1. Refraction at the first surface

A ray enters the glass sphere from air.

• Refractive index of air: $n_1=1$
• Refractive index of glass: $n_2=\mu=\sqrt{3}$
• Angle of incidence at first surface: $i=60^\circ$

Using Snell’s law: $n_1\sin i=n_2\sin r$ $1\cdot \sin 60^\circ=\sqrt{3}\sin r$ $\frac{\sqrt{3}}{2}=\sqrt{3}\sin r$ $\sin r=\frac{1}{2}$ So, $r=30^\circ$

Thus, inside the sphere, the ray makes an angle $30^\circ$ with the normal at the first surface.

2. Angle of incidence at the second surface

For a sphere, the normals at the two points of intersection are along the radii. A standard result for a ray inside a sphere is that the angle the ray makes with the normal at the second surface is the same as the refracted angle at the first surface.

Hence, angle of incidence at the farther surface is: $i_2=30^\circ$

3. Reflection and refraction at the farther surface

At the second surface, the ray goes from glass $(\sqrt{3})$ to air $(1)$.

Using Snell’s law again: $\sqrt{3}\sin 30^\circ = 1\cdot \sin r_2$ $\sqrt{3}\cdot \frac{1}{2}=\sin r_2$ $\sin r_2=\frac{\sqrt{3}}{2}$ So, $r_2=60^\circ$

The reflected ray makes the same angle with the normal as the incident ray: $\text{angle of reflection}=30^\circ$

4. Angle between reflected and refracted rays

At the second surface:

• reflected ray makes $30^\circ$ with the normal inside the glass
• refracted ray makes $60^\circ$ with the normal outside the glass

These two rays lie on opposite sides of the normal, so the angle between them is: $30^\circ + 60^\circ = 90^\circ$

5. Final answer

$\boxed{90}$