When a ray of light is incident at an angle of incidence 60° on one face of an equilateral prism the angle of deviation is found to be minimum. The refractive index of the prism material is:

$\sqrt{3}$

The correct answer is Option (1) → $\sqrt{3}$

Angle of Minimum deviation ($δ_{min}$) through the prism -

$δ_{min}=2i-A$ [i = Angle of incidence, A = Angle of prism]

$=2×60-60$

$=60°$

$μ=\frac{\sin\left(\frac{A+δ_{min}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

$=\frac{\sin\left(\frac{60+60}{2}\right)}{\sin\left(\frac{60}{2}\right)}=\sqrt{3}$