For the angle of minimum deviation of a prism | vedclass

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(B) The refractive index $\mu$ of a prism is given by the formula: $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.Given that the angle of minimum de

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between $2$ and $\sqrt{2}$

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(B) The refractive index $\mu$ of a prism is given by the formula: $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.Given that the angle of minimum deviation $\delta_m$ is equal to the refracting angle $A$,we substitute $\delta_m = A$ into the formula:$\mu = \frac{\sin((A + A)/2)}{\sin(A/2)} = \frac{\sin(A)}{\sin(A/2)}$.Using the trigonometric identity $\sin(A) = 2 \sin(A/2) \cos(A/2)$,we get:$\mu = \frac{2 \sin(A/2) \cos(A/2)}{\sin(A/2)} = 2 \cos(A/2)$.Since the refracting angle $A$ of a prism typically ranges from $0^\circ If $A 	o 0^\circ$,$\mu 	o 2 \cos(0) = 2$.If $A 	o 90^\circ$,$\mu 	o 2 \cos(45^\circ) = 2(1/\sqrt{2}) = \sqrt{2}$.Therefore,the refractive index must be between $\sqrt{2}$ and $2$.

For the angle of minimum deviation of a prism to be equal to its refracting angle,the prism must be made of a material whose refractive index is:

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