From the graph of angle of deviation versus angle of incidence for an equilateral prism, the refractive index of material of prism is

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

From the given graph of angle of deviation ($\delta$) versus angle of incidence ($i$), the minimum angle of deviation is $\delta_m = 60^\circ$ which occurs at $i = 60^\circ$.

The refractive index ($\mu$) of the material of the prism is given by the formula:
$\mu = \dfrac{\sin\left(\dfrac{A + \delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}$

Substituting the values of $A$ and $\delta_m$:
$\mu = \dfrac{\sin\left(\dfrac{60^\circ + 60^\circ}{2}\right)}{\sin\left(\dfrac{60^\circ}{2}\right)}$

$\mu = \dfrac{\sin(60^\circ)}{\sin(30^\circ)}$

$\mu = \dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}$

$\mu = \sqrt{3}$

Answer: $\sqrt{3}$