Suppose we have a mirror shaped like the outside of our last mirror, reflecting light off the outside surface. Such a mirror is known as a convex mirror. Parallel light that approaches the mirror cannot converge to the mirror's focus, because the focus is on the far side of the mirror. Instead, the light reflects as if it has come from the focus, and diverges outward. Notice that if we extend the outgoing rays backward through the mirror they do meet at the focus. Because the rays reflected from this sort of mirror diverge, a convex mirror is also called a diverging mirror. As you might expect from our earlier discussion, rays moving towards the mirror that are heading towards the focal point are parallel to the mirror's axis when they are reflected back by the mirror. Here again, though these rays are directed towards the focus as they approach the mirror, they are reflected before actually reaching the focus. Our principal rays are now First: parallel going in and out as if they come from the focus and through the point of reflection. Second: in towards the focus and out from the point of reflection parallel to the mirror's axis. Third: into the middle of the mirror, reflected back such that angle in equals angle out. Both angles are measured relative to the axis of the mirror. Fourth: towards the center of curvature and straight back out, following the ingoing path backwards. Notice that the outgoing rays do not cross. However, if we extend the lines back behind the mirror from their points of reflection, they do cross. Thus this is a virtual image, because the rays leave as if they had crossed at a point, even though none of them actually reached that point. The image created is virtual, upright, and smaller than the object. In general, images from a concave or diverging mirror will be virtual, upright, and smaller than the original object. So far we have used ray tracing to find and classify images formed by spherical mirrors. However, much can be gained from a mathematical approach to this subject, especially when used in conjunction with the drawings. Ray tracing helps develop physical intuition, while math helps quantify the phenomena. The first equation simply states that the focal length is the half the radius of curvature as we've previously mentioned. The second equation is known as the lens equation, though it works equally well for mirrors. s sub o is the object distance - the distance from the object to the mirror, s sub i is the image distance - the distance from the image to the mirror, and f is the focal length - the distance from the focal point to the mirror. Each of these distances is positive in cases when the light actually reaches the described point (object, image or focus), and negative when reflected rays merely behave as though they had. Thus a virtual image has a negative image distance, because the light rays don't extend to the image. A convex diverging mirror has a negative focal length because the light rays never actually focus at the focal point. The object distance is almost always positive, because the light actually orriginates at the object. Be careful when using this equation. You cannot simply invert each term and get s sub i plus s sub o equals f. It is not! Let's work an example. Suppose an object is placed 10 centimeters in front of a mirror with a radius of curvature of 8 centimeters. Where is the image? First we find the focal length to be 4 centimeters. Next we solve the equation for the image distance, and plug in the values and find that the image is 6.7 centimeters from the mirror. Because the distance is positive, the image is real. Now consider the third equation. The first part simply defines the magnification, m, as the ratio of the image height, h sub i, to the object height, h sub o. The sign of m indicates whether the image is upright or inverted. m is positive for an upright image and negative for an inverted image. The magnitude of m indicates the relative size of the image compared to that of the object. If the image is larger than the object, the absolute value of m is greater than 1. If the image is smaller than the object, the absolute value of m is less than 1. The second part of the equation relates the image magnification to the image distance and the object distance. Notice that if both the image and the object are real, the magnification will be negative and the image will be inverted. Returning to our example, suppose the object is 3 centimeters high. How tall is the image? By rearranging the equations we find that the magnification is negative 0.67 and that the image is 2 centimeters and inverted. These equations can be used together to solve a wide variety of problems. It is often useful to determine quickly where an image will be positioned and what its relative size and orientation will be. Try using ray tracing and the equations to verify this chart. The next section will discuss how the ideas discussed here apply to lenses.

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