Maximum angle of deviation of a prism having refractive index $\sqrt{2}$ and angle of prism 75° will be

$\frac{\pi}{6}$

The correct answer is Option (1) → $\frac{\pi}{6}$

For grazing emergence e = 90°

For Snell's law $\frac{\sin r_2}{\sin e}=\frac{1}{\sqrt{2}}$

$\Rightarrow r_2=45^{\circ}$

$\Rightarrow r_1+r_2=75^{\circ} ~~~~~\Rightarrow r_1=30^{\circ}$

Now $\frac{\sin i}{\sin r_1}=\sqrt{2}$

$\sin i=\frac{1}{2} \times \sqrt{2} ~~~~~\Rightarrow i=45^{\circ}$

As $i+e=A+\delta$

$\delta=45^{\circ}+90^{\circ}-75^{\circ}=60^{\circ}$

$\delta=\frac{\pi}{3}$