A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is $$\sqrt{3}$$, then the angle of incidence is:

We are dealing with an equilateral triangular prism, so its apex angle is $$A = 60^\circ$$ because each angle of an equilateral triangle measures $$60^\circ$$.

For a prism, the condition of minimum deviation is very useful. At minimum deviation the light ray travels symmetrically inside the prism, giving us two important equalities:

$$i = e \quad\text{and}\quad r_1 = r_2 = r$$

where

$$i$$ = angle of incidence,
$$e$$ = angle of emergence,
$$r_1, r_2$$ = angles of refraction at the two surfaces, and
$$r$$ = their common value in the symmetric case.

Because the two refracted angles add up to the apex angle inside the prism, we have

$$r + r = A \Longrightarrow 2r = A.$$

Substituting the value $$A = 60^\circ$$, we get

$$2r = 60^\circ \;\Longrightarrow\; r = \frac{60^\circ}{2} = 30^\circ.$$

Now we invoke Snell’s Law at the first refracting surface. Snell’s law states

$$n = \frac{\sin i}{\sin r},$$

where $$n$$ is the refractive index of the prism material. The problem tells us that

$$n = \sqrt{3}.$$

So we write

$$\sqrt{3} = \frac{\sin i}{\sin 30^\circ}.$$

We know that $$\sin 30^\circ = \dfrac{1}{2}$$, therefore

$$\sqrt{3} = \frac{\sin i}{1/2} = 2 \sin i.$$

Rearranging gives

$$\sin i = \frac{\sqrt{3}}{2}.$$

The angle whose sine equals $$\dfrac{\sqrt{3}}{2}$$ is $$60^\circ$$, because $$\sin 60^\circ = \dfrac{\sqrt{3}}{2}.$$ Hence

$$i = 60^\circ.$$

Therefore, the required angle of incidence is $$60^\circ$$.

Hence, the correct answer is Option C.