Mirage, refraction or reflection

[Andre]

So what is it?

This is not a homework question. But I anticipate a good discussion.

 

[selfAdjoint]

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?

 

[sridhar_n]

Mirages - A Reflection effect! - a consequence of Refraction

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...

 

[Mr. Robin Parsons]

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"??

 

[Andre]

sridhar_n

So i't is reflexion, Right. But how about this?

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?

 

[Andre]

In the mean time I found that other thread about mirages:

https://www.physicsforums.com/showthread.php?s=&threadid=4656

but it is not really adressing the question "refraction or reflection?", so we can continue.

 

[Integral]

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.

 

[krab]

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.

 

[turin]

My vote is that the bending of light is caused by semantics.

 

[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.

 

[Integral]

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.

 

[Andre]

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

 

[Integral]

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.


http://home.comcast.net/~rossgr1/Math/mirage.pdf

edit: spell check

 

[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.

 

[Integral]

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.

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.

 

[Andre]

Well, impressive textbook, 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.

 

[Integral]

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.

 

[Andre]

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 90^o as can be seen in the law of Snellius. The refraction of light grazing at the critical angle is exactly at 90^o. 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/gbssci/phys/Class/refrn/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 90^o. 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 89,9^o. Now turn the figure upside down and you're looking at the normal inferior mirage, famous for deserts.
http://www.glenbrook.k12.il.us/gbssci/phys/Class/refrn/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 0^o or 90^o 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.

 

[Andre]

then please do not continue your criticisms of the general understanding, when it is YOU who is having the trouble.

I do have some problems with this. Would this be the sole reason why it takes a generation before a correct paradigm shift can take place. I think that general understanding should be questioned at all times if not the Earth would still have been flat. Arent those forums to discuss those things. OK most challenges will be plain wrong. Please explain where I am wrong.

 

[turin]

Originally posted by Andre
Please explain where I am wrong.

You're not totally wrong, but I think that you're not considering the appropriate situation. This business with the critical angle and total internal reflection if for a stark boundary between two distinct media. Mirages and the like are phenomena that occur in continuous media (no boundaries, but gradients). You are correct in that total internal reflection occurs when the angle of incidence is greater than the critical angle (by DEFINITION), but you are not correct to assume that the phenomenon of a mirage is a total internal reflection, as there is no distinct medium to which it is internal.

Refraction at a boundary can be considered strictly using trigonometry in the ray theory of optics, but you mustn't forget that this is based on Hyugen's principle. When you shift your consideration to a continuous medium (gradients vs. boundaries), the intamacy to the principle is greatly amplified. You should take a step back and consider the light in terms of wavefronts rather than rays. The rays are like the vectors and the wavefronts are like the 1-forms. You can consider them as identical objects in a uniform medium and then only worry about them at a boundary (and thus use the ray theory), but, in a non-uniform medium, it becomes a bit more tricky.

 

[Andre]

Okay but that still does not legalize the refraction to exceed the critical angle.

Getting back on this

 

[Integral]

Originally posted by Andre
Okay but that still does not legalize the refraction to exceed the critical angle.

Getting back on this

Does it?

Sometimes theory may need revisiting.

Before you can make statements such as this you MUST fully understand the theory to be revisised. It is clear that you have only rudimentry knowledge of this subject. Search out higher level texts (mine is an undergrad Optics text).

 

[turin]

Originally posted by Andre
Okay but that still does not legalize the refraction to exceed the critical angle.

What I was trying to explain is that the critical angle concept is probably not the right way to think about the situation. If you consider the light in terms of phase-fronts, then the part of the phase-front in lower index of refraction will travel faster than the part of the phase-front in the higher index of refraction. For the desert sand in the heat of the day, the air just above the surface is very hot, and the temperature decreases with height above the surface, so the index of refraction is lowest closest to the surface, BUT IT'S ALL AIR, NO BOUNDARY. There isn't a critical angle in the sense that there isn't a boundary, and therefore there is no question whether the light will remain in the medium; it will. If you want to think about the critical angle, then it IS 90 DEGREES less some MINUTE deficit. Why? Because the difference in the index of refraction from infinitesimal layer to adjacent infinitesimal layer is practically zero. That's why it is better to think of the problem in terms of gradients.

 

[ketch]

Why does no body even try to explain the phenomenon from beginning to end? I'm quite ignorent, studied physics 25 years ago and hardly even remember Snellius law. My daughter wanted to make a schoolthesis on the subject and found no literature describing it in detail. We only red a popular but nice text from Minnaert.
Now we are so ignorent/arrogant that after quite some discussion we think we understand it all and then we found this (a bit disappointing)discussion so we give it a try:

1- Mirage concerns a virtual image as a result of a vertical temperature gradiënt and thus a density and thus a n-gradiënt (diffraction index).
2- Light propagating through this gradiënt wil curve towards the higher n-side.
3- Because off the gradiënt no critical angle can be identified (or asymptotic to 90degr.), the medium can be seen as a multi layer with infinetesimal small differences off n.
4- The curving process continues until the light propagation is asymptoticly horrizontal.
5 Then we have a problem. With the preassumption off a ideal medium (exact horrizontal isotherms an consequently exact vertical n-gradiënt the beam is captured in the horrizontal plane because it propagates perpendicular to the n-gradiënt.

6 Up to this point we can draw a first conclusion.
From Snellius it follows that for a multi layer medium, diffraction depends only from n of the first and the last layer. Consider a point object at a certain level where n=n1 and angel Q1 of the departing beam, and consider n=n2 at the level at which the beam is appr. horrizontal.(Q2=90degr) Then we CONCLUDE that:
On the MICRO LEVEL the beam curves increasingly from layer to layer (with infinetisemal small differences of n) until horrizontal, so no critical angle can be identified and no total reflection takes place.
On a MACRO LEVEL the relation between Q1, n1 and n2 is described just AS IF Q1 is the crical angel.
(However if we consider the same point object from which departs a beam with a smaller Q, this Q can also be considered as a Critical angle, be it in a other macro system: the horrizontal will be reached at a lower point with a smaller n. The smallest possible Q is from the beam which reaches horrizontal at ground level.)

7 The first escape from this capture to the horrizontal is to take in account the curving off the Earth and thus the curving of isotherms etc. In this forum this is correctly rejected, it only works with upward bending (positive upward n-gradiënt). With downward curving (in the case of inversion = negative n-gradiënt) the beam will be captured in the horrizontal isotherm and curve with the rounding of the Earth but not curve back to the earth.(It also cannot escape upward because pushed backward by the negative n-gradiënt;would it go in circles around the Earth for ever?)
8 The second escape from the horrizontal capture is to take in account the imperfection of the medium. In real life isotherms are not perfect horrizontal. The temperature gradiënt results in turbulations of the air and fluctuating isotherms. Consider the isotherm as (slightly) sinusoïdal, then also the horrizontal beam meets a gradiënt and tends to bent towards the positive of the n-gradiënt. So it can escape from the horrizontal and then curves further upward. Consequently bending backwards can be explained in both situations: warm earth/cold air as well as in the case of inversion.

8 We can conclude that air turbulations can explain curving back, which then still is a refraction phenomenon.

9 This model of continuïng curving also explains the compression of the virtual image which often takes place. If the observer is at a lower altitude then the object the Qin is smaller then Qout (Qin is the departing beam angle at the top of the object where Qout is the angel at which this refracted beam is observed). In fact the difference between Qin and Qout depends on the difference between n at the level of the object and n at the observers level. If n at the upper part of the object is (about) the same as n at the observer, the bending of the downward curve is (about) the same as the bending of the upward curve, Qin equals (about) Qout and the image is not compressed.