The angle of a prism is 60° and its refractive index is 2 The angle of minimum deviation suffered by a ray of light in passing through it is

μ=sinA+δm2sinA2⇒2;μ=sin60+δm2sin602⇒2×sin30=sin60+δm2⇒sin45o=sin60+δm2⇒dm=30o

The angle of a prism is 60° and its refractive index is 2 The angle of minimum deviation suffered by a ray of light in passing through it is

μ=sinA+δm2sinA2⇒2;μ=sin60+δm2sin602⇒2×sin30=sin60+δm2⇒sin45o=sin60+δm2⇒dm=30o

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The angle of a prism is 60° and its refractive index is $\sqrt{2}$ The angle of minimum deviation suffered by a ray of light in passing through it is

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About 20°

30°

60°

45°

answer is B.

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Detailed Solution

$\begin{array}{l} μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})} \Rightarrow \sqrt{2}; μ = \frac{\sin (\frac{60 + δ_{m}}{2})}{\sin (\frac{60}{2})} \\ \Rightarrow \sqrt{2} \times \sin 30 = \sin (\frac{60 + δ_{m}}{2}) \Rightarrow \sin 45^{o} \\ = \sin (\frac{60 + δ_{m}}{2}) \Rightarrow d_{m} = 30^{o} \end{array}$