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Physics Question on Refraction of Light

The angle of incidence for a ray of light at a refracting surface of a prism is 45∘45^{\circ}. The angle of prism is 60∘60^{\circ }. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are :

A

30∘;230^{\circ} ; \sqrt{2}

B

45∘;245^{\circ} ; \sqrt{2}

C

30∘;1230^{\circ} ; \frac{1}{ \sqrt{2}}

D

45∘;1245^{\circ} ; \frac{1}{ \sqrt{2}}

Answer

30∘;230^{\circ} ; \sqrt{2}

Explanation

Solution

i=45∘;A=60∘;δm=2i−A=30∘i = 45^{\circ} ; A = 60^{\circ} ; \delta_m = 2i - A = 30^{\circ}
μ=sin⁡(A+δm2)sin⁡A/2=sin⁡45∘sin⁡30∘=12.21=2\mu = \frac{\sin\left(\frac{A + \delta_{m}}{2}\right)}{\sin A/2} = \frac{\sin45^{\circ}}{\sin30^{\circ}} = \frac{1}{\sqrt{2}}. \frac{2}{1} = \sqrt{2}