A small air bubble is situated at a distance of 3 cm from the center of a glass sphere of radius 9 cm. When viewed from the nearest side, the bubble appears to be at a distance of 5 cm from the surface. Its apparent distance when viewed from the farthest side is `nxx5` cm where n is?

Answer

To solve the problem, we need to find the apparent distance of the air bubble when viewed from the farthest side of the glass sphere. Let's break down the solution step by step.

### Step 1: Understand the Geometry
1. The air bubble is located 3 cm from the center of a glass sphere with a radius of 9 cm.
2. The distance from the center of the sphere to the surface is 9 cm, and the distance from the bubble to the surface can be calculated as follows:
- Distance from the center to the bubble = 3 cm
- Distance from the center to the surface = 9 cm
- Therefore, the distance from the bubble to the surface = 9 cm - 3 cm = 6 cm.

### Step 2: Determine the Apparent Distance from the Nearest Side
1. When viewed from the nearest side, the bubble appears to be at a distance of 5 cm from the surface.
2. Since the distance from the bubble to the surface is 6 cm, the apparent distance can be represented as:
- Apparent distance = Actual distance - Distance from surface
- Thus, the apparent distance from the nearest side is 6 cm - 5 cm = 1 cm (this is the distance from the bubble to the nearest side).

### Step 3: Apply the Refraction Formula
1. We will use the formula for refraction at a spherical surface:
\[ \frac{n_2 - n_1}{R} = \frac{1}{v} - \frac{1}{u} \]
where:
- \( n_1 = 1 \) (refractive index of air),
- \( n_2 = \mu \) (refractive index of glass),
- \( R = 9 \) cm (radius of the sphere),
- \( u \) is the object distance, and
- \( v \) is the image distance.

### Step 4: Calculate the Refractive Index
1. For the nearest side:
- Object distance \( u = -6 \) cm (negative because it is on the same side as the incoming light).
- Image distance \( v = -5 \) cm (negative because the image is virtual).

Plugging into the formula:
\[ \frac{\mu - 1}{9} = \frac{1}{-5} - \frac{1}{-6} \]
Simplifying the right side:
\[ \frac{1}{-5} + \frac{1}{6} = \frac{-6 + 5}{30} = \frac{-1}{30} \]
Thus, we have:
\[ \frac{\mu - 1}{9} = -\frac{1}{30} \]
Cross-multiplying gives:
\[ 30(\mu - 1) = -9 \implies 30\mu - 30 = -9 \implies 30\mu = 21 \implies \mu = \frac{21}{30} = \frac{7}{10} \]

### Step 5: Calculate the Apparent Distance from the Farthest Side
1. Now, we need to find the apparent distance when viewed from the farthest side.
- Object distance \( u = 12 \) cm (since it is 3 cm from the center and the radius is 9 cm, total distance from the center to the farthest side is 9 cm + 3 cm = 12 cm).
- We need to find the new image distance \( v \).

Using the same refraction formula:
\[ \frac{\mu - 1}{9} = \frac{1}{v} - \frac{1}{12} \]
Substituting \( \mu = \frac{7}{10} \):
\[ \frac{\frac{7}{10} - 1}{9} = \frac{-3}{90} = -\frac{1}{30} \]
Thus:
\[ -\frac{1}{30} = \frac{1}{v} - \frac{1}{12} \]
Rearranging gives:
\[ \frac{1}{v} = -\frac{1}{30} + \frac{1}{12} \]
Finding a common denominator (60):
\[ \frac{1}{v} = -\frac{2}{60} + \frac{5}{60} = \frac{3}{60} = \frac{1}{20} \]
Therefore, \( v = 20 \) cm.

### Step 6: Relate the Apparent Distance to \( n \)
1. The problem states that the apparent distance from the farthest side is \( n \times 5 \) cm.
- We found \( v = 20 \) cm.
- Thus, \( n \times 5 = 20 \) implies \( n = \frac{20}{5} = 4 \).

### Final Answer
The value of \( n \) is **4**.