A liquid is placed in a hollow prism of angle 60∘. If angle of minimum deviation is 30∘, what is the refractive index of liquid

μ=Sin⁡A+D2Sin⁡A2=Sin⁡60+302Sin⁡602=2

μ=Sin⁡A+D2Sin⁡A2=Sin⁡60+302Sin⁡602=2

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A liquid is placed in a hollow prism of angle $60^{\circ} .$ If angle of minimum deviation is $30^{\circ},$ what is the refractive index of liquid

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1.95

answer is A.

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$μ = \frac{Sin (\frac{A + D}{2})}{Sin (\frac{A}{2})} = \frac{Sin (\frac{60 + 30}{2})}{Sin (\frac{60}{2})} = \sqrt{2}$

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