The Science of Mirages

Stories of exhausted and dehydrated travelers mistaking dry land for oases have appeared throughout our collective anthology for centuries, if not millennia. How is it that these poor souls could mistake dry land for water? Well, first, we will think about the properties of bodies of water. When you do see an actual body of water off in the distance, what makes you think it is water? Unless there are people in swimsuits close-by, or animals (or humans!) drinking from it, it may be hard to tell! What you have probably noticed, then, is that the surface of the water appears reflective. This then makes it easy to mistake mirages in the desert for bodies of water, as mirages behave the same way.

In fact, we are actually seeing the exact same thing, a reflection! How does this work though, and what is it reflecting? Let’s answer these questions one-by-one, since you may be able to guess the answer to the second question after I explain the first.

First, we have all heard that hot air rises, which means that any cooler air in the desert (or anywhere) will be closer to the ground. Cooler air will be more dense, as the excited molecules in the air have less energy than in hot air, so they aren’t bouncing around and spreading out as much as in hot air. This in turn means that light passing through cold air will move more slowly, since it will have a lot more obstacles in it’s way. In optics terms, then, this means that it will have a higher index of refraction. However, in order to explain mirages, we also have to consider that in the desert, there will be a thin layer of hot air closest to the surface of the ground/road that will have a lower index of refraction than the cooler air above it. Now’s the time, then, to use our favorite equation in optics: Snell’s Law! Snell’s Law is given in Equation \eqref{Snel}

$$n_1{sin(\theta_1)} = n_2{sin(\theta_2)} \label{Snel}$$

where “$n_1$” is the index of refraction of the first medium (in this case, cold air), “$\theta_1$” is the angle of incidence on the second medium (in this case, the angle from you, in the cold air region, to the mirage you are seeing), “$n_2$” is the index of refraction of the second medium (in this case, hot air), and “$\theta_2$” is the angle of refraction into the second medium. When “$\theta$” is equal to zero, in the case where the refracted light is parallel to the ground, then we have the angle at which total internal reflection occurs. Total internal reflection means that all the light hitting a surface will be reflected back into the medium where it first came. Equation \eqref{Snel} would then become:

$$\frac{n_1}{n_2}{sin(\theta_1)} = 1$$

such that, with the cold air index being larger than the hot air index, we will see total internal reflection as long as we are far way from the mirage! If you are following the math, then you can probably see that as you walk closer to the “water”, the reflection will go away, because now you have increased your angle between you (in the hot air medium) and the cold air on the ground.

Now, I bet you were able to guess what the reflection you are seeing is of, but I will say it explicitly anyways because I like to talk. The reflection is of the sky, which is what we would also see if there was water as well. Thus, although well-intentioned, the assumption of water off in the distance proves once again that there is often more to life than meets the eye!