# Mirage, refraction or reflection

by Andre
Tags: mirage, reflection, refraction
 PF Gold P: 5,450 So what is it? This is not a homework question. But I anticipate a good discussion.
 Emeritus PF Gold P: 8,147 Mostly refraction, but there might be some reflection in it. It's mostly a refraction of light from the sky, which looks like it's coming from the ground ahead, and is taken for a lake or something of the sort. We had a thread on the optical technicalities a while back. Does anyone know where it is?
 P: 85 U see, the air does not get heated uniformly. Instead, it gets heated in layers. Thus the optical properties of the diff layers of air will be diff. The light from a distant object rises up, and moves from an optically denser layer to an optically lighter layer and if the angle of incidence is such that the critical angle is achieved, TOTAL INTERNAL REFLECTION occurs. This causes an image of the object to be created on the surface at diff location which we observe as the mirage... Hope u understood...
P: 1,560

## Mirage, refraction or reflection

 Originally posted by sridhar_n (SNIP) The light from a distant object rises up, and moves from an optically denser layer to an optically lighter layer and if the angle of incidence is such that the critical angle is achieved, (SNoP)
Just out of curiosity, how does "light rise up"??
PF Gold
P: 5,450
sridhar_n

http://mintaka.sdsu.edu/GF/mirages/mirintro.html

 Reflections or Refractions? Pedants and purists insist that all mirages are purely refraction phenomena; and, strictly speaking, they are correct. Nevertheless, it is so useful to regard the classical inferior and superior mirages as approximately due to internal reflection that we cannot avoid considering that point of view — but with the caveat that certain subtle phenomena, which turn out to be essential for green flashes, are not explained by the reflection'' approach. Furthermore, one should bear in mind that the words for mirage'' in the major languages are all derived from terms meaning reflection.'' I have already pointed out that the French term mirage'' is derived from a phrase meaning to be reflected.'' In German and Dutch, the terms used translate literally to air-reflection.'' (Luchtspiegeling -Edit) Thus the notion of reflection is implicit in every discussion of mirages. I think this close association of reflection'' with the classical inferior and superior mirages is a good reason to distinguish them from the pseudomirages,'' which are purely refractive. Still, one can regard all these inverted images as real images (in one dimension) formed by a positive lens. From that point of view, even the classical mirages are refraction, rather than reflection, phenomena.
So is he right or is he wrong?
 PF Gold P: 5,450 In the mean time I found that other thread about mirages: http://www.physicsforums.com/showthr...&threadid=4656 but it is not really adressing the question "refraction or reflection?", so we can continue.
 Mentor P: 7,291 Why the or? A mirage is a reflection produced by refaction. The point is that a reflection is not a mechanism it is a result. When you see a refletion in a mirror there is a different mechanism (free electron mobility) which produces that reflection. Refraction is the apparent bending of light rays due to light passing an interface to a material with a different index of refaction. As far as I know reflection is not a carefully defined term.
P: 905
 Originally posted by Integral As far as I know reflection is not a carefully defined term.
I wanted to say that.

The physics of a mirage is understood. Light bends due to a gradient in the refractive index. The rest is semantics. I like sridhar_n's sig. It's a distinction some of my kids' teachers don't appreciate.
 HW Helper P: 2,327 My vote is that the bending of light is caused by semantics.
 PF Gold P: 5,450 Semantics? I don't think so. I think that sridhar_n is about right. As far as the input and output of a Mirage producing system is concerned I think it is 100% reflection and 0,0% refraction. Not even a little refraction as net result.
Mentor
P: 7,291
 Originally posted by Andre Semantics? I don't think so. I think that sridhar_n is about right. As far as the input and output of a Mirage producing system is concerned I think it is 100% reflection and 0,0% refraction. Not even a little refraction as net result.
Once again, refraction is the MECHANISM which produces a REFLECTION.
 PF Gold P: 5,450 There we are, this is the essential flaw in my humble opinion. Edit made a thinking error here. There is refraction going on but most certainly there are limits to refractions. Why is it that none of these pages explaining mirages do not substantiate it with Snellius law? Because it doesn't work. I knew we would have a good discussion [;)]
Mentor
P: 7,291
 Why is it that none of these pages explaining mirages do not substantiate it with Snellius law? Because it doesn't work.
Perhaps this failure to explain is because you are not reading a real Optics text. I have scanned a few pages from an old text I have, Optics by Rossi.

Warning for the bandwidth challenged this is a 2Meg PDF file.

light propagation in non homogeneous media

edit: spell check
 P: 419 Perhaps not that relevant, but we use a nice experiment at my school to show refraction of light. A long thin tank has extremely saturated salt solution in the bottom, to a depth of about an inch. Tap water is then slowly put on top of this (with a U shaped tube), to a depth of several inches. and allowed to stand for a few hours. A laser beam is then placed at the side of the tank, about midway up, and if you've done it right, the beam takes a beautiful curved path, dipping down to the bottom. Simple to do and VERY impressive. [:D]
Mentor
P: 7,291
 Originally posted by Adrian Baker Perhaps not that relevant, but we use a nice experiment at my school to show refraction of light. A long thin tank has extremely saturated salt solution in the bottom, to a depth of about an inch. Tap water is then slowly put on top of this (with a U shaped tube), to a depth of several inches. and allowed to stand for a few hours. A laser beam is then placed at the side of the tank, about midway up, and if you've done it right, the beam takes a beautiful curved path, dipping down to the bottom. Simple to do and VERY impressive. [:D]
I have done this, only used a sugar solution. If you get the laser beam the correct distance from the bottom of the dish it will be totally reflected from the bottom and mirror its entry path on exit.
 PF Gold P: 5,450 Well, impressive text book, Integral but did you notice that there is always a division by sin(phi). Sinusses can be zero. Let's look at Snellius law. R1*sin(phi-in)=R2*sin(phi-out). In which Ri is the refractive index of the medium and phi-in and phi-out obviously the angle of the light before and after refraction. We want to investigate what happens to the light direction after refraction. Therefore we have to reshuffle that law as Phi-out= arcsine(R1/R2*sin(Phi-in)) (angles in relation to the vertical plane) Now, weren't there some complications for an arcsine? The factor under the brackets must be less than or equal to one. Otherwise there is no solution. So let us take a closer look at that expression. R1 is the refractive index for the incoming light in the heavier colder air and since R2 is a smaller index for lighter warm air, the R1/R2 term is always more than one. Hence sin(Phi-in) must always be less than one or concequently Phi-in must be less than 90 degrees. Also, this means that there is a critical angle for Phi-in beyond which refraction is not possible. The formula also indicated that the maximum angle for the refracted light is 90 degrees. So if a light ray is refracted in another material with a horizontal surface, it will never exceed the horizontal, so obviously there is no such thing as refracting back. Yet we see things at and beyond the horizon apparently refracted. Ever been swimming underwater? When you look up you only see a circle of light directly above you. The more distant water surface around you is dark. Why? Because the critical angle has been exceeded and the light coming in too shallow is not refracting but either absorbed or reflected. Beyond that angle only the dark bottom under the water is reflected. Now look at this impressive picture of a superior (on top) mirage. http://www.thulebageren.dk/gallery/fata%20morgana.htm It's with an inverted mirage, due to very cold air on the surfaces, think the cold inversion being water instead of cold air, with the higher refractive index. The Light at the horizon cannot be refracted due to the critical angle. Instead, you see the bottom reflected and in this case that's the mountains. No wonder that they are upside down. Reflection has that mirroring habit. So this mirage is just the same as happens under water, reflection at the water surface , but now there is a critical layer somewhere where the reflection takes place. How about a regular mirage in the desert or on the road on a sunny day? Apart from turning the scenario upside down, it's exactly the same. Here several examples of inferior (Desert warm type) and superior (artic cold type) mirages: http://virtual.finland.fi/finfo/english/mirage.html So the ultimate trick is real reflection, just like sridhar_n indicated. Now how come that after ages of Snellius law and know mirages, we have always assumed a wrong explanation? Time to rewrite some textbooks.
 Mentor P: 7,291 It is still not clear to me where the incorrect explanation is. Seems to work to me. The angle at which total internal reflection occurs is called the Brewster angle, are you attempting to say that this is incorrectly handled? I think it is correctly handled. Perhaps you are reading different books then I am. You tone seems to indicate that you have read them all. Is this possible? Or are you making some assumptions? Edit: Rereading your post, and my text, it occurs to me that while you are impressed with it, you are ignoring it. Please read, and make an effort to understand the development. If you are unable to understand the approach, then please do not continue your criticisms of the general understanding, when it is YOU who is having the trouble.
PF Gold
P: 5,450
No I admit that I did not study it intensely and I will do it on your recommandation but the only thing that I tried to state that the limit of refraction is 90o as can be seen in the law of Snellius. The refraction of light grazing at the critical angle is exactly at 90o. Beyond the critical angle there is no refraction, just total internal reflection. You can read that in any textbook as well:

http://www.glenbrook.k12.il.us/gbssc...rn/u14l3c.html
 TIR (Total internal reflection) only takes place when both of the following two conditions are met: -a light ray is in the more dense medium and approaching the less dense medium. -the angle of incidence for the light ray is greater than the so-called critical angle.
Now the mirage object is always in the heavier cooler air, the more dense medium, regardless of an inferior or superior mirage. It approaches the warmer air, the less dense medium, either from aloft or below. So the first condition has most certainly been met.
The mirage is always an image of something close to the horizon where the angle of incidence is closing in on the limit of 90o. So for a mirage this conditions looks plausible.

Now please look at the first figure in the same link with the example of refraction at the critical angle (left) and total internal reflection (right) at the water surface from below. Now replace "water" with "denser colder air" and you've got yourself a superior mirage. Total internal reflection beyond the critical angle of probably something like 89,9o. Now turn the figure upside down and you're looking at the normal inferior mirage, famous for deserts.
http://www.glenbrook.k12.il.us/gbssc...rn/u14l3c.html

For a inferior mirage you could argue that the curvature of the Earth would bring the horizontal refracted light into cooler regions again causing it to refract up again, but that does not work for superior mirages where the curvature of the Earth opposes refraction.

edit after studying:
Integral, does your textbook actually say something else? I see equations 2-6 and the final one on chapter 2-2 of the example. Now try and use those in practice to see if you can get refraction beyond 0o or 90o depending upon the definition of phi. Notice that the textbook does not indicate that it has proved the bending back up with refraction.

Sometimes theory may need revisiting.

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