Inside a solid glass sphere of radius `R`, a point source of light lies at a distance `x(xltR)` from centre of the sphere. The solid is surrounded by air of refractive index 1. The maximum angle of incidence for rays incident on the spherical glass air interface directly from the point soure is:

Options

A. `"cos"^(-1)x/R`
B. `"sin"^(-1)x/R`
C. `"cos"^(-1)sqrt(x/R)`
D. `"sin"^(-1)sqrt(x/R)`

Explanation

To solve the problem, we need to find the maximum angle of incidence for rays incident on the spherical glass-air interface directly from a point source of light located inside the glass sphere.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a solid glass sphere with radius \( R \).
- A point source of light is located at a distance \( x \) from the center of the sphere, where \( x < R \).
- The sphere is surrounded by air, which has a refractive index of 1.

2. **Identifying the Geometry**:
- Let \( O \) be the center of the sphere.
- The point source of light is at point \( S \), located at a distance \( x \) from \( O \).
- The maximum angle of incidence occurs when the light rays are tangent to the sphere at the point of incidence.

3. **Drawing the Relevant Triangle**:
- Consider a ray of light traveling from point \( S \) to the point of incidence \( P \) on the surface of the sphere.
- Draw the radius \( OP \) to the point of incidence \( P \).
- The angle of incidence \( i \) is the angle between the incoming ray \( SP \) and the normal \( OP \).

4. **Using Geometry to Find the Angles**:
- The triangle formed by points \( O \), \( S \), and \( P \) can be analyzed.
- The length \( OP \) is equal to the radius \( R \).
- The length \( OS \) is equal to \( x \).
- We want to maximize the angle \( i \).

5. **Applying the Sine Rule**:
- By the sine rule in triangle \( OSP \):
\[ \frac{x}{\sin(i)} = \frac{R}{\sin(\alpha)} \]
- Here, \( \alpha \) is the angle \( OSP \).

6. **Maximizing the Angle**:
- To maximize \( i \), we need to consider when \( \alpha \) is at its maximum.
- The maximum value of \( \sin(\alpha) \) occurs when \( \alpha = 90^\circ \).
- Thus, \( \sin(\alpha) = 1 \).

7. **Finding the Maximum Angle of Incidence**:
- Substituting \( \alpha = 90^\circ \) into the sine rule gives:
\[ \sin(i) = \frac{x}{R} \]
- Therefore, the maximum angle of incidence \( i \) is:
\[ i_{\text{max}} = \sin^{-1}\left(\frac{x}{R}\right) \]

### Final Answer:
The maximum angle of incidence for rays incident on the spherical glass-air interface directly from the point source is:
\[ i_{\text{max}} = \sin^{-1}\left(\frac{x}{R}\right) \]