A converging lens has a focal length of 6 cm. If an object is placed 8 cm from the lens, calculate the position of the image and give an application of this lens arrangement.

The lens formula for a thin lens is given by:

where $f$ is the focal length, $v$ is the image distance, and $u$ is the object distance. For a converging lens, $f$ is positive. For a real object, $u$ is positive.

Find a common denominator, which is 24:

The positive value of $v$ indicates that the image is formed on the opposite side of the lens from the object.
The position of the image formed is $\boxed{\text{24 cm from the lens, on the opposite side of the object}}$.

Since the image distance $v$ is positive ($+24 \text{ cm}$), the image is formed by actual intersection of refracted rays, meaning it is a real image.

For a real image formed by a single converging lens, the image is always inverted.

The magnification $M$ is given by $M = -\frac{v}{u}$.

The magnitude of magnification $|M| = 3$, which is greater than 1, indicating that the image is magnified.

The nature of the image formed is $\boxed{\text{real, inverted, and magnified}}$.