Q161 / 200 · AP EAMCET 2001 · Physics

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
    $42^{\circ}$
  • Correct answer
    B
    $45^{\circ}$
  • C
    $50^{\circ}$
  • D
    $52^{\circ}$

Solution

$A=60^{\circ}, \delta=30^{\circ}$ Refractive index of prism with respect to liquid, $\begin{aligned} \mu & =\frac{\sin \frac{A+\delta_n}{2}}{\sin \frac{A}{2}}=\frac{\sin \frac{\left(60^{\circ}+30^{\circ}\right)}{2}}{\sin \frac{60^{\circ}}{2}} \\ & =\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}=\sqrt{2}\end{aligned}$ $\therefore$ Critical angle $=\sin ^{-1}\left(\frac{1}{\mu}\right)=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$ $=\sin ^{-1}\left(\sin 45^{\circ}\right)=45^{\circ}$

Hence, the correct answer is option: 'B'.

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