Question

When a glass prism of refracting angle $60^{\circ}$ is immersed in a liquid its angle of minimum deviation is $30^{\circ}$. The critical angle of glass with respect to the liquid medium is

• A. $42^{\circ}$
• B. $45^{\circ}$
• C. $50^{\circ}$
• D. $52^{\circ}$

Answer 1

Answer: (B)

$A=60^{\circ}, \delta_{m}=30^{\circ}$ so $\mu=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\mu=\frac{\sin \left(\frac{60^{\circ}+30^{\circ}}{2}\right)}{\sin \left(\frac{60^{\circ}}{2}\right)}=\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\sqrt{2}$
Also $\mu=\frac{1}{\sin C} \Rightarrow C=\sin ^{-1}\left(\frac{1}{\mu}\right) $
$\Rightarrow C=45^{\circ}$