Question

A light wave travels from glass to water. The refractive indices for glass and water are $\frac{3}{2}$ and $\frac{4}{3}$ respectively. The value of the critical angle will be:

• A. ${\sin }^{-1}\left(\frac{1}{2}\right)$
• B. ${\sin }^{-1}\left(\frac{9}{8}\right)$
• C. ${\sin }^{-1}\left(\frac{9}{8}\right)$
• D. ${\sin }^{-1}\left(\frac{5}{7}\right)$

Answer 1

Answer: (C)

From the relation,
$\frac{1}{\sin C}{=}_{\omega }{\mu }_g$ Or $\frac{1}{\sin C}=\frac{{\mu }_g}{{\mu }_{\omega }}$ where $C=$ critical angle ${\mu }_g=$ refractive index of glass $=\frac{3}{2}$ ${\mu }_{\omega }=$ refractive index of water $=\frac{4}{3}$ $\therefore$ $\sin C=\frac{{\mu }_{\omega }}{{\mu }_g}$ $\sin C=\frac{4/3}{3/2}$ $\sin C=\frac{8}{9}$ Hence, $C={\sin }^{-1}\left(\frac{8}{9}\right)$