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A concave lens is a fascinating optical device. It diverges light rays and forms images in interesting ways. To mathematically describe this behavior, we use the lens formula. The formula for a concave lens is given by: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] Here,

• \( f \) is the focal length of the lens.
• \( v \) represents the image distance, or how far the image appears from the lens.
• \( u \) stands for the object distance, the distance from the object to the lens.

When it comes to a concave lens, the focal length \( f \) is negative. This signifies that the focus is on the same side as the incoming light. For instance, if a focal length is \(-21 \text{ cm}\), this means the lens will focus diverging light at a point 21 cm in front of it on the same side as the incoming light.

Now that we understand the lens formula, let's apply it to calculate the image distance, \( v \). The image distance tells us where the image of the object will form relative to the lens. For a concave lens, this distance is particularly intriguing, as it often results in a virtual image on the same side as the object. Using the formula, we rearrange it to find \( v \):

• Substitute the known values into the lens formula.
• For the given exercise, with a focal length of \(-21 \text{ cm}\), and an object distance \(-14 \text{ cm}\), the equation becomes: \[ \frac{1}{-21} = \frac{1}{v} + \frac{1}{-14} \]
• By solving this, \( v \) is found to be \(42 \text{ cm}\).

This result indicates a virtual image positioned 42 cm away from the lens on the same side as the object.

Magnification describes how much larger or smaller the image is compared to the object. It combines beautifully with our understanding of image distance to fully describe the image properties. The magnification formula is:

• \[ m = \frac{v}{u} \]
• \( m \) also equals \( \frac{h'}{h} \), where \( h' \) is the image height, and \( h \) is the height of the object.

Imagine our earlier situation. With an image distance \( v = 42 \text{ cm} \) and object distance \( u = -14 \text{ cm} \), we solve for magnification:\[ m = \frac{42}{-14} = -3 \]This outcome clues us into a few essential facts:

• A negative magnification (-3) suggests the image is inverted relative to the object.
• The absolute value \( |3| \) tells us the image is three times larger than the object.

In optics, a virtual image is quite captivating because it can't be projected onto a screen. A virtual image forms in such a way that it appears on the same side of the lens as the object. With a concave lens, this happens consistently. Our calculations showed that the image distance \( v = 42 \text{ cm} \), which being positive (considering the sign conventions used) indicates a virtual image. This means the light appears to be coming from 42 cm away but on the object's same side of the lens.Moreover, examining the magnification shows us the image not only is virtual but also inverted and enlarged compared to the original object. Understanding virtual images helps in grasping the overall image formation process through different lenses and the varied applications they hold in everyday optics.