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(C) For an equilateral prism, the angle of the prism $A = 60^{\circ}$.Given that the angle of minimum deviation $\delta_m = A = 60^{\circ}$.The formul

Vedclass · Accepted Answer

$1.732$

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(C) For an equilateral prism, the angle of the prism $A = 60^{\circ}$.Given that the angle of minimum deviation $\delta_m = A = 60^{\circ}$.The formula for the refractive index $n$ of a prism is given by $n = \frac{\sin((A + \delta_m) / 2)}{\sin(A / 2)}$.Substituting the values, we get $n = \frac{\sin((60^{\circ} + 60^{\circ}) / 2)}{\sin(60^{\circ} / 2)}$.$n = \frac{\sin(120^{\circ} / 2)}{\sin(30^{\circ})} = \frac{\sin(60^{\circ})}{\sin(30^{\circ})}$.Since $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$ and $\sin(30^{\circ}) = \frac{1}{2}$, we have $n = \frac{\sqrt{3} / 2}{1 / 2} = \sqrt{3}$.The value of $\sqrt{3}$ is approximately $1.732$.

If the angle of minimum deviation produced by an equilateral prism is equal to the angle of the prism, then the refractive index of the material of the prism is nearly

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