Lectures on Physics has been derived from Benjamin Crowell's Light and Matter series of free introductory textbooks on physics. See the editorial for more information.... 
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Other Cases With Curved MirrorsThe equation d_{i} = (1/f  1/d_{o})^{1} can easily produce a negative result, but we have been thinking of di as a distance, and distances can't be negative. A similar problem occurs with θ_{i}=θ_{f}θ_{o} for θ_{o}>θ_{f}. What's going on here? The interpretation of the angular equation is straightforward. As we bring the object closer and closer to the image, θ_{o} gets bigger and bigger, and eventually we reach a point where θ_{o}=θ_{f} and θ_{i}=0. This large object angle represents a bundle of rays forming a cone that is very broad, so broad that the mirror can no longer bend them back so that they reconverge on the axis. The image angle θ_{i}=0 represents an outgoing bundle of rays that are parallel. The outgoing rays never cross, so this is not a real image, unless we want to be charitable and say that the rays cross at infinity. If we go on bringing the object even closer, we get a virtual image. To analyze the distance equation, let's look at a graph of d_{i} as a function of d_{o}. The branch on the upper right corresponds to the case of a real image. Strictly speaking, this is the only part of the graph that we've proven corresponds to reality, since we never did any geometry for other cases, such as virtual images. As discussed in the previous section, making d_{o }bigger causes d_{i} to become smaller, and viceversa. Letting d_{o} be less than f is equivalent to θ_{o}>θ_{f}: a virtual image is produced on the far side of the mirror. This is the first example of Wigner's "unreasonable effectiveness of mathematics" that we have encountered in optics. Even though our proof depended on the assumption that the image was real, the equation we derived turns out to be applicable to virtual images, provided that we either interpret the positive and negative signs in a certain way, or else modify the equation to have different positive and negative signs.
It turns out that for any of the six possible combinations of real or virtual images formed by converging or diverging lenses or mirrors, we can apply equations of the form θ_{f} = θ_{i}+θ_{o} and
with only a modification of plus or minus signs. There are two possible approaches here. The approach we have been using so far is the more popular approach in American textbooks: leave the equation the same, but attach interpretations to the resulting negative or positive values of the variables. The trouble with this approach is that one is then forced to memorize tables of sign conventions, e.g. that the value of d_{i} should be negative when the image is a virtual image formed by a converging mirror. Positive and negative signs also have to be memorized for focal lengths. Ugh! It's highly unlikely that any student has ever retained these lengthy tables in his or her mind for more than five minutes after handing in the final exam in a physics course. Of course one can always look such things up when they are needed, but the effect is to turn the whole thing into an exercise in blindly plugging numbers into formulas. As you have gathered by now, there is another method which I think is better, and which I'll use throughout the rest of this book. In this method, all distances and angles are positive by definition, and we put in positive and negative signs in the equations depending on the situation. (I thought I was the first to invent this method, but I've been told that this is known as the European sign convention, and that it's fairly common in Europe.) Rather than memorizing these signs, we start with the generic equations
and then determine the signs by a twostep method that depends on ray diagrams. There are really only two signs to determine, not four; the signs in the two equations match up in the way you'd expect.The method is as follows: 1. Use ray diagrams to decide whether θ_{o} and θ_{i} vary in the same way or in opposite ways. (In other words, decide whether making θ_{o} greater results in a greater value of θ_{i} or a smaller one.) Based on this, decide whether the two signs in the angle equation are the same or opposite. If the signs are opposite, go on to step 2 to determine which is positive and which is negative. 2. If the signs are opposite, we need to decide which is the positive one and which is the negative. Since the focal angle is never negative, the smaller angle must be the one with a minus sign. In step 1, many students have trouble drawing the ray diagram correctly. For simplicity, you should always do your diagram for a point on the object that is on the axis of the mirror, and let one of your rays be the one that is emitted along the axis and reflect straight back on itself, as in the figures in section 3.1. As shown in figure (d) in section 3.1, there are four angles involved: two at the mirror, one at the object (θ_{o}), and one at the image (θ_{i}). Make sure to draw in the normal to the mirror so that you can see the two angles at the mirror. These two angles are equal, so as you change the object position, they fan out or fan in, like opening or closing a book. Once you've drawn this effect, you should easily be able to tell whether θ_{o} and θ_{i} change in the same way or in opposite ways. Although focal lengths are always positive in the method used in this book, you should be aware that diverging mirrors and lenses are assigned negative focal lengths in the other method, so if you see a lens labeled f = 30 cm, you'll know what it means.


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