Two thin converging lenses are placed on a common axis, so that the centre of one of them coincides with the focus of the other. An object is placed at a distance twice the focal length from the left hand lens. Where will its image be? What is the lateral magnification? The focal of each lens is f.

Explanation

To solve the problem step by step, we will analyze the situation with two converging lenses placed on a common axis, with the center of one lens coinciding with the focus of the other.

### Step 1: Understand the Configuration
We have two converging lenses, both with a focal length \( f \). The first lens has an object placed at a distance of \( 2f \) from it. The second lens is positioned such that its center coincides with the focus of the first lens.

### Step 2: Determine Object Distance for the First Lens
The object distance \( u_1 \) for the first lens is given as:
\[ u_1 = -2f \]
(Note: The negative sign indicates that the object is on the same side as the incoming light.)

### Step 3: Use the Lens Formula for the First Lens
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
For the first lens, substituting \( u_1 \) and \( f \):
\[ \frac{1}{f} = \frac{1}{v_1} - \frac{1}{-2f} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{v_1} + \frac{1}{2f} \]
Combining the terms gives:
\[ \frac{1}{v_1} = \frac{1}{f} - \frac{1}{2f} = \frac{2 - 1}{2f} = \frac{1}{2f} \]
Thus, we find:
\[ v_1 = 2f \]

### Step 4: Calculate Magnification for the First Lens
The magnification \( m_1 \) for the first lens is given by:
\[ m_1 = -\frac{v_1}{u_1} = -\frac{2f}{-2f} = 1 \]

### Step 5: Determine Object Distance for the Second Lens
The image formed by the first lens acts as a virtual object for the second lens. The distance from the first lens to the second lens is \( f \) (the focal length of the first lens). Therefore, the object distance \( u_2 \) for the second lens is:
\[ u_2 = -f \]

### Step 6: Use the Lens Formula for the Second Lens
Using the lens formula again for the second lens:
\[ \frac{1}{f} = \frac{1}{v_2} - \frac{1}{u_2} \]
Substituting \( u_2 \):
\[ \frac{1}{f} = \frac{1}{v_2} - \frac{1}{-f} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{v_2} + \frac{1}{f} \]
Rearranging gives:
\[ \frac{1}{v_2} = 0 \implies v_2 = \infty \]
This indicates that the image formed by the second lens is at infinity.

### Step 7: Calculate Magnification for the Second Lens
The magnification \( m_2 \) for the second lens is given by:
\[ m_2 = -\frac{v_2}{u_2} = -\frac{\infty}{-f} = \infty \]

### Step 8: Calculate Total Magnification
The total magnification \( m \) is the product of the individual magnifications:
\[ m = m_1 \times m_2 = 1 \times \infty = \infty \]

### Final Result
The final image is formed at infinity, and the lateral magnification is also infinite.