Angle of minimum deviation is equal to the half of the angle of prism in an equilateral prism. The refractive index of the prism is __________.

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: \(\sqrt{2}\)