The angle of a prism is A and its refractive | vedclass

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(D) The formula for the refractive index $\mu$ of a prism in terms of the prism angle $A$ and the angle of minimum deviation $\delta_m$ is given by:$\

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$180^\circ - 2A$

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(D) The formula for the refractive index $\mu$ of a prism in terms of the prism angle $A$ and the angle of minimum deviation $\delta_m$ is given by:$\mu = \frac{\sin \left( \frac{A + \delta_m}{2} ight)}{\sin \left( \frac{A}{2} ight)}$Given $\mu = \cot \frac{A}{2}$,we substitute this into the formula:$\cot \frac{A}{2} = \frac{\sin \left( \frac{A + \delta_m}{2} ight)}{\sin \left( \frac{A}{2} ight)}$Using the trigonometric identity $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$:$\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}} = \frac{\sin \left( \frac{A + \delta_m}{2} ight)}{\sin \left( \frac{A}{2} ight)}$Since $\cos 	heta = \sin(90^\circ - 	heta)$,we have:$\sin \left( 90^\circ - \frac{A}{2} ight) = \sin \left( \frac{A + \delta_m}{2} ight)$Comparing the angles:$90^\circ - \frac{A}{2} = \frac{A + \delta_m}{2}$$180^\circ - A = A + \delta_m$$\delta_m = 180^\circ - 2A$

The angle of a prism is $A$ and its refractive index is $\cot \frac{A}{2}$. The angle of minimum deviation is:

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