For a prism of refractive index \(\sqrt{3}\), the angle of minimum deviation is equitation is equal to the angle of prism, then angle of the prism is (A) \(60^{\circ}\) (B) \(90^{\circ}\) (C) \(45^{\circ}\) (D) \(180^{\circ}\)

Snell's law states that the ratio of the sines of the angles of incidence and refraction is equal to the refractive index. Let's denote the angle of incidence as \(i_1\) and the angle of refraction as \(r_1\): \[\frac{\sin{ i_1 }}{\sin{ r_1 }} = \sqrt{3}\]

The second refraction occurs when the light ray exits the prism. Let's denote the angle of incidence as \(i_2\) and the angle of refraction as \(r_2\): \[\frac{\sin{ i_2 }}{\sin{ r_2 }} = \frac{1}{\sqrt{3}}\]

We now combine the two Snell's law equations. Since the angle of minimum deviation is equal to the angle of the prism, we can use the following relationships: \(i_1 + i_2 = A + \delta\), \(r_1 + r_2 = A\), and since the angle formed by the refracted rays inside the prism at the minimum deviation is supplementary to the angle of the prism, \(r_1 = i_2\). Substitute \(i_2 = r_1\) in the equation: \[\frac{\sin{ i_1 }}{\sin{ r_1 }} = \frac{\sin{ r_1 }}{\sin{ r_2 }}\]

To find the angle of the prism, we need to solve the equation above for \(r_1\): \[\sin^2{ r_1} = \sin{ i_1 } \sin{ r_2 }\] Using the fact that \(\sin^2{ u } =\frac{1- \cos{ 2u }}{2}\), we rewrite the equation: \[\frac{1- \cos{ 2r_1 }}{2} = \frac{\sin{ i_1 } ( 1- \cos{ 2r_1 })}{2}\] \[\cos{ 2r_1 } = \sin{ i_1 } - 1\] Now, for the minimum deviation condition to hold, we have \(r_1 = A\). Thus, \[\cos{ 2A } = \sin{ i_1 } - 1\] Since we know \(\sin^2{ i_1} = 3\sin^2{ A }\), we can replace \(\sin^2{ i_1 }\) with \(3\sin^2{ A }\) and re-write the equation: \[\cos{ 2A } = \sqrt{ 3\sin^2{ A }} - 1\] \[\cos{ 2A } = \sqrt{ 3(1- \cos^2{ A })} - 1\] Solving this equation, we find the angle of the prism A: \(A = 60^{\circ}\) So, the correct answer is: (A) \(60^{\circ}\)