Consider light travelling from a medium $A$ to medium $B$ separated by a plane interface. If the lig

Question

Consider light travelling from a medium $A$ to medium $B$ separated by a plane interface. If the light undergoes total internal reflection during its travel from medium $A$ to $B$ and the speed of light in media $A$ and $B$ are $2.4 	imes 10^8 \mathrm{~m} / \mathrm{s}$ and $2.7 	imes 10^8 \mathrm{~m} / \mathrm{s}$, respectively, then the value of critical angle is :

Professor PC Thomas Classes · Answer

Total internal reflection occurs when light travels from a denser medium to a rarer medium. Here, $v_A < v_B$, so medium A is denser. The critical angle $	heta_c$ is given by $\sin	heta_c = \frac{\mu_B}{\mu_A} = \frac{v_A}{v_B}$. Substituting the speeds: $\sin	heta_c = \frac{2.4 	imes 10^8}{2.7 	imes 10^8} = \frac{8}{9}$. If $\sin	heta_c = \frac{8}{9}$, we can draw a right triangle with opposite side 8 and hypotenuse 9. The adjacent side is $\sqrt{9^2 - 8^2} = \sqrt{81 - 64} = \sqrt{17}$. Therefore, $	an	heta_c = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{8}{\sqrt{17}}$, giving $	heta_c = 	an^{-1}(\frac{8}{\sqrt{17}})$.