An object is placed at a distance of `f/2` from a convex lens of focal length f. The image will be

Options

A. At one of the foci, virtual and double its size
B. Is greater than 1.5 but less than 2.0
C. At 2f , virtual and erect
D. None of the above

Answer

To solve the problem, we will use the lens formula and the concept of magnification for a convex lens.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Focal length of the convex lens, \( f \)
- Object distance, \( u = -\frac{f}{2} \) (the object distance is taken as negative in lens formula convention)

2. **Use the Lens Formula:**
The lens formula is given by:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
Here, \( v \) is the image distance.

3. **Substitute the Values:**
Substitute \( u = -\frac{f}{2} \) into the lens formula:
\[ \frac{1}{v} - \left(-\frac{2}{f}\right) = \frac{1}{f} \]
This simplifies to:
\[ \frac{1}{v} + \frac{2}{f} = \frac{1}{f} \]

4. **Rearranging the Equation:**
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{f} - \frac{2}{f} = -\frac{1}{f} \]

5. **Calculate the Image Distance \( v \):**
Taking the reciprocal:
\[ v = -f \]

6. **Determine the Nature of the Image:**
The negative sign indicates that the image is formed on the same side as the object, which means it is a virtual image.

7. **Calculate the Magnification:**
The magnification \( m \) is given by:
\[ m = \frac{v}{u} \]
Substituting the values:
\[ m = \frac{-f}{-\frac{f}{2}} = 2 \]
This indicates that the image is double the size of the object.

### Conclusion:
The image formed is virtual, located at a distance of \( -f \) from the lens, and is double the size of the object.