An obejct is placed at a distance `2f` from the pole of a convex mirror of focal length `f`. The linear magnification is

Options

A. `(1)/(3)`
B. `(2)/(3)`
C. `(3)/(4)`
D. 1

Explanation

To find the linear magnification of an object placed at a distance of `2f` from the pole of a convex mirror with a focal length of `f`, we can follow these steps:

### Step 1: Identify the given values
- Focal length of the convex mirror, \( f \) (positive value)
- Object distance, \( u = -2f \) (negative because the object is in front of the mirror)

### Step 2: Use the mirror formula
The mirror formula for a convex mirror is given by:
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]
where:
- \( v \) is the image distance,
- \( u \) is the object distance,
- \( f \) is the focal length.

### Step 3: Substitute the known values into the mirror formula
Substituting \( u = -2f \) and \( f = f \):
\[ \frac{1}{v} + \frac{1}{-2f} = \frac{1}{f} \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ \frac{1}{v} - \frac{1}{2f} = \frac{1}{f} \]

### Step 5: Solve for \( \frac{1}{v} \)
Adding \( \frac{1}{2f} \) to both sides:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{2f} \]
Finding a common denominator (which is \( 2f \)):
\[ \frac{1}{v} = \frac{2}{2f} + \frac{1}{2f} = \frac{3}{2f} \]

### Step 6: Find \( v \)
Taking the reciprocal gives:
\[ v = \frac{2f}{3} \]

### Step 7: Calculate the linear magnification \( m \)
The linear magnification \( m \) is given by the formula:
\[ m = -\frac{v}{u} \]
Substituting the values of \( v \) and \( u \):
\[ m = -\frac{\frac{2f}{3}}{-2f} = \frac{2f}{3} \cdot \frac{1}{2f} = \frac{1}{3} \]

### Final Answer
The linear magnification is \( \frac{1}{3} \).

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