The angle of minimum deviation for an inciden | vedclass

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(B) For an equilateral prism,the angle of prism or refracting angle is $A = 60^{\circ}$.Given that the angle of minimum deviation $\delta_{m}$ is equa

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$\sqrt{3}$

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(B) For an equilateral prism,the angle of prism or refracting angle is $A = 60^{\circ}$.Given that the angle of minimum deviation $\delta_{m}$ is equal to the refracting angle $A$,so $\delta_{m} = A = 60^{\circ}$.The formula for the refractive index $\mu$ of a prism is given by $\mu = \frac{\sin((A + \delta_{m})/2)}{\sin(A/2)}$.Substituting the values,we get $\mu = \frac{\sin((60^{\circ} + 60^{\circ})/2)}{\sin(60^{\circ}/2)}$.$\mu = \frac{\sin(60^{\circ})}{\sin(30^{\circ})}$.Since $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$ and $\sin(30^{\circ}) = \frac{1}{2}$,we have $\mu = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.

The angle of minimum deviation for an incident light ray on an equilateral prism is equal to its refracting angle. The refractive index of its material is

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