91Ó°ÊÓ

Magnification helps us understand how the size of an image relates to the size of the object itself. When dealing with lenses, magnification can tell us whether the image will appear larger or smaller than the object. In mathematical terms, it is defined as the ratio of the image height to the object height, but it can also be expressed as:

• \( m = \frac{d_i}{d_o} \)

where \( d_i \) is the image distance and \( d_o \) is the object distance. In our problem, since the image's size is reduced by a factor of 2, the magnification \( m \) is \(-\frac{1}{2}\). The negative sign indicates that the image formed is inverted relative to the object. Magnification not only tells us about the size change but also provides information on the orientation of the image formed by a lens.

The focal length of a lens is a crucial concept in optics and determines how strongly a lens converges or diverges light. It is the distance between the lens and the focal point, where light rays converge or appear to diverge from. For a diverging lens, like in this exercise, the focal length is negative.

• A negative focal length signifies that the lens will spread out light rays.
• In our exercise, the focal length of the lens is \(-12\, \text{cm}\).

Using the lens formula:

• \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)

we can calculate how the object and image distances change with respect to the focal length. This formula connects object distance \( d_o \), image distance \( d_i \), and the focal length \( f \) of the lens. Understanding the focal length helps predict the nature of the images produced by different types of lenses.

A diverging lens, often known as a concave lens, causes light rays to spread out or diverge. This type of lens is characterized by having a focal length that is negative. In contrast to converging lenses, which bring light rays to a focal point, diverging lenses have a focal point from which light rays appear to disperse.

• Common examples include glasses for nearsightedness and certain camera lenses.
• The diverging lens always produces virtual, upright, and smaller images compared to the object size.

In our specific problem, the diverging lens has a focal length of \(-12\, \text{cm}\), showing that it moves light rays outward, making objects placed in front of it appear smaller and farther than they truly are. The diverging lens formula used earlier helps us to locate where this reduced image forms relative to the object's distance.