12.3.2 Other cases with curved mirrors
The equationdi= (1/f− 1 /do)−^1 can easily produce a negative
result, but we have been thinking ofdias a distance, and distances
can’t be negative. A similar problem occurs withθi=θf−θofor
θo> θf. What’s going on here?
The interpretation of the angular equation is straightforward.
As we bring the object closer and closer to the image,θogets bigger
and bigger, and eventually we reach a point whereθo = θf and
θi= 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
angleθi= 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.f/A graph of the image distance
dias a function of the object dis-
tancedo.
To analyze the distance equation, let’s look at a graph ofdias
a function ofdo. 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, makingdobigger causesdito become smaller,
and vice-versa.792 Chapter 12 Optics