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

$$ A = 60^\circ $$

$$ \delta_m = \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ $$

$$ \mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} $$

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

$$ \mu = \frac{\sin\left(\frac{90^\circ}{2}\right)}{\sin(30^\circ)} $$

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

Substitute the standard trigonometric values ($$ \sin(45^\circ) = \frac{1}{\sqrt{2}} $$ and $$ \sin(30^\circ) = \frac{1}{2} $$):

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

$$ \mu = \frac{1}{\sqrt{2}} \times 2 $$

$$ \mu = \frac{2}{\sqrt{2}} $$

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