A simple way to understand the distance effect is to think of rays with constant curvature; then the vertical deviation of each ray grows quadratically with distance from the observer. You can see this effect in the ray diagrams: near the observer, the rays diverge from the eye at first nearly proportionally to the distance; but then the deviations become large, and ray-crossings and mirages appear rather suddenly beyond a certain distance. The simulations show this effect for both the classical inferior and superior mirages, as well as for mock mirages.

A different way to think about this is to regard the atmosphere as a gradient-index lens. Then the power of the lens is proportional to the thickness of the gradient-index medium — which is just the path length from observer to object. An object twice as far away as a nearby one is seen through twice as powerful a lens.

But, in the atmospheric-lens model, mirages appear when the object is
more than 4 focal lengths away, so that an inverted image can be formed
in front of the observer. As we regard more and more distant objects,
the focal length of the atmospheric lens shrinks *inversely *
with the distance; but the separation of object and observer grows
*directly * with the distance between them. So, once again,
we see that the measure of mirage formation — in this case, the
separation of observer and object, measured in atmospheric-lens focal
lengths — grows with the **square of the distance**.

This model is more abstract than simple ray-tracing; but, for someone with a little familiarity with optics, I think it provides more insight into the rapidity of the transition from erect to inverted images with increasing distance.

Copyright © 2008 – 2012 Andrew T. Young

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