An object is placed at a distance u from an equiconvex lens such that the distannce between the object and its real image is minimum. The focal length of the lens is f. The value of u is

Options

A. `oo`
B. 1.5 f
C. 2f
D. 4f

Explanation

To solve the problem, we need to find the distance \( u \) at which an object should be placed in front of an equiconvex lens such that the distance between the object and its real image is minimized. The focal length of the lens is denoted as \( f \).

### Step-by-Step Solution:

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

2. **Identify the Conditions for Minimum Distance**:
We need to find \( u \) such that the distance between the object and the image \( |v - u| \) is minimized.

3. **Consider the Object at Different Positions**:
- If the object is placed at the center of curvature \( C \) (which is at a distance \( 2f \) from the lens), the image will also form at \( C \) on the opposite side of the lens.
- Thus, when \( u = 2f \), we have:
\[ v = 2f \]
- The distance between the object and the image is:
\[ |v - u| = |2f - 2f| = 0 \]
- This indicates that the object and image coincide, which gives a minimum distance of zero.

4. **Calculate the Value of \( u \)**:
From the above analysis, we find that the value of \( u \) that minimizes the distance between the object and its real image is:
\[ u = 2f \]

5. **Conclusion**:
Therefore, the value of \( u \) is \( 2f \).

### Final Answer:
\[ u = 2f \]