ =\frac{n_{1}L}{c{n}_3}-\frac{n_{1}L}{c} \\ =\frac{n_{1}L}{c}\left(\frac{n_1}{n_2}-1\right) \end{gathered}$$

With$n_1=1.58,n_2=1.53andL=300 \text{m}$gives

$\Delta t=\frac{n_{1}L}{c}\left(\frac{n_1}{n_2}-1\right)$

$=\frac{\left(1.58\right)\left(800m\right)}{3\times {10}^{8}m/s}\left(\begin{array}{c}\frac{1.58}{1.53}-1\\ \end{array}\right)$

$=5.16\times {10}^{-8} \text{s}$

$=51.6 \text{ns}$

Hence, the difference in the travel times along two routes is $51.6ns$.