When an object is at distances x and y from a lens, a real image and a virtual image is formed respectively having same magnification. The focal length of the lens is

Options

A. `(x+y)/(2)`
B. `x-y`
C. `sqrt(x+y)`
D. `x+y`

Answer

To solve the problem step by step, we need to analyze the situation where an object is placed at two different distances from a lens, resulting in a real image and a virtual image with the same magnification.

### Step-by-Step Solution:

1. **Understand the Problem**:
- We have an object placed at two distances, \( x \) and \( y \), from a lens.
- At distance \( x \), a real image is formed.
- At distance \( y \), a virtual image is formed.
- Both images have the same magnification.

2. **Magnification Formula**:
- The magnification \( m \) for a lens is given by:
\[ m = \frac{v}{u} \]
- Where \( v \) is the image distance and \( u \) is the object distance.

3. **Real Image Case**:
- For the real image formed at distance \( x \):
- The object distance \( u_1 = -x \) (negative as per sign convention).
- The image distance \( v_1 \) is positive.
- Thus, the magnification is:
\[ m = \frac{v_1}{-x} \implies v_1 = -mx \]

4. **Virtual Image Case**:
- For the virtual image formed at distance \( y \):
- The object distance \( u_2 = -y \).
- The image distance \( v_2 \) is negative.
- Thus, the magnification is:
\[ m = \frac{v_2}{-y} \implies v_2 = -my \]

5. **Lens Formula**:
- According to the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

6. **Applying Lens Formula for Real Image**:
- For the real image at distance \( x \):
\[ \frac{1}{f} = \frac{1}{v_1} - \frac{1}{-x} = \frac{1}{v_1} + \frac{1}{x} \]
- Substituting \( v_1 = -mx \):
\[ \frac{1}{f} = \frac{1}{-mx} + \frac{1}{x} \]
- Simplifying:
\[ \frac{1}{f} = \frac{-1 + m}{mx} \]

7. **Applying Lens Formula for Virtual Image**:
- For the virtual image at distance \( y \):
\[ \frac{1}{f} = \frac{1}{v_2} - \frac{1}{-y} = \frac{1}{v_2} + \frac{1}{y} \]
- Substituting \( v_2 = -my \):
\[ \frac{1}{f} = \frac{1}{-my} + \frac{1}{y} \]
- Simplifying:
\[ \frac{1}{f} = \frac{-1 + m}{my} \]

8. **Equating the Two Expressions for \( \frac{1}{f} \)**:
- Since both expressions equal \( \frac{1}{f} \):
\[ \frac{-1 + m}{mx} = \frac{-1 + m}{my} \]
- Cross-multiplying gives:
\[ (-1 + m)y = (-1 + m)x \]
- Rearranging leads to:
\[ y + x = 2f \]

9. **Finding the Focal Length**:
- Thus, the focal length \( f \) is:
\[ f = \frac{x + y}{2} \]

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