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. (a)`(x+y)/2`
B. (b)x-y
C. (c)`sqrt(xy)`
D. (d)x+y

Explanation

To solve the problem, we need to find the focal length of the lens given that a real image and a virtual image are formed at distances \( x \) and \( y \) respectively, and both images have the same magnification.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two scenarios: one where the object is at distance \( x \) from the lens and a real image is formed, and another where the object is at distance \( y \) from the lens and a virtual image is formed.
- Both images have the same magnification.

2. **Using the Lens Formula**:
- The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
- For the real image formed at distance \( x \):
- Let \( v_1 \) be the image distance for the real image.
- The object distance \( u_1 = -x \) (negative because the object is on the same side as the incoming light).
- Thus, the lens formula becomes:
\[ \frac{1}{f} = \frac{1}{v_1} + \frac{1}{-x} \]
Rearranging gives us:
\[ \frac{1}{v_1} = \frac{1}{f} + \frac{1}{x} \quad \text{(Equation 1)} \]

3. **For the Virtual Image**:
- For the virtual image formed at distance \( y \):
- Let \( v_2 \) be the image distance for the virtual image.
- The object distance \( u_2 = -y \).
- The lens formula becomes:
\[ \frac{1}{f} = \frac{1}{v_2} + \frac{1}{-y} \]
Rearranging gives us:
\[ \frac{1}{v_2} = \frac{1}{f} + \frac{1}{y} \quad \text{(Equation 2)} \]

4. **Using the Magnification Condition**:
- The magnification \( m \) for the real image is given by:
\[ m_1 = \frac{v_1}{-x} \]
- The magnification for the virtual image is given by:
\[ m_2 = \frac{v_2}{y} \]
- Since both magnifications are equal, we have:
\[ \frac{v_1}{-x} = \frac{v_2}{y} \]
- Rearranging gives:
\[ \frac{x}{v_1} = \frac{y}{v_2} \]
- Cross-multiplying gives:
\[ x \cdot v_2 = y \cdot v_1 \quad \text{(Equation 3)} \]

5. **Substituting Equations**:
- From Equations 1 and 2, we can express \( v_1 \) and \( v_2 \) in terms of \( f \):
- From Equation 1:
\[ v_1 = \frac{xf}{x - f} \]
- From Equation 2:
\[ v_2 = \frac{yf}{y + f} \]

6. **Substituting into Equation 3**:
- Substitute \( v_1 \) and \( v_2 \) into Equation 3:
\[ x \left(\frac{yf}{y + f}\right) = y \left(\frac{xf}{x - f}\right) \]
- Cross-multiplying and simplifying leads to:
\[ xyf = y(xf - f^2) \quad \text{and} \quad xyf = x(yf + f^2) \]

7. **Solving for Focal Length**:
- After simplification, we find:
\[ f = \frac{x + y}{2} \]

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