To find the angle of minimum deviation for an equilateral prism, we need to understand the relationship between the prism's angle and its refractive index. Here are the steps involved in solving this problem:
- Given that the refractive index \(n\) of the material of the prism is \(\sqrt{3}\).
- An equilateral prism has an angle \(A\) of \(60^{\circ}\), since all its angles are equal and sum up to \(180^{\circ}\).
- For a prism, the relationship between the angle of minimum deviation \(\delta_m\), the refractive index \(n\), and the angle \(A\) of the prism is given by the formula: \(n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\)
- Substituting the known values:
- \(n = \sqrt{3}\)
- \(A = 60^{\circ}\)
- Solve for \(\frac{A}{2}\), we get \(\frac{60}{2} = 30^{\circ}\). Thus, \(\sin(30^{\circ}) = \frac{1}{2}\).
- Substitute into the equation: \(\sqrt{3} = \frac{\sin\left(\frac{60 + \delta_m}{2}\right)}{\frac{1}{2}}\)
- Simplifying gives: \(2\sqrt{3} = \sin\left(\frac{60 + \delta_m}{2}\right)\)
- For \(\sin\theta = 2\sqrt{3}\), the angle \(\theta = 60^{\circ}\), so: \(\frac{60 + \delta_m}{2} = 60^{\circ}\)
- Solving for \(\delta_m\), we find: \(\delta_m = 60^{\circ}\)
Thus, the angle of minimum deviation of the prism is \(60^{\circ}\). Therefore, the correct answer is 60o.