In his famous Feynman Lectures on Physics, Richard Feynman provided an intuitive explanation of how a ‘mirage on a hot road’ can arise due to the bending of light rays from the sky in accordance with Fermat’s Principle (see The Feynman Lectures on Physics, Volume I, Chapter 26). Feynman wrote the following:
I was discussing this with a beginning engineering student who did not quite understand why the mirage makes it look as if the water is actually on the road. I explained this by augmenting Feynman’s Fig. 26-8 above as follows:
The bent light ray starting at point A and entering the observer’s eye at point B is interpreted by the observer as having followed a straight line path emanating from the road, as indicated in the diagram. Thus, the observer sees the image of the sky on the road surface and interprets it as a shimmering pool of water.
Having done this, the question then arose as to how one could go about constructing an explicit mathematical model of the above scenario, yielding a suitable equation for the curved light ray from A to B, a linear equation for the apparent straight line path seen by the observer, and explicit coordinates for the point on the road where the image of the sky is seen by the observer. This turned out to be an interesting exercise involving Fermat’s Principle and the Calculus of Variations and is what I want to record here.
Suppose the light ray begins at point at time , and enters the observer’s eye at point at time . Fermat’s Principle (see, e.g., this Wikipedia article) says that the path followed by the light ray is such as to make the optical length functional
stationary, where is the refractive index of the medium through which the light passes, is the speed of light in a vacuum and is the speed of light in the medium. This functional can be derived (up to a multiplicative constant) from the ‘Principle of Least Time’ by noting that the time taken by the light ray is
The light ray will find the path that minimises this time of travel.
To apply this setup to the mirage in Feynman’s lecture we need to model the refractive index as a function of the -coordinate in my amended diagram above, which measures the height above the road. As Feynman says, light goes faster in the hot region near the road than in the cooler region higher up. Thus, since the refractive index is inversely proportional to , it should be an increasing function of the height above the road . To get a toy model for the scenario in Feynman’s lecture let us make the simplest possible assumption that the refractive index is a simple linear function of , namely
with and both positive. Then since the arc-length element is
we can write the optical length functional as
We find the stationary path for this functional using the Calculus of Variations. Let
Since this does not depend directly on , the problem admits a first-integral of the form
where is a constant. We have
Therefore the first-integral for this problem is
Multiplying through by , absorbing into the constant term, and writing we get
This is a first-order differential equation for which can be solved by separation of variables. We get the integral equation
To solve the integral on the left-hand side, make the change of variable
For the integral on the right-hand side of the integral equation we get
Therefore the integral equation reduces to
This seems to represent two possible solutions for the first-integral equation, which we may write as
However, for the curved light ray in my amended diagram above we must have as . This condition is not satisfied by either of or on their own, but it is satisfied by their sum. We will therefore take the solution of the first integral equation to be
Furthermore, we have and therefore we require
These cannot be equal unless . Thus, our solution for reduces to
with the constant determined in terms of and by
This is the equation of the curved path of the light ray from the sky in Feynman’s diagram. The slope of at point is
The straight line with this gradient passing through the point has equation
This is the equation of the straight line emanating from the -axis to the observer’s eye in my amended diagram above. On the -axis we have in the straight-line equation so
This is the point on the -axis at which the observer in my amended diagram will see the mirage.