A point source is situated at a distance `x lt f` from the pole of the concave mirror of focal length f. At time t=0 the point source starts moving away from the mirror with constant velocity. Which of the graphs below represents best, variation of image distance |v| with distance x between the pole of mirror and the source.

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to analyze the behavior of the image distance \( |v| \) as the object distance \( x \) changes when the point source moves away from the concave mirror. We will use the mirror formula and the properties of concave mirrors.

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a concave mirror with a focal length \( f \).
- The point source (object) is initially located at a distance \( x \) from the mirror, where \( x < f \). This means the object is located between the mirror's pole and its focus.

2. **Use the Mirror Formula**:
- The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
- Here, \( u \) is the object distance (which is negative for a concave mirror), and \( v \) is the image distance (which can be positive or negative depending on the position of the image).

3. **Determine the Initial Conditions**:
- Since the object is between the mirror's pole and focus, the object distance \( u = -x \) (negative because it's in front of the mirror).
- When the object is at this position, the image formed will be virtual and located behind the mirror.

4. **Calculate the Image Distance**:
- Rearranging the mirror formula gives us:
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \]
- Substituting \( u = -x \):
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{x} \]
- This implies:
\[ v = \frac{fx}{x + f} \]
- Since the image is virtual, we take the modulus \( |v| \).

5. **Analyze the Behavior as the Object Moves**:
- As the object moves away from the mirror (increasing \( x \)), we note that \( |v| \) will change.
- When \( x \) approaches \( f \), \( |v| \) approaches infinity (since the rays converge at the focus).
- When \( x \) is greater than \( f \) (the object moves beyond the focus), the image distance will start decreasing as the object distance increases further.

6. **Graphical Representation**:
- The graph of \( |v| \) vs. \( x \) will show that as \( x \) increases from a value less than \( f \) to \( f \), \( |v| \) will increase towards infinity.
- After \( x \) exceeds \( f \), \( |v| \) will decrease as the object moves further away from the mirror.

### Conclusion:
The graph that best represents the variation of image distance \( |v| \) with the distance \( x \) is one that starts from a finite value (as \( x \) approaches \( f \)) and goes to infinity, then decreases as \( x \) continues to increase beyond \( f \).