Simple Nature - Light and Matter

12.3.2 Other cases with curved mirrorsThe equation d i = (1/f − 1/d o ) −1 can easily produce a negativeresult, but we have been thinking of d i as a distance, and distancescan’t be negative. A similar problem occurs with θ i = θ f − θ o forθ 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 biggerand bigger, and eventually we reach a point where θ o = θ f andθ i = 0. This large object angle represents a bundle of rays forminga cone that is very broad, so broad that the mirror can no longerbend them back so that they reconverge on the axis. The imageangle 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 wewant to be charitable and say that the rays cross at infinity. If wego on bringing the object even closer, we get a virtual image.f / A graph of the image distanced i as a function of the object distanced o .To analyze the distance equation, let’s look at a graph of d i asa function of d o . The branch on the upper right corresponds to thecase of a real image. Strictly speaking, this is the only part of thegraph that we’ve proven corresponds to reality, since we never didany geometry for other cases, such as virtual images. As discussed inthe previous section, making d o bigger causes d i to become smaller,and vice-versa.762 Chapter 12 Optics