(a) Refractive index of outer covering,
$\mathrm{n}_1=1.44$
refractive index of glass fibre, $\mathrm{n}_2=1.68$
At the interface of the glass fibre and covering
As, $\mathrm{n}=\frac{\mathrm{n}_2}{\mathrm{n}_1}$, also the critical angle,
$\sin i_c^{\prime}=\frac{1}{n}\left(i_c^{\prime}=\right.$ critical angle $)$ $\therefore \sin i_c^{\prime}=\frac{n_1}{n_2}=\frac{1.44}{1.68}=0.8571$
or, $\sin \mathrm{i}_{\mathrm{c}}^{\mathrm{c}_{\mathrm{c}}}=\sin 59^{\circ} \quad \therefore \mathrm{i}_{\mathrm{c}}^{\prime}=59^{\circ}$
$\therefore \angle \mathrm{r}=90^{\circ}-\angle i_{\mathrm{c}}^{\prime}=90^{\circ}-59^{\circ}=31^{\circ}$
At the air - glassfibre interface
as, $\mathrm{n}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}, 1.68=\frac{\sin \mathrm{i}}{\sin 31^{\circ}}$
$\therefore \sin \mathrm{i}=1.68 \times \sin 31^{\circ}=1.68 \times 0.5156$
$=0.8662$
$\therefore \sin \mathrm{i}=\sin 60^{\circ} \quad \therefore \mathrm{i}=60^{\circ}$
(b) If there were no outer covering of the pipe,
for air $\mathrm{n}_1=1$, for glass fibre $\mathrm{n}_2=1.68$
$\sin \mathrm{i}_{\mathrm{c}}=\frac{1}{\mathrm{n}}=\frac{\mathrm{n}_1}{\mathrm{n}_2}$
$$
=\frac{1}{1.68}=0.5952 \therefore \mathrm{i}_{\mathrm{c}}=36.5^{\circ} \text {. }
$$
Now, $\mathrm{i}=90^{\circ}$,
$$
\therefore \mathrm{n}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}, 1.68=\frac{\sin 90^{\circ}}{\sin \mathrm{r}}=\frac{1}{\sin \mathrm{r}}
$$
$$
\begin{aligned}
&\therefore \sin \mathrm{r}=\frac{1}{1.68}=0.5952 \quad \therefore \mathrm{r}=36.5^{\circ} \\
&\therefore \mathrm{i}^{\prime}=90^{\circ}-36.5^{\circ}=53.5^{\circ},
\end{aligned}
$$
Which is greater than $\mathrm{i}_{\mathrm{c}}^{\prime}$.
therefore all incident rays (in range $53.5^{\circ} < \mathrm{i} < 90^{\circ}$ )
will undergo total internal reflection.