Figure shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68.
The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflection inside the pipe take place, as shown in the figure.
(b) What is the answer if there is no outer covering of the pipe?

(a) Given, a cross-section of a light pipe. 

$$\begin{gathered} \mathrm{Refractive} \mathrm{index} \mathrm{of} \mathrm{glass}-\mathrm{fibre}, {\mathrm{\mu }}_2 = 1.68 \\ \mathrm{Refractive} \mathrm{index} \mathrm{of} \mathrm{outer} \mathrm{covering} {\mathrm{\mu }}_1 = 1.44 \end{gathered}$$ 

$\therefore$ $\mathrm{\mu } = \frac{{\mathrm{\mu }}_2}{{\mathrm{\mu }}_1} = \frac{\mathrm{i}}{\sin {\mathrm{i}}_{\mathrm{c}}}$

$\therefore$ Critical angle ${\mathrm{i}}_{\mathrm{c}}\text{'}$ is given by

$\sin \mathrm{i}{\text{'}}_{\mathrm{c}} = \frac{{\mathrm{\mu }}_1}{{\mathrm{\mu }}_2} = \frac{1.44}{1.68} = 0.8571$
$\Rightarrow$ $\mathrm{i}{\text{'}}_{\mathrm{c}} = 59°$ 

Total internal reflection will occur if angle of incidence is greater than the critical angle, i’ > i_c’
i.e., if i’ > 59° or,
when r < r_max, where r = 90°– 59° = 31°, which is the angle of reflection. 

Now using Snell's law 

                 $\frac{\sin {\mathrm{i}}_{\max }}{\sin {\mathrm{r}}_{\max }} = 1.68$

$\Rightarrow$               $\sin {\mathrm{i}}_{\max } = 1.68 \times \sin {\mathrm{r}}_{\max }$ 

                              $$\begin{gathered} = 1.68 \times \sin 31° \\ = 1.68 \times 0.5150 = 0.8662 \end{gathered}$$

$\therefore$                    ${\mathrm{i}}_{\max } = 60°.$ 

Therefore,  in the pipe the rays having incident angles in the range 0 < i < 60° with the axis of the pipe will suffer total internal reflection. 
For the finite length of the pipe, the lower limit on i is determined by the ratio of the diameter to the length of the pipe. The lower limit of angle of incidence is not 0^o. 

(b) If there is no outer covering of the pipe then,

                     ${\mathrm{\mu }}_2 = 1.68 , {\mathrm{\mu }}_1 = 1$ 
Hence,
            $\sin \mathrm{i}{\text{'}}_{\mathrm{c}} = \frac{{\mathrm{\mu }}_1}{{\mathrm{\mu }}_2} = \frac{1}{1.68} = 0.5952$
$\Rightarrow$            $\mathrm{i}{\text{'}}_{\mathrm{c}} = 36.5 °$ 

Now, for angle of incidence, i = 36.5° 

              $$\begin{gathered} \frac{\sin \mathrm{i}}{\sin \mathrm{r}} =\mathrm{\mu } = 1.68 \\ \Rightarrow \frac{\sin {90}^{\mathrm{o}}}{1.68} = \sin \mathrm{r} \\ \Rightarrow \frac{1}{1.68} = \sin \mathrm{r} \\ \Rightarrow \mathrm{r} = 36.5^{\mathrm{o}} \end{gathered}$$ 

we have r = 36.5° and,
i’ = 90° – 36.5° = 53.5°, which is greater than the critical angle, i_c. 

Thus all rays which are incident at angle in the range 0 < i < 90° will suffer total internal reflection.