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A diverging lens, also known as a concave lens, is a lens that causes light rays to spread apart as they pass through it. This type of lens is thinner at the center than at the edges. Due to its shape, parallel rays of light that enter the lens appear to diverge from a point. This point is known as the focal point, and for a diverging lens, it is virtual and located on the same side as the incoming light.

Diverging lenses are characterized by their negative focal lengths, indicating that they do not bring light rays to a real focus. Instead, the image formed is usually virtual, upright, and smaller than the object. Because of these properties, diverging lenses are often used in applications such as eyeglasses for nearsightedness and certain optical devices.

The focal length of a lens is a critical parameter that describes how strongly the lens converges or diverges light. In mathematical terms, it is the distance from the lens to the focal point, where light rays either meet (in converging lenses) or appear to diverge from (in diverging lenses).

For a diverging lens, the focal length is assigned a negative value. This negative sign is a mathematical representation that the focal point is virtual and located on the same side of the lens as the incoming light rays. In our problem, the diverging lens has a focal length of \[f = -20.0 \, \text{cm}\]This indicates that the lens is meant to spread light out, rather than focus it.

• The greater the absolute value of the focal length, the weaker the lens's ability to diverge light.

Image distance refers to the distance from the lens to the location where the image is formed. For diverging lenses, this distance is typically negative, corresponding to the virtual nature of the image, as it forms on the same side as the object.

Using the lens equation: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]where

• \(f\) is the focal length,
• \(d_o\) is the object distance,
• \(d_i\) is the image distance,

We can solve for the image distance \(d_i\). Substituting the values from the exercise:\[ \frac{1}{-20.0} = \frac{1}{20.0} + \frac{1}{d_i} \] Solving this equation will give us the value of \(d_i\), indicating the position of the virtual image. It is important to remember that for diverging lenses, the image formed is usually smaller and upright compared to the actual object.

Magnification in optics refers to how much larger or smaller the image is compared to the object. It is defined as the ratio of the image height to the object height, or equivalently, the image distance to the object distance. Mathematically, it is expressed as:\[ m = \frac{d_i}{d_o} \]For a diverging lens, since the image distance is usually negative, the magnification results in a positive but smaller-than-one value, indicating that the image is upright and reduced in size. In our equation setup, using \(d_i\) that we have calculated and using the known \(d_o = 20.0\, \text{cm}\), we can find:

• \(m < 1\): image is smaller
• \(m > 0\): image is upright

Thus, the magnification provides not only information about the size of the image but also about its orientation relative to the object.