Find the nature of image when an object is placed at 2f from the pole of a convex mirror of focal length f

Options

A. Real, inverted and of same size
B. Virtual, erect and of `1/3` times
C. Virtual, erect and 3 times
D. Real, inverted and `1/3` times

Answer

To find the nature of the image when an object is placed at a distance of 2f from the pole of a convex mirror with focal length f, we can follow these steps:

### Step 1: Understand the Sign Convention
In a convex mirror:
- The focal length (f) is positive.
- The object distance (u) is negative since the object is placed on the same side as the incoming light.

Given that the object is placed at a distance of 2f from the pole of the mirror, we can write:
\[ u = -2f \]

### Step 2: Use the Mirror Formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
Where:
- \( f \) is the focal length,
- \( u \) is the object distance,
- \( v \) is the image distance.

Substituting the values into the formula:
\[ \frac{1}{f} = \frac{1}{-2f} + \frac{1}{v} \]

### Step 3: Solve for Image Distance (v)
Rearranging the equation to find \( \frac{1}{v} \):
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{2f} \]
Finding a common denominator:
\[ \frac{1}{v} = \frac{2}{2f} + \frac{1}{2f} = \frac{3}{2f} \]
Thus, we can find \( v \):
\[ v = \frac{2f}{3} \]

### Step 4: Determine the Nature of the Image
Since \( v \) is positive, this indicates that the image is virtual.

### Step 5: Calculate Magnification (M)
Magnification (M) is given by the formula:
\[ M = -\frac{v}{u} \]
Substituting the values:
\[ M = -\frac{\frac{2f}{3}}{-2f} = \frac{2f}{3 \cdot 2f} = \frac{1}{3} \]

### Step 6: Analyze the Magnification
Since the magnification \( M \) is positive, this indicates that the image is erect.

### Conclusion
The image formed is virtual, erect, and reduced in size (1/3 times the height of the object).

### Summary of Results
- **Nature of Image**: Virtual
- **Orientation**: Erect
- **Size**: Reduced (1/3 of the object height)