A light ray passing through a prism having a prism angle of 60^ ^ ^...

The minimum deviation \delta_\min for a prism is always less than or equal to any observed deviation $\delta$.

Given the prism angle A = 60^{}^{{}^{\circ}} and an observed deviation \delta = 30^{}^{{}^{\circ}}, we have \delta_\min \leq 30^{}^{{}^{\circ}}.

The refractive index of the material of the prism is given by:

\mu = \dfrac{\sin\left(\dfrac{A + \delta_\min}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}

Since \delta_\min \leq 30^{}^{{}^{\circ}} and the sine function is increasing in this interval:

\mu \leq \dfrac{\sin\left(\dfrac{60^{}^{{}^{\circ}} + 30^{}^{{}^{\circ}}}{2}\right)}{\sin\left(\dfrac{60^{}^{{}^{\circ}}}{2}\right)}

\mu \leq \dfrac{\sin 45^{}^{{}^{\circ}}}{\sin 30^{}^{{}^{\circ}}}

$\mu \leq \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{2}}$

$\mu \leq \sqrt{2}$