The refractive index of a prism with apex ang | vedclass

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

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$\delta_{m}=180^{\circ}-2A$

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(D) The formula for the refractive index $\mu$ of a prism in terms of the apex angle $A$ and the angle of minimum deviation $\delta_m$ is given by:$\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$Given $\mu = \cot(A/2) = \frac{\cos(A/2)}{\sin(A/2)}$,we equate the two expressions:$\frac{\cos(A/2)}{\sin(A/2)} = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$Canceling $\sin(A/2)$ from both sides,we get:$\cos(A/2) = \sin((A + \delta_m)/2)$Using the trigonometric identity $\cos(	heta) = \sin(90^{\circ} - 	heta)$,we write:$\sin(90^{\circ} - A/2) = \sin((A + \delta_m)/2)$Equating the angles:$90^{\circ} - A/2 = (A + \delta_m)/2$$180^{\circ} - A = A + \delta_m$$\delta_m = 180^{\circ} - 2A$

The refractive index of a prism with apex angle $A$ is $\cot(A/2)$. The angle of minimum deviation is:

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