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 $\mu =\frac{1}{\sin C}\Rightarrow C={\sin }^{-1}\left(\frac{1}{\mu }\right)$
$\Rightarrow C=45^{\circ }$