Maximum angle of deviation of a prism having refractive index $\sqrt{2}$ and angle of prism 75° will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{\pi}{3}$
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 $i+e=A+\delta$
$\delta=45^{\circ}+90^{\circ}-75^{\circ}=60^{\circ}$
$\delta=\frac{\pi}{3}$