An object is moving away from concave mirror of focal length f starting from focus. The distance of an object from pole of mirror is x. The correct graph of magnitude of magnification(m) verses distance x is:

Options

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Answer

To solve the problem of finding the correct graph of the magnitude of magnification (m) versus the distance (x) for an object moving away from a concave mirror, we can follow these steps:

### Step 1: Understand the Mirror Formula
The mirror formula for a concave mirror is given by:
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]
where:
- \( v \) = image distance
- \( u \) = object distance (negative for concave mirrors)
- \( f \) = focal length (negative for concave mirrors)

### Step 2: Set the Initial Conditions
The object starts at the focus of the mirror, so:
- At the focus, \( u = -f \) (since object distances are taken as negative in mirror convention).
- Therefore, at this point, we can calculate the image distance \( v \).

### Step 3: Calculate Magnification at the Focus
The magnification \( m \) is defined as:
\[ m = -\frac{v}{u} \]
At \( u = -f \):
- From the mirror formula, substituting \( u = -f \):
\[ \frac{1}{v} + \frac{1}{-f} = \frac{1}{f} \]
This simplifies to:
\[ \frac{1}{v} = \frac{2}{f} \implies v = \frac{f}{2} \]
Now substituting \( v \) into the magnification formula:
\[ m = -\frac{\frac{f}{2}}{-f} = \frac{1}{2} \]
However, as the object is at the focus, the image distance approaches infinity, leading to infinite magnification.

### Step 4: Calculate Magnification at \( x = 2f \)
Now, let’s consider when \( x = 2f \):
- Here, \( u = -2f \).
- Using the mirror formula:
\[ \frac{1}{v} + \frac{1}{-2f} = \frac{1}{f} \]
This simplifies to:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{2f} = \frac{3}{2f} \implies v = \frac{2f}{3} \]
Now substituting \( v \) into the magnification formula:
\[ m = -\frac{\frac{2f}{3}}{-2f} = \frac{1}{3} \]

### Step 5: Analyze the Behavior of Magnification as \( x \) Increases
As the object moves further away from the mirror (increasing \( x \)):
- The image distance \( v \) will decrease in magnitude.
- Consequently, the magnification \( m \) will approach zero but remain negative, indicating that the image is inverted and diminishing in size.

### Step 6: Determine the Graph Characteristics
- At \( x = f \), \( m \) is infinite.
- At \( x = 2f \), \( m = -1 \).
- As \( x \) increases beyond \( 2f \), \( m \) approaches zero.

### Conclusion
The correct graph of the magnitude of magnification \( |m| \) versus distance \( x \) will show:
- A vertical asymptote at \( x = f \) (where \( m \) is infinite).
- A value of \( |m| = 1 \) at \( x = 2f \).
- A decreasing curve approaching zero as \( x \) increases.