A

Step 1: Given the refractive index \( \mu = \sqrt{3} \) and the angle of incidence \( \theta_1 = 60^{\circ} \).
Step 2: Use Snell's law at the interface between air and glass for refraction:
\[ \mu_1 \sin(\theta_1) = \mu_2 \sin(\theta_2) \]
Here, \( \mu_1 = 1 \) (air) and \( \mu_2 = \sqrt{3} \).
\[ 1 \cdot \sin(60^{\circ}) = \sqrt{3} \cdot \sin(\theta_2) \]
Step 3: Calculate \( \sin(60^{\circ}) = \frac{\sqrt{3}}{2} \):
\[ \frac{\sqrt{3}}{2} = \sqrt{3} \cdot \sin(\theta_2) \]
Step 4: Rearranging gives:
\[ \sin(\theta_2) = \frac{1}{2} \]
Thus, \( \theta_2 = 30^{\circ} \).
Step 5: At the farther surface of the sphere, the angle of incidence is \( \theta_2 = 30^{\circ} \) and the angle of reflection (by the law of reflection) is also \( \theta_r = 30^{\circ} \).
Step 6: The angle between the reflected and refracted rays is given by:
\[ \theta_{r} + \theta_{t} = 30^{\circ} + \theta_{3} \] where \( \theta_{t} \) is the angle of refraction into the external medium (assumed to be air).
Step 7: Now, calculating the angle of refraction back to air using Snell's Law again gives us \( \theta_{3} = \text{arcsin}(\frac{1}{\sqrt{3}}) \approx 60^{\circ} \).
The angle between the reflected ray (30 degrees) and the refracted ray (60 degrees) is:
\[ 60^{\circ} - 30^{\circ} = 30^{\circ} \]
Therefore, the total angle between the reflected and refracted rays is \( 30^{\circ} + 30^{\circ} = 60^{\circ} \).
Thus the final answer is 60 degrees.