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If the minimum deviation produced by a prism of refracting angle 60 is \( 30, \) then calculate the refractive index of the material of the prism.

1. 2
2. 1.414
3. \( \sqrt{13} \)
4. 4

Correct answer: 1.414

Solution

For a prism at minimum deviation, the refractive index is given by \(n=\sin\left(\frac{A+D_m}{2}\right)/\sin\left(\frac{A}{2}\right)\). Substituting \(A=60^\circ\) and \(D_m=30^\circ\) gives \(n=\sin 45^\circ/\sin 30^\circ=\frac{\sqrt2/2}{1/2}=\sqrt2\approx1.414\).

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