Explanation

(c) Here, $i = 45{^\circ}$
Applying Snell’s law at air-glass surface, we get
$\mu_{a}\sin i = \mu_{g}\sin{r''}$
$1\sin i = \sqrt{2}\sin{r''}$
$\sin{r''} = \frac{1}{\sqrt{2}}\sin i$
$= \frac{1}{\sqrt{2}}\sin{45{^\circ}}$
[image:9_image88.emf]
$\sin{r''} = \frac{1}{2}$
$r^{''} = \sin^{- 1}{\left( \frac{1}{2} \right) = 30{^\circ}}$
From figure, $r + \theta + r^{''} = 180{^\circ}$
$i + \theta + 30^{o} = 180^{o}\ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack\therefore i = r\rbrack$
$i + \theta + 30{^\circ} = 180{^\circ} \Rightarrow \theta = 180{^\circ} - 75{^\circ} = 105{^\circ}$
Hence, the angle between reflected and refracted rays is $105{^\circ}$