The refracting angle of a prism is A and the | vedclass

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

Vedclass · Accepted Answer

$180^\circ - 2A$

Vedclass · Answer

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

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