Explanation

(b) For total internal reflection from glass-air interface, critical angle $C$ must be less than angle of incidence.
$ie\ \ \ C < i$
$or\ C < 45{^\circ}\ \ \ \ \ \ \ \ (\because\ \angle i = 45{^\circ})$
$but\ \ \ \ n = \frac{1}{\sin C} \Longrightarrow C = \sin^{- 1}\left( \frac{1}{n} \right)$
[image:9_image13.emf]
$\therefore\ \sin^{- 1}\left( \frac{1}{n} \right) < 45{^\circ}$
$\Longrightarrow \frac{1}{n} < \sin{45{^\circ}}$
$\Longrightarrow n > \frac{1}{\sin{45{^\circ}}}$
$\Longrightarrow n > \frac{1}{\left( \frac{1}{\sqrt{2}} \right)}$
$\Longrightarrow n > \sqrt{2}$