The minimum deviation produced by a hollow prism filled with a certain liquid is found to be $30^{\circ}$.…

The minimum deviation produced by a hollow prism filled with a certain liquid is found to be $30^{\circ}$. The light ray is also found to be refracted at an angle of $30^{\circ}$. Then the refractive index of the liquid is
  1. $\sqrt{2}$
  2. $\sqrt{3}$
  3. $\sqrt{\frac{3}{2}}$
  4. $\frac{3}{2}$

Solution

For prism, $\mathrm{s}_{\mathrm{m}}=30^{\circ}, \mathrm{r}=30^{\circ}$ $\therefore \quad A=2 r=2 \times 30^{\circ}=60^{\circ}$ $\therefore$ Refractive index, $\mu=\frac{\sin \left(\frac{A+S_m}{2}\right)}{\sin \frac{A}{2}}$ $=\frac{\sin \left(\frac{60+30}{2}\right)}{\sin \left(\frac{60}{2}\right)}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}=\sqrt{2}$

Asked in: AP EAMCET 2024 (19 May Shift 2)

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