Q. 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:
MGIMS WardhaMGIMS Wardha 2003
Solution:
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) $
