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 shows the relationship between the distance from the object to the mirror, the distance from the image to the mirror, and the focal length of the mirror. Each of these distances is positive in cases when the light actually reaches the described point, whether it be the object, image or focus, and negative when, instead, reflected rays converge behind the mirror to form a virtual image. Thus, a virtual image has a negative image distance, because the light rays don't extend to a real image. A convex diverging mirror has a negative focal length because the light rays never actually reach the focal point. The object distance is almost always positive, because the light actually originates at the object. Be careful when using this equation. Solving for the focal length cannot be achieved by simply invert the object distance and image distance terms to solve for the focal length. 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 six 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 4 centimeters and inverted. This chart can be used to determine quickly where an image will be positioned and what its relative size and orientation will be. This chart can be used to determine quickly where an image will be positioned and what its relative size and orientation will be.

Copyright 2006 The Regents of the University of California and Monterey Institute for Technology and Education