Ray Optics

JEE Main level, single correct, medium

233 of 295

Question 233 of 295

Angle of minimum deviation is equal to the half of the angle of prism in an equilateral prism. The refractive index of the prism is __________.

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Answer: (C) 2\sqrt{2}

For an equilateral prism, the angle of the prism is A=60∘A = 60^{\circ}.

Given that the angle of minimum deviation is half of the angle of the prism:

δm=A2=60∘2=30∘\delta_{m} = \dfrac{A}{2} = \dfrac{60^{\circ}}{2} = 30^{\circ}

The refractive index of the prism is given by:

μ=sin⁡(A+δm2)sin⁡(A2)\mu = \dfrac{\sin\left(\dfrac{A + \delta_{m}}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}

Substituting the values of AA and δm\delta_{m}:

μ=sin⁡(60∘+30∘2)sin⁡(60∘2)\mu = \dfrac{\sin\left(\dfrac{60^{\circ} + 30^{\circ}}{2}\right)}{\sin\left(\dfrac{60^{\circ}}{2}\right)}

μ=sin⁡(45∘)sin⁡(30∘)\mu = \dfrac{\sin(45^{\circ})}{\sin(30^{\circ})}

μ=1212=2\mu = \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{2}} = \sqrt{2}

Answer: 2\sqrt{2}

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