Assume that a certain spherical mirror has a focal length of -10.0 cm. Locate and describe the
image for object distances of (a) 25.0 cm (b) 10.0 cm (c) 5.0 cm.

Explanation

To solve the problem, we will use the mirror formula, which is given by:

\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]

where:
- \( f \) is the focal length of the mirror,
- \( u \) is the object distance (always negative for real objects),
- \( v \) is the image distance.

Given that the focal length \( f = -10.0 \, \text{cm} \), we will analyze three different object distances: \( u = -25.0 \, \text{cm} \), \( u = -10.0 \, \text{cm} \), and \( u = -5.0 \, \text{cm} \).

### Step-by-Step Solution

#### Part (a): Object distance \( u = -25.0 \, \text{cm} \)

1. **Substitute values into the mirror formula**:
\[ \frac{1}{-10} = \frac{1}{-25} + \frac{1}{v} \]

2. **Rearranging the equation**:
\[ \frac{1}{v} = \frac{1}{-10} - \frac{1}{-25} \]

3. **Finding a common denominator (LCM of 10 and 25 is 50)**:
\[ \frac{1}{v} = -\frac{5}{50} + \frac{2}{50} = -\frac{3}{50} \]

4. **Calculating \( v \)**:
\[ v = -\frac{50}{3} \, \text{cm} \approx -16.67 \, \text{cm} \]

5. **Interpretation of the result**:
The negative sign indicates that the image is formed on the same side as the object, and since \( |v| \) is less than \( |u| \), the image is virtual and upright.

#### Part (b): Object distance \( u = -10.0 \, \text{cm} \)

1. **Substitute values into the mirror formula**:
\[ \frac{1}{-10} = \frac{1}{-10} + \frac{1}{v} \]

2. **Rearranging the equation**:
\[ \frac{1}{v} = \frac{1}{-10} - \frac{1}{-10} = 0 \]

3. **Calculating \( v \)**:
\[ v = \infty \]

4. **Interpretation of the result**:
The image is formed at infinity, indicating that the rays are parallel after reflection, and thus the image is highly diminished and cannot be captured on a screen.

#### Part (c): Object distance \( u = -5.0 \, \text{cm} \)

1. **Substitute values into the mirror formula**:
\[ \frac{1}{-10} = \frac{1}{-5} + \frac{1}{v} \]

2. **Rearranging the equation**:
\[ \frac{1}{v} = \frac{1}{-10} - \frac{1}{-5} \]

3. **Finding a common denominator (LCM of 10 and 5 is 10)**:
\[ \frac{1}{v} = -\frac{1}{10} + \frac{2}{10} = \frac{1}{10} \]

4. **Calculating \( v \)**:
\[ v = 10 \, \text{cm} \]

5. **Interpretation of the result**:
The positive sign indicates that the image is formed on the opposite side of the mirror, and since \( |v| \) is equal to \( |u| \), the image is real and inverted.

### Summary of Results
- For \( u = -25.0 \, \text{cm} \): Image at \( v \approx -16.67 \, \text{cm} \) (virtual, upright).
- For \( u = -10.0 \, \text{cm} \): Image at \( v = \infty \) (image at infinity).
- For \( u = -5.0 \, \text{cm} \): Image at \( v = 10.0 \, \text{cm} \) (real, inverted).