If refractive index, $μ$, is property of the material of the prism and not of the prism itself, then why do deviation the prism angle, α, and minimum deviation Dm, both of which are characteristic properties. of the prism, enter into the equation for the refractive index?All I want to know is, that is there any way by which I can proof that μ is independent of α & Dm through rigorous math?
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$\begingroup$ Perhaps see the answers here: physics.stackexchange.com/q/145372/313612 and look around this stack exchange for other prism-relevant questions and answers. $\endgroup$– Ed VAug 25 at 15:46
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That the minimum ray deviation, see, must also depend on the geometry can be appreciated by a symmetry argument that the minimum must be such that the incoming refracted angle be the same as the outgoing one.
If you agree to that then now imagine that you symmetrically and gradually increase the prism angle $\alpha$. For each angle the symmetry is maintained, thus, the deviation angle is still a minimum for that particular $\alpha$. But the deviation angle while still a minimum must also increase, otherwise the prism could not refract at all as $\alpha \to \pi - \epsilon$ for some arbitrarily small $\epsilon>0$.
