An object 9.0 mm tall is placed 12.0 cm to the left of the vertex of a convex spherical mirror
whose radius of curvature has a magnitude of 20.0 cm.
(a) Draw a principal ray diagram showing formation of the image.
(b) Determine the position, size, orientation, and nature (real or vertual) of the image.

Explanation

To solve the problem step by step, we will follow the instructions given in the question regarding the convex mirror.

### Step 1: Understand the Given Data
- Height of the object (h_o) = 9.0 mm
- Object distance (u) = -12.0 cm (negative because the object is placed in front of the mirror)
- Radius of curvature (R) = 20.0 cm
- Focal length (F) = R/2 = 20.0 cm / 2 = 10.0 cm (positive for convex mirror)

### Step 2: Draw the Ray Diagram
1. Draw a convex mirror and label its principal axis.
2. Mark the focal point (F) at +10 cm from the mirror's vertex.
3. Place the object (a vertical arrow representing the object) 12 cm to the left of the vertex (at -12 cm).
4. Draw two principal rays:
- **Ray 1**: A ray parallel to the principal axis that reflects and appears to diverge from the focal point.
- **Ray 2**: A ray directed towards the mirror's pole that reflects back along the same path.
5. Extend the reflected rays backward to find the point where they appear to converge (the image).

### Step 3: Use the Mirror Formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Where:
- \( f \) = focal length
- \( v \) = image distance
- \( u \) = object distance

Substituting the known values:
\[ \frac{1}{10} = \frac{1}{v} + \frac{1}{-12} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{10} + \frac{1}{12} \]
Finding a common denominator (60):
\[ \frac{1}{v} = \frac{6}{60} + \frac{5}{60} = \frac{11}{60} \]
Thus,
\[ v = \frac{60}{11} \approx 5.45 \text{ cm} \]

### Step 4: Determine the Magnification
The magnification (m) is given by:
\[ m = -\frac{v}{u} \]
Substituting the values:
\[ m = -\frac{60/11}{-12} = \frac{60}{132} = \frac{5}{11} \approx 0.45 \]

### Step 5: Analyze the Image Characteristics
1. **Position**: The image is located at \( v \approx 5.45 \) cm behind the mirror (positive value indicates virtual).
2. **Size**: The size of the image can be calculated using:
\[ h_i = m \cdot h_o = \frac{5}{11} \cdot 9 \text{ mm} \approx 4.09 \text{ mm} \]
3. **Orientation**: Since the magnification is positive, the image is erect.
4. **Nature**: The image is virtual because it is formed behind the mirror.

### Final Summary
- **Position of the image**: \( \approx 5.45 \) cm behind the mirror
- **Size of the image**: \( \approx 4.09 \) mm
- **Orientation**: Erect
- **Nature**: Virtual