91Ó°ÊÓ

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${\mathbf{v}}_{\mathbf{1}}\mathbf{³¦´Ç²õθ}$ , so the time of travel for this ray is:

$$\begin{gathered} {\mathbf{t}}_{\mathbf{zig}\mathbf{-}\mathbf{zag}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{v}}_{\mathbf{1}}\mathbf{³¦´Ç²õθ}} \\ \mathbf{=}\frac{{\mathbf{n}}_{\mathbf{1}}\mathbf{L}}{\mathbf{c}\sqrt{\mathbf{1}\mathbf{-}{\left(\frac{\mathrm{²õ¾\pm ²Ôθ}}{n_1}\right)}^{\mathbf{2}}}} \end{gathered}$$

Using results from the previous solution. Plugging in $\mathbf{²õ¾\pm ²Ôθ}\mathbf{=}\sqrt{{\mathbf{n}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{-}{\mathbf{n}}_{\mathbf{2}}^{\mathbf{2}}}$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 is written as follows:

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

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

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

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

$=5.16×{10}^{-8} \text{s}$

$=51.6 \text{ns}$

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