An optical fibre ( μ = 1.72 ) is enclosed by a glass coating ( μ =...

The critical angle $\theta_c$ for total internal reflection at the interface of two media is given by $\sin \theta_c = \dfrac{\mu_2}{\mu_1}$, where $\mu_2 is the refractive index of the rarer medium and$\mu_1$is the refractive index of the denser medium.<br><br>Given$\mu_1 = 1.72$(optical fibre)$ and \mu_2 = 1.50$(glass coating).<br><br>Substituting the values:<br><br>$\sin \theta_c = \dfrac{1.50}{1.72}$<br><br>$\sin \theta_c = \dfrac{150}{172}$<br><br>$\sin \theta_c = \dfrac{75}{86}$<br><br>$\theta_c = \sin^{-1} \left(\dfrac{75}{86}\right)