The minimum deviation produced by a hollow prism filled with a certain liquid to be 30⁰ . The light ray is also found to be refracted at angle 30⁰ . The refractive index of the liquid is

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3/2

$\sqrt{3}$

$\sqrt{2}$

$\sqrt{\frac{3}{2}}$

answer is A.

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Detailed Solution

$\begin{array}{l} A = 2 r \Rightarrow A = 60^{\circ}; ∵ r_{1} = r_{2} = r a n d r_{1} + r_{2} = A \\ μ = \frac{\sin (\frac{60 + 30}{2})}{\sin [\frac{60}{2}]} \Rightarrow μ = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \sqrt{2} \end{array}$

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