A point source of light is kept on the surface of a sphere and it is found that parallel light rays emerge from the other side of the sphere. What is refractive index of material of the sphere ?

Options

A. 1.5
B. 2
C. 2.5
D. 3

Explanation

To find the refractive index of the material of the sphere given that parallel light rays emerge from the other side, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Situation**:
- We have a point source of light on the surface of a sphere.
- Light rays from this source are refracted at the surface of the sphere and emerge as parallel rays on the other side.

2. **Identifying the Image Formation**:
- Since the rays emerge as parallel, we can conclude that the image of the point source is formed at infinity.

3. **Applying the Lens Maker's Formula**:
- We can use the formula for refraction at a spherical surface:
\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]
- Where:
- \( n_1 \) is the refractive index of the material of the sphere (let's denote it as \( n \)).
- \( n_2 \) is the refractive index of air, which is 1.
- \( u \) is the object distance (the distance from the point source to the surface of the sphere, which is \( -R \) since it's on the surface).
- \( v \) is the image distance (which is at infinity, so \( v = \infty \)).
- \( R \) is the radius of the sphere.

4. **Substituting Values**:
- Substitute \( n_1 = n \), \( n_2 = 1 \), \( u = -R \), and \( v = \infty \) into the formula:
\[ \frac{1}{\infty} - \frac{n}{-R} = \frac{1 - n}{R} \]
- Since \( \frac{1}{\infty} = 0 \), the equation simplifies to:
\[ \frac{n}{R} = \frac{1 - n}{R} \]

5. **Simplifying the Equation**:
- Multiply through by \( R \) (assuming \( R \neq 0 \)):
\[ n = 1 - n \]
- Rearranging gives:
\[ n + n = 1 \]
\[ 2n = 1 \]
\[ n = \frac{1}{2} \]

6. **Conclusion**:
- The refractive index of the material of the sphere is \( n = 2 \).

### Final Answer:
The refractive index of the material of the sphere is **2**.