An object 6.0 mm is placed 16.5 cm to the left of the vertex of a concave spherical mirror having
a radius of curvature of 22.0 cm.
(a) Draw principal ray diagram showing formation of the image.
(b) Determine the position, size, orientation, and nature (real or virtual) of the image.

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the given data
- Object height (h) = 6.0 mm = 0.6 cm (convert to cm for consistency)
- Object distance (u) = -16.5 cm (negative because the object is on the same side as the incoming light)
- Radius of curvature (R) = 22.0 cm

### Step 2: Calculate the focal length (F)
The focal length (F) of a concave mirror is given by the formula:
\[ F = \frac{R}{2} \]
Substituting the given radius of curvature:
\[ F = \frac{22.0 \text{ cm}}{2} = 11.0 \text{ cm} \]

### Step 3: Use the mirror formula to find the image distance (v)
The mirror formula is:
\[ \frac{1}{F} = \frac{1}{v} + \frac{1}{u} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{F} - \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{v} = \frac{1}{11.0} - \frac{1}{-16.5} \]
Calculating:
\[ \frac{1}{v} = \frac{1}{11.0} + \frac{1}{16.5} \]
Finding a common denominator and calculating:
\[ \frac{1}{v} = \frac{16.5 + 11.0}{11.0 \times 16.5} = \frac{27.5}{181.5} \]
So,
\[ v = \frac{181.5}{27.5} \approx 6.6 \text{ cm} \]

### Step 4: Determine the nature of the image
Since the value of v is positive, the image is formed on the same side as the object, indicating that it is a real image.

### Step 5: Calculate the magnification (m)
The magnification (m) is given by:
\[ m = -\frac{v}{u} \]
Substituting the values:
\[ m = -\frac{6.6}{-16.5} = 0.4 \]

### Step 6: Determine the size of the image (h')
The size of the image can be calculated using:
\[ m = \frac{h'}{h} \]
Where h is the height of the object. Rearranging gives:
\[ h' = m \times h \]
Substituting the values:
\[ h' = 0.4 \times 0.6 \text{ cm} = 0.24 \text{ cm} \]

### Step 7: Summarize the results
- **Position of the image (v)**: 6.6 cm (to the left of the mirror)
- **Size of the image (h')**: 0.24 cm
- **Orientation**: Since the magnification is positive, the image is upright.
- **Nature of the image**: Real and inverted.