In the above question, if the rays were to converge between F and C of mirror, then find the nature of final image formed

Options

A. real
B. virtual
C. erect
D. inverted

Explanation

To solve the problem of determining the nature of the final image formed when rays converge between the focal point (F) and the center of curvature (C) of a mirror, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a concave mirror with a focal point (F) and a center of curvature (C).
- The rays converge at a point between F and C.

2. **Using the Mirror Formula**:
- The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
- Rearranging gives:
\[ v = \frac{u \cdot f}{u - f} \]
- Here, \(u\) is the object distance (which is negative for real objects in mirror conventions), and \(f\) is the focal length (negative for concave mirrors).

3. **Identifying the Sign of Distances**:
- Since the rays converge between F and C, we know that \(u\) (the object distance) is greater than \(f\) (the focal length) but less than \(C\) (the center of curvature).
- Thus, \(u < C\) and \(u > f\).

4. **Calculating the Image Distance (v)**:
- Since \(u\) is greater than \(f\), substituting into the rearranged mirror formula shows that \(v\) will be greater than 0 (indicating a real image).
- This means that the image is formed on the same side as the object.

5. **Determining the Nature of the Image**:
- Since \(v > 0\), the image is real and inverted.
- The image is also formed between F and C, which indicates it is larger than the object.

6. **Calculating the Magnification (M)**:
- The magnification formula for mirrors is given by:
\[ M = -\frac{v}{u} \]
- Since \(v > 0\) and \(u < 0\) (as per mirror conventions), the magnification \(M\) will be negative, indicating that the image is inverted.

### Conclusion:
- The final image formed is **real and inverted**.