The minimum distance between a real object and its real image formed by a thin converging lens of focal length `f` is

Options

A. `4f`
B. `2f`
C. `f`
D. `f//2`

Explanation

To find the minimum distance between a real object and its real image formed by a thin converging lens of focal length \( f \), we can follow these steps:

### Step 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:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
For a real object, \( u \) is negative, and for a real image, \( v \) is positive.

### Step 2: Express the Distance Between Object and Image
The distance \( D \) between the object and the image can be expressed as:
\[ D = v - (-u) = v + u \]
This simplifies to:
\[ D = v + u \]

### Step 3: Rearrange the Lens Formula
Rearranging the lens formula gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]
This can be rewritten as:
\[ v = \frac{uf}{u + f} \]

### Step 4: Substitute \( v \) into the Distance Equation
Substituting \( v \) into the distance equation gives:
\[ D = \frac{uf}{u + f} + u \]
Now, we can combine the terms:
\[ D = \frac{uf}{u + f} + \frac{u(u + f)}{u + f} = \frac{uf + u^2 + uf}{u + f} = \frac{u^2 + 2uf}{u + f} \]

### Step 5: Differentiate to Find Minimum Distance
To find the minimum distance, we differentiate \( D \) with respect to \( u \) and set the derivative equal to zero:
\[ \frac{dD}{du} = \frac{(2u + 2f)(u + f) - (u^2 + 2uf)}{(u + f)^2} = 0 \]
This simplifies to:
\[ 2u^2 + 2uf + 2fu + 2f^2 - u^2 - 2uf = 0 \]
\[ u^2 + 2f^2 = 0 \]
From this, we can solve for \( u \).

### Step 6: Solve for Minimum Distance
Setting the derivative to zero leads to:
\[ u = -2f \]
Thus, the minimum distance \( D \) can be calculated as:
\[ D = v + u = \frac{(-2f)f}{-2f + f} - 2f = 2f \]

### Conclusion
The minimum distance between the real object and its real image formed by a thin converging lens of focal length \( f \) is:
\[ \boxed{2f} \]