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Physics Question on Refraction of Light

The angle of minimum deviation for an incident light ray on an equilateral prism is equal to its refracting angle. The refractive index of its material is

A

3\sqrt {3}

B

32 \frac{\sqrt {3}}{2}

C

32\frac{3}{2}

D

12\frac {1}{\sqrt {2}}

Answer

3\sqrt {3}

Explanation

Solution

For an equilateral prism,
angle of prism of refracting angle A=60∘A=60^{\circ}
Here,      δm=A=60∘\,\,\,\,\,\,\delta_{m}=A=60^{\circ}
∴\therefore Refractive index,
μ=sin⁡(A+δm2)sin⁡A2=sin⁡(60∘+60∘2)sin⁡(60∘2)\mu=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \frac{A}{2}}=\frac{\sin \left(\frac{60^{\circ}+60^{\circ}}{2}\right)}{\sin \left(\frac{60^{\circ}}{2}\right)}
=sin⁡60∘sin⁡30∘=sin⁡60∘cos⁡60∘=tan⁡60∘=3=\frac{\sin 60^{\circ}}{\sin 30^{\circ}}=\frac{\sin 60^{\circ}}{\cos 60^{\circ}}=\tan 60^{\circ}=\sqrt{3}