Explanation

(d) By Brewster’s law, $\mu = {\tan\theta}_{p}$
Or $\mu = \tan{54.74{^\circ}}$
Or $\mu = 1.414$
For an equilateral prism, $\angle A = 60{^\circ}$
$\therefore\ \mu = \frac{\sin\left( \frac{A + \delta_{m}}{2}\ \right)\ }{\sin\left( \frac{A}{2} \right)}$
Or $1.414 = \frac{\sin\left( \frac{60{^\circ} + \delta_{m}}{2} \right)}{\sin\left( \frac{60{^\circ}}{2}\ \right)}$
Or $\frac{1.414\ \times 1}{2} = \sin\left( \frac{60{^\circ} + \delta_{m}}{2} \right)\ \ \ \ \ \ \ \ \ \ \ \left\lbrack \because 1.414 = \sqrt{2} \right\rbrack$
Or $\frac{\sqrt{2}}{2} = \sin\left( \frac{60{^\circ} + \delta_{m}}{2} \right)\ $
Or $\frac{1}{\sqrt{2}} = \sin\left( \frac{60{^\circ} + \delta_{m}}{2} \right)$
Or $45{^\circ} = \left( \frac{60{^\circ} + \delta_{m}}{2} \right)$
Or $\delta = 30{^\circ}$