8.6 The Classification of Images

An image can be considered to be like a painting - it is a representation of an object. In physics terms, however, an image is necessarily formed by some sort of an optical device, not by a paintbrush. Images are classified in several categories in order to better describe their appearance. This is important because in Optics the transformations an object has undergone are more important than its appearance.

A first classification deals simply with the magnification of the object, which can be found by comparing the image height to the height of the object. If the ratio is greater than one, the image can be said to have been magnified, or enlarged. If the ratio is less than one, the image is said to have been diminished, or reduced. Secondly, the orientation of the image can be taken into account. In the case of lenses, images are often flipped over the principal axis, and are therefore described as being vertically inverted. In a mirror, however, objects become laterally inverted, that is, flipped from side to side.

A slightly more complicated classification of images has to do with how they can be observed. A real image is one that can be projected onto or intercepted by a screen. This means that rays of light, for example in a lens diagram, actually have to pass through the image. A real image is created on the film inside a camera, for instance after light rays bouncing off your dog enter through the lens and are bent. Out of convention, the distance to any real image is considered to be positive.

The opposite of a real image, as you might imagine, is a virtual one. In fact, this name suits this kind of image quite well, because a virtual image exists only with perception. These images cannot be placed on a screen, they can only be perceived by an optical receptor, such as your eye system. For instance, if you stand next to your friend in front of a plane mirror, and look at him in the mirror, he will appear to be standing behind the mirror, diagonally across from and facing you. The virtual image is seen by you to be at the point where the rays seem to originate (i.e. you see your friend as where he appears to be standing - behind the mirror), though the rays do not actually come from that point (after all, your friend is standing beside you). The distance from the optical device to a virtual image, by convention, is regarded as negative.

Let’s look at what kinds of images can be made by the lenses that have been mentioned so far. A diverging lens, as we found out in the previous unit, always produces an image that has a value for v that is smaller than u, and negative.

In this diagram, we can see that the image of a pencil formed by a diverging lens is upright, diminished, and virtual. If you were to look through the lens at the pencil, what you would in fact see is a much, much smaller pencil.

Converging lenses, on the other hand, are capable of producing two kinds of images. We can see this from the lens equation, which we will change slightly for easier interpretation: 1/v = 1/f - 1/u. When u is greater than f, that is, when the object is farther from the lens than the focal point, 1/v, and therefore v are positive. The image is therefore real, and also it is inverted (though this cannot be found using the equation). The magnification of the object cannot be stated right away, as it still depends on the variables u and v. However, we can come up with some general guidelines for determining whether an image is magnified or diminished, simply by using the equation. When u = v, the lensequation becomes 1/u + 1/u = 1/f, or 2/u = 1/f. This means that when u is equal to twice the focal length, the object and the image are the same size. When the object is beyond this point, the image is reduced, and when it is in front of this point (relative to the lens), the image is magnified.

Now consider the equation, 1/v = 1/f - 1/u, when u is less than f, or when the object is closer to the lens than the focal length,. 1/v and v become negative, which means that a virtual image is formed. 1/v will always be less than 1/u, in absolute terms, but this means that v is greater than u. In return, this means that the ratio of image distance to object distance is greater than one, and therefore that the image is magnified, as, for example, in this diagram: