A concave mirror has a radius of curvature of 24 cm. How far is
an object from the mirror if an image is formed that is:
(a) virtual and 3.0 times the size of the object,
(b) real and 3.0 times the size of the object and
(c) real and `1//3` the size of the object?.

Explanation

To solve the problem, we will use the mirror formula and the magnification formula. The mirror formula is given by:

\[ \frac{1}{V} + \frac{1}{U} = \frac{1}{F} \]

Where:
- \( V \) = image distance
- \( U \) = object distance
- \( F \) = focal length

For a concave mirror, the focal length \( F \) is negative and is given by:

\[ F = -\frac{R}{2} \]

Where \( R \) is the radius of curvature. Given that \( R = 24 \, \text{cm} \), we find:

\[ F = -\frac{24}{2} = -12 \, \text{cm} \]

Now, we will solve for each case:

### (a) Virtual Image, Magnification = 3

1. **Use the magnification formula**:
\[ M = -\frac{V}{U} \]
For a virtual image, \( M \) is positive, so:
\[ 3 = -\frac{V}{U} \implies V = -3U \]

2. **Substitute \( V \) in the mirror formula**:
\[ \frac{1}{-3U} + \frac{1}{U} = \frac{1}{-12} \]

3. **Solve the equation**:
\[ \frac{-1 + 3}{3U} = \frac{1}{-12} \]
\[ \frac{2}{3U} = \frac{-1}{12} \]
Cross-multiplying gives:
\[ 2 \cdot (-12) = -3U \implies -24 = -3U \implies U = 8 \, \text{cm} \]

### (b) Real Image, Magnification = 3

1. **Use the magnification formula**:
\[ M = -\frac{V}{U} = -3 \implies V = -3U \]

2. **Substitute \( V \) in the mirror formula**:
\[ \frac{1}{-3U} + \frac{1}{U} = \frac{1}{-12} \]

3. **Solve the equation**:
\[ \frac{-1 + 3}{3U} = \frac{1}{-12} \]
\[ \frac{2}{3U} = \frac{-1}{12} \]
Cross-multiplying gives:
\[ 2 \cdot (-12) = -3U \implies -24 = -3U \implies U = 8 \, \text{cm} \]

### (c) Real Image, Magnification = 1/3

1. **Use the magnification formula**:
\[ M = -\frac{V}{U} = -\frac{1}{3} \implies V = -\frac{1}{3}U \]

2. **Substitute \( V \) in the mirror formula**:
\[ \frac{1}{-\frac{1}{3}U} + \frac{1}{U} = \frac{1}{-12} \]

3. **Solve the equation**:
\[ \frac{-3 + 1}{U} = \frac{1}{-12} \]
\[ \frac{-2}{U} = \frac{1}{-12} \]
Cross-multiplying gives:
\[ -2 \cdot (-12) = U \implies U = 24 \, \text{cm} \]

### Summary of Results:
- (a) Object distance \( U = 8 \, \text{cm} \) (virtual image)
- (b) Object distance \( U = 8 \, \text{cm} \) (real image)
- (c) Object distance \( U = 24 \, \text{cm} \) (real image)