The internal surface of the walls of a sphere is specular. The radius of the sphere is `R=36cm.` A point source S is placed at a distance `R//2` from the cneter of the sphere and sends light to the remote part of the sphere. Where will the image of the source be after two successive reflections from the remote and then nearest wall of the sphere? How will the position of the image change if the source sends light to the nearest wall first?Consider paraxial rays.

Explanation

To solve the problem, we will analyze the situation step by step, considering the reflections from the walls of the sphere.

### Given:
- Radius of the sphere, \( R = 36 \, \text{cm} \)
- Distance of the point source \( S \) from the center of the sphere, \( d = \frac{R}{2} = 18 \, \text{cm} \)

### Step 1: Identify the positions
1. The center of the sphere is at point \( O \).
2. The source \( S \) is located at a distance of \( 18 \, \text{cm} \) from \( O \) towards one side of the sphere.

### Step 2: Determine the distances for the remote wall reflection
1. The distance from the source \( S \) to the remote wall of the sphere (the wall opposite to the source) is:
\[ d_{\text{remote}} = R - d = 36 \, \text{cm} - 18 \, \text{cm} = 18 \, \text{cm} \]
2. The object distance \( u \) for the remote wall (considering the sign convention) is:
\[ u = -d_{\text{remote}} = -18 \, \text{cm} \]

### Step 3: Calculate the image position after the first reflection (remote wall)
1. The focal length \( f \) of the concave mirror (remote wall) is:
\[ f = -\frac{R}{2} = -18 \, \text{cm} \]
2. Using the mirror formula:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
Substituting the values:
\[ \frac{1}{v} - \frac{1}{-18} = \frac{1}{-18} \]
This simplifies to:
\[ \frac{1}{v} + \frac{1}{18} = -\frac{1}{18} \]
\[ \frac{1}{v} = -\frac{2}{18} = -\frac{1}{9} \]
Therefore, the image distance \( v \) is:
\[ v = -9 \, \text{cm} \]
This means the image is located \( 9 \, \text{cm} \) from the center of the sphere towards the source.

### Step 4: Determine the distances for the nearest wall reflection
1. The distance from the image (after the first reflection) to the nearest wall is:
\[ d_{\text{nearest}} = R - |v| = 36 \, \text{cm} - 9 \, \text{cm} = 27 \, \text{cm} \]
2. The object distance \( u \) for the nearest wall (considering the sign convention) is:
\[ u = -d_{\text{nearest}} = -27 \, \text{cm} \]

### Step 5: Calculate the image position after the second reflection (nearest wall)
1. The focal length \( f \) of the nearest wall (concave mirror) is:
\[ f = -\frac{R}{2} = -18 \, \text{cm} \]
2. Using the mirror formula again:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
Substituting the values:
\[ \frac{1}{v} - \frac{1}{-27} = \frac{1}{-18} \]
This simplifies to:
\[ \frac{1}{v} + \frac{1}{27} = -\frac{1}{18} \]
\[ \frac{1}{v} = -\frac{1}{18} - \frac{1}{27} \]
Finding a common denominator (54):
\[ \frac{1}{v} = -\frac{3}{54} - \frac{2}{54} = -\frac{5}{54} \]
Therefore, the image distance \( v \) is:
\[ v = -\frac{54}{5} \approx -10.8 \, \text{cm} \]
This means the final image position is approximately \( 10.8 \, \text{cm} \) from the center of the sphere towards the source.

### Step 6: Change in image position if the source sends light to the nearest wall first
1. If the source sends light to the nearest wall first, the image will be formed at infinity since the source is at the focal point of the concave mirror.
2. The image will then reflect back to the remote wall, which will create an image at the center of the sphere.

### Summary of Results:
- **First Case (Remote Wall then Nearest Wall)**: Final image position is approximately \( 10.8 \, \text{cm} \) from the center towards the source.
- **Second Case (Nearest Wall then Remote Wall)**: Final image position is at the center of the sphere.