Angle of minimum deviation is equal to the half of the angle of...

For an equilateral prism, the angle of the prism is $A = 60^{\circ}$.

Given that the angle of minimum deviation is half of the angle of the prism:

$\delta_{m} = \dfrac{A}{2} = \dfrac{60^{\circ}}{2} = 30^{\circ}$

The refractive index of the prism is given by:

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

Substituting the values of $A$ and $\delta_{m}$:

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

$\mu = \dfrac{\sin(45^{\circ})}{\sin(30^{\circ})}$

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

Answer: $\sqrt{2}$