Question

The minimum deviation produced by a hollow prism filled with a certain liquid to be ${30}^{\circ }$. If the refracting angle is ${60}^{\circ }$, then refractive index of the liquid is
(Neglect the refraction at the glass surface)

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\frac{3}{2}$

Answer 1

The correct option is A $\sqrt{2}$

Given that,
Minimum deviation produced by the prism filled by the liquid, ${\delta }_{min}={30}^{\circ }$
Referacting angle of prism, $A={60}^{\circ }$

The refractive index of the liquid in the prism is given by
$\mu =\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

$\Rightarrow \mu =\frac{\sin \left[\frac{{60}^{\circ }+{30}^{\circ }}{2}\right]}{\sin \left[\frac{{60}^{\circ }}{2}\right]}$

$\Rightarrow \mu =\frac{\sin {45}^{\circ }}{\sin {30}^{\circ }}$

$\Rightarrow \mu =\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}$

$\therefore \mu =\sqrt{2}$

Why this question? Tip: The liquid of refractive index $\mu$, will behave as the material of prism (when the hollow prism is filled with liquid)