When an small object is placed at a distance `x_(1)` and `x_(2)` from a lens on its principal axis, then real image and a virtual image are formed respectively having same magnitude of transverse magnification. Then the focal length of the lens is :

Options

A. `x_(1) -x_(2)`
B. `(x_(1)-x_(2))/(2)`
C. `(x_(1) +x_(2))/(2)`
D. `x_(1) +x_(2)`

Explanation

To solve the problem, we need to analyze the situation involving a lens and the formation of images. We have two distances from the lens, \( x_1 \) and \( x_2 \), where a real image and a virtual image are formed, respectively. Both images have the same magnitude of transverse magnification.

### Step-by-Step Solution:

1. **Understanding Magnification**:
- For a real image formed by a lens, the magnification \( m_1 \) is given by:
\[ m_1 = \frac{h'}{h} = -\frac{v}{u_1} \]
where \( v \) is the image distance, \( u_1 \) is the object distance (which is \( x_1 \)), and \( h' \) and \( h \) are the heights of the image and object, respectively.

- For a virtual image, the magnification \( m_2 \) is given by:
\[ m_2 = \frac{h'}{h} = \frac{v'}{u_2} \]
where \( v' \) is the image distance for the virtual image and \( u_2 \) is the object distance (which is \( x_2 \)).

2. **Setting Up the Equations**:
- Since both images have the same magnitude of magnification, we can write:
\[ |m_1| = |m_2| \]
- This leads to:
\[ \frac{v}{x_1} = \frac{v'}{x_2} \]

3. **Using the Lens Formula**:
- The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
- For the real image:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{x_1} \quad \text{(1)} \]
- For the virtual image:
\[ \frac{1}{f} = \frac{1}{v'} - \frac{1}{x_2} \quad \text{(2)} \]

4. **Equating the Two Expressions for Focal Length**:
- From equations (1) and (2), we can set them equal to each other:
\[ \frac{1}{v} - \frac{1}{x_1} = \frac{1}{v'} - \frac{1}{x_2} \]

5. **Eliminating \( v \) and \( v' \)**:
- Rearranging gives:
\[ \frac{1}{v} + \frac{1}{x_2} = \frac{1}{v'} + \frac{1}{x_1} \]
- We can express \( v \) and \( v' \) in terms of \( x_1 \) and \( x_2 \) using the magnification relationships.

6. **Finding the Focal Length**:
- After eliminating magnification \( m \) from the equations, we find:
\[ f = \frac{x_1 + x_2}{2} \]

### Final Answer:
The focal length of the lens is:
\[ f = \frac{x_1 + x_2}{2} \]