For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the ref

Question

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index $n$ of the prism satisfies.

Professor PC Thomas Classes · Answer

At minimum deviation, the refractive index is $n = \frac{\sin((A+D_m)/2)}{\sin(A/2)}$. Given $D_m = A$, this simplifies to $n = \frac{\sin((A+A)/2)}{\sin(A/2)} = \frac{\sin A}{\sin(A/2)} = \frac{2\sin(A/2)\cos(A/2)}{\sin(A/2)} = 2\cos(A/2)$. For a physically viable transparent prism, the refracting angle $A$ must be $0^\circ < A < 90^\circ$ (for the emerging ray not to suffer total internal reflection at grazing incidence). Substituting the limits: if $A 	o 0^\circ$, $n 	o 2\cos(0) = 2$; if $A 	o 90^\circ$, $n 	o 2\cos(45^\circ) = \sqrt{2}$. Therefore, the refractive index must satisfy $\sqrt{2} < n < 2$.