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,

\[\frac{\sin  i}{\sin  r} = \frac{\sin  \theta_c}{\sin  90^\circ } = \frac{\mu_g}{\mu_0}\] 

\[ \Rightarrow \frac{\sin  \theta_c}{\sin  90^\circ } = \frac{1 . 50}{1 . 72} = \frac{75}{86}\] 

\[ \Rightarrow    \theta_c  =  \sin^{- 1} \left( \frac{75}{86} \right)\]
Hence, the required critical angle is \[\sin^{- 1} \left( \frac{75}{86} \right)\]