Explanation

(b) Here, angle of incidence $i = 45{^\circ}$
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$\frac{Lateral\ shift(d)}{Thickness\ of\ glass\ slab\ (t)} = \frac{1}{\sqrt{3}}$
Lateral shift $d = \frac{t\sin\delta}{\cos r} = \frac{t\sin{(i - r)}}{\cos r}$
$\Rightarrow \frac{d}{t} = \frac{sin(i - r)}{\cos r}$
or $\frac{d}{t} = \frac{\sin i\cos r - \cos i\sin r}{\cos r}$
or $\frac{d}{t} = \frac{\sin{45{^\circ}}\cos r - \cos{45{^\circ}}\sin r}{\cos r} = \frac{\cos r - \sin r}{\sqrt{2}\cos r}$
or $\frac{d}{t} = \frac{1}{\sqrt{2}}\left( 1 - \tan r \right)$
or $\frac{1}{\sqrt{3}} = \frac{1}{\sqrt{2}}\left( 1 - \tan r \right)$
or $\tan{r = 1 - \frac{\sqrt{2}}{\sqrt{3}}}$
or $r = \tan^{- 1}\left( 1 - \frac{\sqrt{2}}{\sqrt{3}} \right)$