Why does no body even try to explain the phenomenon from beginning to end? I'm quite ignorent, studied physics 25 years ago and hardly even remember Snellius law. My daughter wanted to make a schoolthesis on the subject and found no literature describing it in detail. We only red a popular but nice text from Minnaert.
Now we are so ignorent/arrogant that after quite some discussion we think we understand it all and then we found this (a bit disappointing)discussion so we give it a try:
1 Mirage concerns a virtual image as a result of a vertical temperature gradiënt and thus a density and thus a ngradiënt (diffraction index).
2 Light propagating through this gradiënt wil curve towards the higher nside.
3 Because off the gradiënt no critical angle can be identified (or asymptotic to 90degr.), the medium can be seen as a multi layer with infinetesimal small differences off n.
4 The curving process continues until the light propagation is asymptoticly horrizontal.
5 Then we have a problem. With the preassumption off a ideal medium (exact horrizontal isotherms an consequently exact vertical ngradiënt the beam is captured in the horrizontal plane because it propagates perpendicular to the ngradiënt.
[o)] 6 Up to this point we can draw a first conclusion.
From Snellius it follows that for a multi layer medium, diffraction depends only from n of the first and the last layer. Consider a point object at a certain level where n=n1 and angel Q1 of the departing beam, and consider n=n2 at the level at which the beam is appr. horrizontal.(Q2=90degr) Then we CONCLUDE that:
On the MICRO LEVEL the beam curves increasingly from layer to layer (with infinetisemal small differences of n) until horrizontal, so no critical angle can be identified and no total reflection takes place.
On a MACRO LEVEL the relation between Q1, n1 and n2 is described just AS IF Q1 is the crical angel.
(However if we consider the same point object from which departs a beam with a smaller Q, this Q can also be considered as a Critical angle, be it in a other macro system: the horrizontal will be reached at a lower point with a smaller n. The smallest possible Q is from the beam which reaches horrizontal at ground level.)
7 The first escape from this capture to the horrizontal is to take in account the curving off the earth and thus the curving of isotherms etc. In this forum this is correctly rejected, it only works with upward bending (positive upward ngradiënt). With downward curving (in the case of inversion = negative ngradiënt) the beam will be captured in the horrizontal isotherm and curve with the rounding of the earth but not curve back to the earth.(It also cannot escape upward because pushed backward by the negative ngradiënt;would it go in circles around the earth for ever?)
8 The second escape from the horrizontal capture is to take in account the imperfection of the medium. In real life isotherms are not perfect horrizontal. The temperature gradiënt results in turbulations of the air and fluctuating isotherms. Consider the isotherm as (slightly) sinusoïdal, then also the horrizontal beam meets a gradiënt and tends to bent towards the positive of the ngradiënt. So it can escape from the horrizontal and then curves further upward. Consequently bending backwards can be explained in both situations: warm earth/cold air as well as in the case of inversion.
[o)] 8 We can conclude that air turbulations can explain curving back, which then still is a refraction phenomenon.
9 This model of continuïng curving also explains the compression of the virtual image which often takes place. If the observer is at a lower altitude then the object the Qin is smaller then Qout (Qin is the departing beam angle at the top of the object where Qout is the angel at which this refracted beam is observed). In fact the difference between Qin and Qout depends on the difference between n at the level of the object and n at the observers level. If n at the upper part of the object is (about) the same as n at the observer, the bending of the downward curve is (about) the same as the bending of the upward curve, Qin equals (about) Qout and the image is not compressed.
