91Ó°ÊÓ

Optical magnification is a key concept in understanding how lenses magnify objects. It is defined as the ratio of the image height to the object height, which can be expressed as \( m = \frac{h_i}{h_o} \). When dealing with a converging lens, if the image appears larger than the object, the magnification is greater than one, and if the image appears smaller, the magnification is less than one.

An inverted image will have a negative magnification, indicating that the image height \( h_i \) and the object height \( h_o \) have opposite signs. In our problem, the magnification is determined to be 1.40625, indicating that the image appears larger than the object by roughly 40.625%. Since the image height is given as a negative value, we know that the image is inverted with respect to the object.

The thin lens formula is an essential equation used to locate images formed by lenses. It is given by \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( f \) represents the focal length of the lens, \( v \) is the image distance from the lens, and \( u \) is the object distance from the lens. For a converging lens, the focal length is positive, and for a diverging lens, it is negative.

To solve for either \( u \) or \( v \) when the other distance is known, we rearrange the formula accordingly. In the given exercise, the thin lens formula was used along with the magnification to find that the object distance \( u \) is -57.53 cm, which implies that it's located 57.53 cm from the lens on the side from where the light originates. Since focal length \( f \) is positive for converging lenses, the image distance \( v \) is found to be negative, indicating an image formed on the same side as the light enters the lens.

The distinction between real and virtual images is fundamental in optics. A real image is formed when rays of light converge at a point after passing through the lens and can be projected onto a screen. Real images are always inverted relative to the object. In contrast, a virtual image occurs when rays of light diverge and cannot be projected; instead, they seem to originate from a particular location behind the lens. Virtual images are always upright relative to the object.

In the problem at hand, since the image distance \( v \) is negative, we learned that the image formed is real and inverted. This aligns perfectly with the fact that the image height is given with a negative value, solidifying it as an inverted, real image.

Understanding the lens sign convention is crucial for solving problems in optics correctly. In the sign convention for lenses, distances are positive if they measure in the direction of the light's travel after it passes through the lens, and negative if they measure against this direction. The focal length of a converging lens is positive, and for a diverging lens, it is negative. When using the thin lens formula, it's important to maintain these conventions for accuracy.

For our exercise, the object distance \( u \) came out as negative, meaning the object is located on the opposite side of the lens' incoming light. The image distance \( v \) result is also negative indicating the image is formed on the same side as the input light, characteristic of real images. By respecting this convention, one can accurately determine the image and object positions as well as the type of image formed by a lens.