When the object is at distance `u_1 and u_2` from the optical centre of a convex lens, a real and a virtual image of the same magnification are obtained. The focal length of the lens is

Options

A. `(u_1-u_2)/2`
B. `u_1+u_2`
C. `sqrt(u_1u_2)`
D. `(u_1+u_2)/2`

Explanation

To solve the problem, we need to find the focal length \( f \) of a convex lens when the object is placed at two different distances \( u_1 \) and \( u_2 \) from the optical center, producing a real and a virtual image with the same magnification.

### Step-by-Step Solution:

1. **Understanding Magnification**:
- For a convex lens, the magnification \( m \) is given by the formula:
\[ m = \frac{h'}{h} = \frac{v}{u} \]
- Where \( h' \) is the height of the image, \( h \) is the height of the object, \( v \) is the image distance, and \( u \) is the object distance.

2. **Lens Formula**:
- The lens formula relates the object distance \( u \), the image distance \( v \), and the focal length \( f \):
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
- Rearranging gives:
\[ v = \frac{fu}{u - f} \]

3. **Setting Up the Equations**:
- For the object at distance \( u_1 \) (real image):
\[ m_1 = -\frac{v_1}{u_1} \]
- For the object at distance \( u_2 \) (virtual image):
\[ m_2 = \frac{v_2}{u_2} \]
- Since the magnifications are equal in magnitude but opposite in sign:
\[ -\frac{v_1}{u_1} = \frac{v_2}{u_2} \]

4. **Using the Lens Formula for Both Cases**:
- For \( u_1 \):
\[ \frac{1}{f} = \frac{1}{v_1} - \frac{1}{u_1} \implies v_1 = \frac{fu_1}{u_1 - f} \]
- For \( u_2 \):
\[ \frac{1}{f} = \frac{1}{v_2} - \frac{1}{u_2} \implies v_2 = \frac{fu_2}{u_2 - f} \]

5. **Equating the Magnifications**:
- From the magnification equations:
\[ -\frac{v_1}{u_1} = \frac{v_2}{u_2} \]
- Substituting the expressions for \( v_1 \) and \( v_2 \):
\[ -\frac{\frac{fu_1}{u_1 - f}}{u_1} = \frac{\frac{fu_2}{u_2 - f}}{u_2} \]

6. **Cross Multiplying and Simplifying**:
- After cross-multiplying and simplifying, we will find:
\[ -fu_2(u_1 - f) = fu_1(u_2 - f) \]
- This leads to:
\[ 2f = u_1 + u_2 \]

7. **Finding the Focal Length**:
- Finally, we can solve for \( f \):
\[ f = \frac{u_1 + u_2}{2} \]

### Final Answer:
The focal length of the lens is:
\[ f = \frac{u_1 + u_2}{2} \]