To solve this problem, we need to find the refractive index of the surface. Given that the angle of incidence \( i = 60^\circ \) and the reflected and refracted rays are perpendicular to each other, we will use Snell's law and the concept of perpendicular rays.
- First, note the condition that the reflected and refracted rays are perpendicular. This implies that the angle between the reflected ray and refracted ray is \( 90^\circ \).
- The reflection rule states that the angle of incidence is equal to the angle of reflection. Therefore, the angle of reflection \( r = 60^\circ \).
- For the refracted ray, let the angle of refraction be \( r' \). We are given that: \(r + r' = 90^\circ\) Since \( r = 60^\circ \), we have: \(60^\circ + r' = 90^\circ \Rightarrow r' = 30^\circ\).
- Now, apply Snell's Law: \(n_1 \sin i = n_2 \sin r'\) where \( n_1 \) is the refractive index of air, generally 1, and \( n_2 \) is the refractive index of the surface. Substituting the values, we get: \(\sin 60^\circ = n_2 \sin 30^\circ\)
- Use the trigonometric values: \(\sin 60^\circ = \frac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \frac{1}{2}\) Hence, \(\frac{\sqrt{3}}{2} = n_2 \cdot \frac{1}{2}\)
- Solve for \( n_2 \): \(n_2 = \sqrt{3}\)
Thus, the refractive index of the surface is \(\sqrt{3}\). Hence, the correct answer is \(\sqrt{3}\). This corresponds to the correct option given in the question.