In case of a concave mirror, when we move the object from infinity to focus of the mirror (i.e., u decreases from `oo` to f), the image moves from focus to `oo`. So, the correct graph is shown by

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Answer

To solve the question regarding the behavior of the image formed by a concave mirror as the object moves from infinity to the focus, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Mirror Formula**:
The mirror formula for concave mirrors is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance.

2. **Identify the Movement of the Object**:
As the object moves from infinity to the focus, the object distance \( u \) decreases from \( -\infty \) (considering the convention that distances measured against the direction of incident light are negative) to \( -f \).

3. **Determine the Corresponding Image Distance**:
- When \( u \) is at \( -\infty \), substituting in the mirror formula gives:
\[ \frac{1}{f} = \frac{1}{v} + 0 \implies v = f \]
- When \( u \) is at \( -f \):
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{(-f)} \implies \frac{1}{v} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f} \implies v = \frac{f}{2} \]
- As \( u \) approaches \( -f \), \( v \) approaches \( -\infty \).

4. **Analyze the Graph**:
- As \( u \) decreases from \( -\infty \) to \( -f \), \( v \) moves from \( f \) to \( -\infty \).
- This indicates that the graph of \( v \) versus \( u \) will not be linear, and will instead be a curve.

5. **Identify the Correct Graph**:
- Since the relationship is not linear and \( v \) approaches \( -\infty \) as \( u \) approaches \( -f \), the correct graph will show a curve that starts at \( f \) when \( u \) is \( -\infty \) and moves towards \( -\infty \) as \( u \) approaches \( -f \).

### Conclusion:
The correct graph is the one that shows a curve starting from the focus and moving towards infinity as the object distance decreases from infinity to the focus.