An object is placed `12 cm` to the left of a diverging lens of focal length `-6.0 cm.` A converging lens with a focal length of `12.0 cm` is placed at a distance d to the right of the diverging lens. Find the distance d that corresponds to a final image at infinity.

Explanation

To solve the problem step-by-step, we will use the lens formula and the concept of image formation by lenses.

### Step 1: Understand the Problem
We have a diverging lens with a focal length of \( f_1 = -6.0 \, \text{cm} \) and an object placed \( u_1 = -12 \, \text{cm} \) (to the left of the lens, hence negative). We need to find the distance \( d \) such that the final image formed by a converging lens (with a focal length of \( f_2 = 12.0 \, \text{cm} \)) is at infinity.

### Step 2: Use the Lens Formula for the Diverging Lens
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
For the diverging lens:
\[ \frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1} \]
Substituting the known values:
\[ \frac{1}{-6} = \frac{1}{v_1} - \frac{1}{-12} \]

### Step 3: Solve for \( v_1 \)
Rearranging the equation:
\[ \frac{1}{v_1} = \frac{1}{-6} + \frac{1}{12} \]
Finding a common denominator (which is 12):
\[ \frac{1}{v_1} = \frac{-2 + 1}{12} = \frac{-1}{12} \]
Thus,
\[ v_1 = -12 \, \text{cm} \]
This means the image formed by the diverging lens is \( 12 \, \text{cm} \) to the left of the lens.

### Step 4: Determine the Position of the Converging Lens
Let the distance between the diverging lens and the converging lens be \( d \). The image formed by the diverging lens acts as the object for the converging lens. The object distance for the converging lens \( u_2 \) is given by:
\[ u_2 = -(d - 12) \]
This is negative because the object (image from the diverging lens) is to the left of the converging lens.

### Step 5: Use the Lens Formula for the Converging Lens
For the converging lens, we want the final image to be at infinity, which means the object must be at the focal point of the converging lens:
\[ f_2 = 12 \, \text{cm} \]
Thus, we set:
\[ u_2 = -f_2 = -12 \, \text{cm} \]
Substituting for \( u_2 \):
\[ -(d - 12) = -12 \]
This simplifies to:
\[ d - 12 = 12 \]
Thus,
\[ d = 24 \, \text{cm} \]

### Conclusion
The distance \( d \) that corresponds to a final image at infinity is \( 24 \, \text{cm} \).

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