A dust particle is inside a sphere of refractive index `4/3.` If the dust particle is `10.0 cm` from the wall of the `15.0 cm` radius bowl, where does it appear to an observer outside the bowl.

Explanation

To solve the problem of where the dust particle appears to an observer outside the bowl, we will use the refraction formula for a spherical surface. Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Refractive index of the medium (bowl) \( \mu_2 = \frac{4}{3} \)
- Refractive index of air (outside the bowl) \( \mu_1 = 1 \)
- Distance of the dust particle from the wall of the bowl \( d = 10.0 \, \text{cm} \)
- Radius of the bowl \( R = 15.0 \, \text{cm} \)

2. **Calculate the Distance from the Center of the Bowl:**
- The distance of the dust particle from the center of the bowl is given by:
\[ U = R - d = 15.0 \, \text{cm} - 10.0 \, \text{cm} = 5.0 \, \text{cm} \]
- Since the object (dust particle) is inside the bowl, we take \( U \) as negative:
\[ U = -5.0 \, \text{cm} \]

3. **Apply the Spherical Surface Refraction Formula:**
- The formula for refraction at a spherical surface is given by:
\[ \frac{\mu_2}{V} - \frac{\mu_1}{U} = \frac{\mu_2 - \mu_1}{R} \]
- Substituting the known values:
\[ \frac{\frac{4}{3}}{V} - \frac{1}{-5} = \frac{\frac{4}{3} - 1}{-15} \]

4. **Simplify the Equation:**
- Calculate \( \mu_2 - \mu_1 \):
\[ \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3} \]
- Substitute this back into the equation:
\[ \frac{\frac{4}{3}}{V} + \frac{1}{5} = \frac{\frac{1}{3}}{-15} \]
- Simplifying the right side:
\[ \frac{1}{3} \div -15 = -\frac{1}{45} \]
- Now the equation becomes:
\[ \frac{\frac{4}{3}}{V} + \frac{1}{5} = -\frac{1}{45} \]

5. **Solve for \( V \):**
- Rearranging gives:
\[ \frac{\frac{4}{3}}{V} = -\frac{1}{45} - \frac{1}{5} \]
- Finding a common denominator for the right side (which is 45):
\[ -\frac{1}{45} - \frac{9}{45} = -\frac{10}{45} = -\frac{2}{9} \]
- Thus:
\[ \frac{\frac{4}{3}}{V} = -\frac{2}{9} \]
- Cross-multiplying gives:
\[ 4 \cdot 9 = -2 \cdot 3V \implies 36 = -6V \implies V = -6 \, \text{cm} \]

6. **Interpret the Result:**
- The negative sign indicates that the image appears on the same side as the object, which means it appears to be located \( 6 \, \text{cm} \) from the center of the bowl towards the outside.

### Final Answer:
The dust particle appears to be located \( 6 \, \text{cm} \) from the center of the bowl, which means it appears to be \( 9 \, \text{cm} \) from the wall of the bowl.