This is an interesting thought which I had when driving home today looking in my wing mirrors.

If you are driving a car and looking in your, say, right wing mirror, you see an image of the car behind you and to the right. The image is reasonably crisp, clear, and not distorted in any way. This is presumably as a consequence of the density of the air being low, and lack of interaction of the photons with particles in the air.

However, if you're looking out across at something and the surface beneath your line of sight is quite hot, we've all seen the effect where what you see appears somehow rippled/shimmering. For example, looking out along a road on a very hot day.

The obvious answer is that, in the example with the road, the road is hot, and as the heat rises, it plays havoc with the light waves.

I ask why the heated air interacts with the light and changes the image you see, whilst in other circumstances the light doesn't interact with the air at all?

I hope this question is not a duplicate, I've looked around and found nothing the same.


The reason is that the index of refraction changes with the temperature. Due to the very hot asphalt, there is a strong temperature gradient in the air. This gradient leads to an changing index of refraction with the distance to the road. This is what you see.

  • $\begingroup$ This is a bit muddled. In fact, what you see is the reflected sky, the sky itself being "shiny" . And the reason you get that reflection is because the hot road has a less dense layer of air adjacent to it with a more dense layer above, leading to total internal reflection at the boundary. $\endgroup$ – Carl Witthoft Oct 4 '15 at 17:08

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