The refractive index of a prism for a monochromatic wave is 2 and its refracting angle is 60°. For minimum deviation, the angle of incidence will be

μ=sin⁡isin⁡A/2⇒2=sin⁡isin⁡602⇒2×sin⁡30=sin⁡i⇒i=45∘

The refractive index of a prism for a monochromatic wave is 2 and its refracting angle is 60°. For minimum deviation, the angle of incidence will be

μ=sin⁡isin⁡A/2⇒2=sin⁡isin⁡602⇒2×sin⁡30=sin⁡i⇒i=45∘

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The refractive index of a prism for a monochromatic wave is $\sqrt{2}$ and its refracting angle is 60°. For minimum deviation, the angle of incidence will be

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30°

45°

60°

75°

answer is B.

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Detailed Solution

$\begin{array}{l} μ = \frac{\sin i}{\sin A / 2} \Rightarrow \sqrt{2} = \frac{\sin i}{\sin (\frac{60}{2})} \\ \Rightarrow \sqrt{2} \times \sin 30 = \sin i \Rightarrow i = 45^{\circ} \end{array}$

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The refractive index of a prism for a monochromatic wave is 2 and its refracting angle is 60°. For minimum deviation, the angle of incidence will be

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The refractive index of a prism for a monochromatic wave is $\sqrt{2}$ and its refracting angle is 60°. For minimum deviation, the angle of incidence will be