Optical Illusion of Water on Pavement

[starzero]

When we drive a car on a hot day we have all experienced what appears to be water on the road. I believe that this is related to the facts that 1) the air near the hot pavement is less dense than the air a few feet above and 2) light will travel from point A to B along the fastest path (in this case the hot air near the pavement).

So here is the questions...Why does the result appear to the eye to look like water?
Can someone provide a really good explanation of the above?

Thanks.

 

[zhermes]

Because the air above the pavement is hotter, it is less dense (as you say), and therefore has a different 'index of refraction' which tells you how fast light travels through the medium (again, as you said). The key is that the hot pocket of air acts like a prism, and reflects light coming from the sky, towards you. Because the heat/air is rising and curling (etc), it gives the appear of flowing or rippling.
Overall you have a reflection of the sky, coming from a region that looks like its rippling... same as looking at water.

 

[K^2]

That's not quite right. When light travels from higher index of refraction to lower, there is an angle called total internal reflection angle under which no light is transmitted. So at a very shallow angle, if you look at a pocket of hot air, it is almost perfectly reflective. What it actually reflects doesn't matter. The only other surface you normally observe to behave this way is water, so that's the connection your brain makes.

 

[cragar]

Why does the angle of light change when it goes from the same medium just the density changes. And is the light getting absorbed and re-emitted . I mean I would think that it wouldn't change, the light is traveling through molecules and it hits a barrier where the molecules are farther apart so it just changes angle, why is this.

 

[Born2bwire]

Why does the angle of light change when it goes from the same medium just the density changes. And is the light getting absorbed and re-emitted . I mean I would think that it wouldn't change, the light is traveling through molecules and it hits a barrier where the molecules are farther apart so it just changes angle, why is this.

The propagation of light through a medium is dependent upon the bulk behavior of the medium's constituents. Just looking at a solid, the atoms that make up a solid material are arrayed in a lattice. They are held in place by bonds and intermolecular forces. The properties of the lattice, like the nature of the restorative forces and the distances between the atoms or molecules, affect its bulk properties. It affects the bulk energy modes and dispersion curves that dictate how light can travel through the bulk. So keeping that in mind we can come to understand that if we were to make the material less dense by allowing for larger intermolecular spacings then this will effect the dispersion curve that we get for light. This in turn will change the speed of propagation and thus its index of refraction. Now this is a little difficult to apply to a gas since we no longer have a lattice and so forth. But an overall density of molecules can still give rise to a dispersion curve (although I guess we notice more individual effects like Rayleigh scattering which occurs prominantly in our atmosphere). Although, come to think of it I don't think I've seen a calculation for the transmission of light through a gas. I would expect though that it would mainly differ in the lack of phonon modes.

One thing to keep in mind is that with a mirage the temperature of the air, ignoring mixing due to turbulence and so forth, gradually changes as we move toward or away from the surface. So the index of refraction is gradually changing and this causes a continuous curvature of the transmitted light. So do not think of it like a single interface like light hitting water but think of it as a layering of interfaces with each interface just changing the angle of propagation of the light by a small amount. If we can achieve a large enough gradient then the distortion of the light's path can be so great that it is bent back upwards and we see our mirage.

EDIT: As being absorbed and reemitted. Not really mainly because absorption and emmission of the light with gaseous molecules usually results in things like Rayleigh scattering. It is much more narrowband and dependent upon the frequency of the light. To whit, the difference between the blue sky and the red dawn/sunset has to do with the path of the light and Rayleigh scattering. So it would not be able to retain a coherent image across the visual spectrum like we see in a mirage. My recollection is that the propagation of the light through the gas has more to do with the interaction of the fields of the light with the molecules and how this affects the energy of the system (since even as a gas the molecules experience intermolecular forces with their nearby neighbors) or how it induces oscillations that give rise to additional fields.

 

[cragar]

thanks for your reply , that helps .

 

[Borek]

Why does the angle of light change when it goes from the same medium just the density changes.

To add to Born2bwire answer, for a given gas:

$$\frac {n_1 - 1} {n_2 - 1} = \frac {d_1} {d_2}$$

where n & d are refractive index and density.

 

[K^2]

And is the light getting absorbed and re-emitted.

Light gets absorbed and re-emitted as a spherical wave at every single point in space. So light, effectively, takes all possible paths between source and destination. However, in most of the points, phases of different paths are canceling each other out. The only place where this isn't so are points along such paths that functional derivative of time it takes to traverse the path with respect to small deviation in path is zero. That is, time is extremized (typically minimized).

 

[Borek]

That is, time is extremized (typically minimized).

So called Fermat's principle or the principle of least time.