Two converging lenses have focal length `f_1 and f_2` The optical axes of the two lenses coincide. This lens system is used to form an image of an object . It is observed that the size of the image does not depend on the distance between the system and object . The distance between lenses is

Options

A. `(f_1f_2)/(f_1+f_2)`
B. `f_1+f_2`
C. `|f_1-f_2|`
D. `((f_1-f_2)^2)/f_1`

Answer

To solve the problem, we will analyze the two converging lenses and derive the relationship between their focal lengths and the distance between them based on the given conditions.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two converging lenses (lens 1 with focal length \( f_1 \) and lens 2 with focal length \( f_2 \)).
- The optical axes of both lenses coincide.
- An object is placed at a distance \( x \) from lens 1.

2. **Using the Lens Formula for Lens 1**:
- The lens formula is given by:
\[ \frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1} \]
- Here, \( u_1 = -x \) (object distance is negative as per sign convention).
- Rearranging gives:
\[ \frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{x} \]
- Therefore, we can express \( v_1 \) (the image distance from lens 1) as:
\[ v_1 = \frac{f_1 x}{x - f_1} \]

3. **Finding Object Distance for Lens 2**:
- The image formed by lens 1 acts as the object for lens 2.
- The object distance for lens 2, \( u_2 \), is given by:
\[ u_2 = D - v_1 \]
- Substituting \( v_1 \):
\[ u_2 = D - \frac{f_1 x}{x - f_1} \]

4. **Using the Lens Formula for Lens 2**:
- For lens 2, we apply the lens formula:
\[ \frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2} \]
- Rearranging gives:
\[ \frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{u_2} \]
- Substituting for \( u_2 \):
\[ v_2 = \frac{f_2 (D - v_1)}{D - v_1 - f_2} \]

5. **Finding Magnifications**:
- The magnification \( m_1 \) for lens 1 is:
\[ m_1 = \frac{v_1}{u_1} = \frac{f_1}{x - f_1} \]
- The magnification \( m_2 \) for lens 2 is:
\[ m_2 = \frac{v_2}{u_2} \]

6. **Condition for Independence of Image Size**:
- The problem states that the size of the image does not depend on the distance between the system and the object. This means the overall magnification \( m = m_1 \times m_2 \) must be independent of \( x \).
- For this to hold true, the term \( D - f_1 - f_2 \) must equal zero, leading to:
\[ D = f_1 + f_2 \]

### Final Result:
The distance between the two lenses is:
\[ \boxed{f_1 + f_2} \]