A magnifying glass forms an image \(10 \mathrm{cm}\) on the opposite side of the lens from the object, which is \(10 \mathrm{cm}\) away. What is the magnification of this lens for a person with a normal near point if their eye \(12 \mathrm{cm}\) from the object?

Understanding the focal length of a lens is crucial for figuring out how it focuses light. The focal length, usually represented by the symbol \( f \), is the distance from a lens where it converges the light to a single point. In simple terms, it's the distance from the lens to the sharp image of an object that is very far away.

The focal length can be positive for converging lenses, often used in magnifying glasses, or negative for diverging lenses. In the context of our magnification problem, we calculated a negative focal length, \( f = -20 \text{cm} \), which indicates that we are dealing with a diverging lens and that the image formed is virtual, meaning it appears to come from a location behind the lens rather than in front of it.

The lens formula is a simple equation that relates the focal length (\( f \)) of a lens to the distances from the object (\( u \)) to the image (\( v \)). It is given by \( 1/f = 1/v - 1/u \). This equation helps to locate the image produced by a lens and is essential for understanding how images are formed by lenses.

In the given exercise, we applied the lens formula to find the focal length with known object and image distances. Since the image is virtual, the image distance was taken as negative. Remember that in the lens formula, conventionally, the object distance (\( u \)) is negative if it is on the same side as the incident light (real object) and positive if it is on the opposite side (virtual object). Conversely, the image distance (\( v \)) is positive for real images and negative for virtual images.

The image distance, denoted by \( v \), is the distance from the lens to the image it forms. It plays a pivotal role when utilizing the lens formula and calculating magnification. Depending on where the image forms in relation to the lens, \( v \) can be either positive or negative.

In our example, the image distance was initially \( -10 \text{cm} \), indicating a virtual image on the same side of the lens as the object. It is essential to include the correct sign when using the lens formula, as it affects the calculation of the focal length. Later, for magnification, we use \( v = -25 \text{cm} \), representing the distance from the observer's eye to where they perceive the image near their near point.

The object distance, indicated by \( u \), is the measure from the lens to the object being viewed. In the case of lenses, this distance affects how the image will be formed, including the image size, location, and whether the image is real or virtual.

In this specific problem, \( u \) was given to be \( 10 \text{cm} \), and we used it to calculate the focal length of the lens. Furthermore, when the magnification process involves a human observer, their near point, or the closest distance at which they can comfortably see objects clearly, comes into play. In such scenarios, the object distance may be altered in the magnification formula to reflect this personal visual distance, as was done in the exercise solution, adjusting it from \( 10 \text{cm} \) to \( 12 \text{cm} \), the distance between the eye and the object.