Question

An optical fibre (μ = 1.72) is surrounded by a glass coating (μ = 1.50). Find the critical angle for total internal reflection at the fibre-glass interface.

Answer 1

Given,
Refractive index of the optical fibre is represented by μ_o = 1.72
Refractive index of glass coating is represented by μ_g= 1.50
Let the critical angle for glass be θ_c

Using the Snell's law,




$$\begin{gathered} \frac{\sin i}{\sin r}=\frac{\sin {\theta }_{\mathrm{c}}}{\sin 90°}=\frac{{\mathrm{\mu }}_{\mathrm{g}}}{{\mathrm{\mu }}_0} \\ \Rightarrow \frac{\sin {\theta }_{\mathrm{c}}}{\sin 90°}=\frac{1.50}{1.72}=\frac{75}{86} \\ \Rightarrow {\theta }_{\mathrm{c}}={\sin }^{-1}\left(\frac{75}{86}\right) \end{gathered}$$

Hence, the required critical angle is ${\sin }^{-1}\left(\frac{75}{86}\right)$