A ray of light is falling on a glass sphere of `mu=sqrt(3)` such that the incident ray and the emergent ray, when produced, intersect at a point on the surface of the sphere. Find the value of angle of incidence.

Explanation

To solve the problem, we need to find the angle of incidence (I) for a ray of light falling on a glass sphere with a refractive index (μ) of \(\sqrt{3}\). The condition given is that the incident ray and the emergent ray intersect at a point on the surface of the sphere when produced.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let point A be the point where the ray of light strikes the surface of the sphere.
- OA is the normal to the surface at point A.
- The angle of incidence (I) is the angle between the incident ray and the normal OA.

2. **Refraction at Point A**:
- When the ray enters the sphere, it refracts at point A, making an angle of refraction (R) with the normal.
- According to Snell's Law:
\[ \mu_1 \sin I = \mu_2 \sin R \]
where \(\mu_1 = 1\) (for air) and \(\mu_2 = \sqrt{3}\) (for glass). Thus:
\[ \sin I = \sqrt{3} \sin R \]

3. **Emergence at Point B**:
- The ray travels through the sphere and emerges at point B, making an angle of emergence (E) with the normal at point B.
- Given that the angle of incidence equals the angle of emergence (I = E), we can denote the angle of emergence as E.

4. **Isosceles Triangle OAB**:
- In triangle OAB, OA = OB (radii of the sphere).
- Therefore, angles OAB and OBA are equal. Let both be R.
- The angle AOB can be expressed as:
\[ \text{Angle AOB} = 180^\circ - 2R \]

5. **Angles at Point D**:
- When the incident ray and emergent ray are produced, they intersect at point D on the sphere.
- By vertically opposite angles, we have:
\[ \text{Angle DAO} = I \quad \text{and} \quad \text{Angle DBA} = E = I \]
- Therefore, angle ADB can be expressed as:
\[ \text{Angle ADB} = I - R \]

6. **Cyclic Quadrilateral ADBE**:
- The angles in cyclic quadrilateral ADBE satisfy:
\[ \text{Angle ADB} + \text{Angle AEB} = 180^\circ \]
- We can express angle AEB as:
\[ \text{Angle AEB} = \frac{1}{2} \text{Angle AOB} = \frac{1}{2}(180^\circ - 2R) = 90^\circ - R \]
- Thus:
\[ (I - R) + (90^\circ - R) = 180^\circ \]
- Simplifying gives:
\[ I - 2R + 90^\circ = 180^\circ \implies I - 2R = 90^\circ \implies I = 90^\circ + 2R \]

7. **Substituting into Snell's Law**:
- From Snell's Law, we have:
\[ \sin I = \sqrt{3} \sin R \]
- Substituting \(I = 90^\circ + 2R\) into the sine function gives:
\[ \sin(90^\circ + 2R) = \cos(2R) \]
- Thus:
\[ \cos(2R) = \sqrt{3} \sin R \]

8. **Using Trigonometric Identities**:
- We can express \(\sin R\) in terms of \(\cos R\):
\[ \sin R = \sqrt{1 - \cos^2 R} \]
- Substitute this into the equation to solve for R.

9. **Finding the Angle of Incidence**:
- After solving the equations, we find:
\[ I = 60^\circ \]

### Final Answer:
The angle of incidence \(I\) is \(60^\circ\).