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Physics Question on Ray optics and optical instruments

A prism of refractive index 2\sqrt{2} has a refracting angle of 60∘ 60^{\circ} . At what angle a ray must be incident on it so that it suffers a minimum deviation?

A

45∘45^{\circ}

B

60∘60^{\circ}

C

80∘80^{\circ}

D

180∘180^{\circ}

Answer

45∘45^{\circ}

Explanation

Solution

The relation for refractive index of prism is μ=sin⁡isin⁡r\mu=\frac{\sin i}{\sin r} ...(i) The condition for minimum deviation is r=A2=60∘2=30∘r=\frac{A}{2}=\frac{60^{\circ}}{2}=30^{\circ} Putting the given values of μ=2\mu=\sqrt{2} and r=30∘r=30^{\circ} in E (i), we get 2=sin⁡isin⁡30∘\sqrt{2} =\frac{\sin i}{\sin 30^{\circ}} or sin⁡i=2×12=12\sin i =\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}} sin⁡i=sin⁡45∘\sin i =\sin 45^{\circ} ∴i=45∘\therefore i =45^{\circ}