Question

an optical fibre has core of refractive index 2 and cladding of a refractive index 1.732. At what angle should light enter the fibre so that it suffers TIR at the core cladding interface.

Answer 1

Dear Student

n₁ =2
n₂ =1.732
for TIR $$\begin{gathered} {\theta }_2={90}^0 \\ \sin {\theta }_2=\sin {90}^0=1 \\ \frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{n_2}{n_1} \\ \frac{\sin {\theta }_1}{1}=\frac{1.732}{2} \\ \sin {\theta }_1=0.866 \\ {\theta }_1={\sin }^{-1}(0.866) \\ {\theta }_1=59.{99}^0={60}^0 \\ Henceangleofincidence{\theta }_1={60}^0 \end{gathered}$$

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