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 angle of minimum deviation $\delta_m$ and the refracting angle $A$ 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 angle of minimum deviation $\delta_m$ and the refracting angle $A$ is given by:$\mu = \frac{\sin[(\delta_m + A)/2]}{\sin(A/2)}$Given that $\mu = \cot(A/2) = \frac{\cos(A/2)}{\sin(A/2)}$,we substitute this into the formula:$\frac{\cos(A/2)}{\sin(A/2)} = \frac{\sin[(\delta_m + A)/2]}{\sin(A/2)}$Canceling $\sin(A/2)$ from both sides,we get:$\cos(A/2) = \sin[(\delta_m + A)/2]$Using the trigonometric identity $\cos(	heta) = \sin(90^\circ - 	heta)$,we can write:$\sin(90^\circ - A/2) = \sin[(\delta_m + A)/2]$Equating the angles:$90^\circ - A/2 = (\delta_m + A)/2$Multiplying by $2$:$180^\circ - A = \delta_m + A$Therefore,$\delta_m = 180^\circ - 2A$.

The refracting angle of a prism is $A$ and the refractive index of the material of the prism is $\cot(A/2)$. Then the angle of minimum deviation is

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