91Ó°ÊÓ

Magnification is a crucial concept when dealing with optical lenses. It essentially tells us how much larger or smaller the image of an object appears, when viewed through a lens, compared to the object's actual size. To find magnification, we use the formula \( m = \frac{h_i}{h_o} \), where \( h_i \) is the image height and \( h_o \) is the object height. But that’s not all! Magnification is also related to the distances in the setup. Another equation used is \( m = \frac{d_i}{d_o} \), with \( d_i \) being the image distance and \( d_o \) the object distance. Combining these two equations allows us to solve exercises where either the height or the distances are unknown.

• If the magnification is positive, the image is upright in relation to the object.
• If it is negative, the image is inverted.

For instance, if the object height is less than the image height, as in our problem, we get a magnification greater than 1, indicating the image is larger than the object.

The object distance \( d_o \) is the distance between the object and the lens. It's an essential part of the lens formula and can be determined by rearranging the magnification formula: \( m = \frac{d_i}{d_o} \). In the solved exercise, knowing the magnification from the heights, the object distance is found by plugging in the known values. It tells us where exactly the real object is placed relative to the lens for the image to form as described. The object distance plays a pivotal role as it directly influences the size and orientation of the image.

• For real objects (those that exist in a definite place), \( d_o \) is positive.
• A smaller object distance generally leads to a larger image distance, significantly affecting magnification as seen in the exercise.

Image distance \( d_i \) is how far the image is located from the lens. The sign of \( d_i \) indicates whether an image is real or virtual. In a converging lens setup,
such as the one discussed in our exercise, the calculated image distance was -12 cm. This negative value signals that the image is virtual, and appears on the same side of the lens as the object. It is this parameter, together with object distance, that impacts image size and clarity. When applying the formula for image distance, it becomes pivotal in using the lens equation to find unknowns in the problem - especially when verified or derived from other values like focal length or magnification. Consider that:

• A positive \( d_i \) represents a real image that appears on the opposite side of the lens from the object.
• A negative \( d_i \) signifies a virtual image on the same side as the object, not directly visible, except through the lens.

A converging lens is a type of lens that bends light rays toward a specific point, called the focal point. This kind of lens is thicker in the center than at the edges. Such lenses are commonly used in corrective glasses, projectors, and cameras. A converging lens has a specific focal length which determines where the parallel light rays meet after passing through the lens. In the context of our exercise, a focal length of approximately 3.69 cm was calculated using the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). By knowing the focal length, it's possible to predict how the lens will form an image of an object placed at various distances.

• When the object is beyond the focal point, the lens creates a real and inverted image.
• If the object is at the focal point, no image is formed as the rays emerge parallel.
• When within the focal point, as seen in the exercise, a virtual, upright, and magnified image appears.