Understanding Refraction and Snel's Law: Common Questions Answered

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In summary: The constant for light entering a medium at an angle less than 90 degrees is called the "Snell angle". The constant for light entering a medium at an angle greater than or equal to 0 degrees is called the "Rayleigh angle".In summary, Snell's law states that the speed of light in a medium is less than the speed of light in a vacuum. The Snell angle is the constant for light entering a medium at an angle less than 90 degrees. The Rayleigh angle is the constant for light entering a medium at an angle greater than or equal to 0 degrees.
  • #1
Nacho
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I've seen the diagrams they write up talking about refraction for ages .. where a light ray goes into a different medium of higher index of refraction at an angle, and then the light ray is bent. And that is used to illustrate Snel's law. I guess I haven't paid close enough attention!

I'm reading this again, in Mark P Silverman's book "Waves and Grains", and paying attention, but I got some questions .. if I'm interpreting things right.

Mark shows Snel's law is actually a series of laws:

sin phi/sin theta = n1/n2 = v1/v2 = y1/y2 = p1/p2

phi = angle of incident to normal
theta = angle of refraction to normal
(1) = incident side (linear propogation)
(2) = refracted side (linear propogation)
n = index of refraction
v = velocity of the light
y = wavelength
p = momentum

I got 2 questions. Assume you have a pool of water, and shining light through the pool, the light entering the water at the top at an angle and going out the bottom of the pool.

Q1) Since water bends the light towards the normal, it has a higher index of refraction, so the light slows down in the water. If you take the time a ray was in the water, and divide that by the depth of the water (not the distance the light moved through the water), wouldn't you come up with the speed of light in a vacumn? I mean, if you just consider the depth the light moved in the water, not its distance, it wouldn't have seemed to slow down any. Is that right?

Q2) Now suppose the light goes straight down into the water. I was going to say a lot here .. but let me just ask: What would that do to Snel's ratio? Wouldn't it be sin(0)/sin(0) (maybe sin(90)/sin(90))? Wouldn't that also mean that the velocity of the light would not change while the light was in the water in this instance?

I'm real confused here ..
 
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  • #2
Concerning question 2, when light is normally incident on a change in medium it does not bend but it does change speed. So when light goes from air to water at normal incidence (0 degrees from normal) it does not bend when it travels through the water but does slow down.
 
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  • #3
mmwave,

Thanks for your response. But I got a question. You said:

Concerning question 2, when light is normally incident on a change in medium it does not bend but it does change speed. So when light goes from air to water at normal incidence (0 degrees from normal) it slows down when it travels through the water but does slow down.

What I bolded .. I'm not sure what you meant, "it slows down .. but does slow down". Any clarification?
 
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  • #4
If you look through the fourms you will find many threads addressing the propogation of light through a medium. In reality light does not "slow down" in water. Up on intering the water a photon is adsorped by a water molecule leaving the molecule in an excied state. A short time later a photon is reemitted. Each emission is in a random direction, perhaps with some favor in the original direction due to momentum consideraions.

The net effect is a delay in the passage of a photon through a medium, this delay depends on the atomic structure of the medium, thus differing index of refractions.

Ok for your quesions, for Snells law measure the angles from a Normal line (this is a line perpendicular to the surface in all directions) A ray entering on the normal passes through the material undeflected, though it does "slow down" as discribed above. If you use the time required to pass through the material and the thickness of the material to compute the speed you should arrive at a velocity in the material which should be less then the speed of light in vacuum. Not sure what you mean.

Ok, let's express Snells law as N1Sin(Θ)=N2Sin{φ)
Now if your angle of entry is Θ =0 then only φ=0 can hold.

Edit: I forgot to put in my plug for Fynman's book QED. If you are interested in how light works find and read a copy of this little book. It is a collection of lectures he gave which are pointed at the layperson.
 
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  • #5
Integral,

Ok, let's express Snells law as N1Sin(È)=N2Sin{ö)

Thanks, you cleared one thing up for me .. by your cross product you got rid of the division by zero (Sin(È)), which was throwing me, but it still has a problem (or, a problem in my mind). That makes the constant zero, if the light goes in through the normal. That constant is not zero if the light goes into the medium at any angle less than 90 degrees and greater than 0 degrees to the normal. This has me stumped .. the only way out I see is that if the light goes in through the normal, it's not refraction anymore, so Snel's law doesn't apply anymore?, i.e., I'm would be looking to the wrong thing for answers?

If you use the time required to pass through the material and the thickness of the material to compute the speed you should arrive at a velocity in the material which should be less then the speed of light in vacuum. Not sure what you mean.

I didn't make that plain enough, I guess. What I meant, and it's important to remember that the light is entering the water and exiting it at the bottom, sometime later, though that time is just mili/micro seconds. Here goes: Measure the amount of time it took the light to travel through the water, on the path it took through the water, that path being at an angle to the normal. Divide the distance it traveled by that time and you're going to get a meters/second somewhat less than the speed of light in a vacumn. That is just a statement of what happens .. no question/disagreement here.

Next, measure the distance the light would have traveled if it had traveled on the normal. Divide that distance by the time it took the light to travel through the water on the actual path it took through the water, measured above. That I'm saying (asking), would be the same as the speed of light in a vacumn. That's what it says in the book I referenced, but I was wanting to make sure that I wasn't just mis-reading it, which is entirely possible. Because, in the book he (Mark Silverman) is hopping around, talking about this setup in how it would be (wrongly) interpreted during Newton's time, and how it was eventually interpreted during Fresnell's time. I'm not at all sure he wasn't talking about the wrong interpretation when he wrote that. I can give you a quote to the book if it would help.

Edit: I forgot to put in my plug for Fynman's book QED.

I have that book and have read it a couple of times. In fact, I made a reference to it a couple days ago on another topic that had something in it about refraction, possibly the one you mentioned. I think it is a horrible little book! But, I think you are talking about towards the end of the book where Feynman goes into the molecular/atomic reasons for refraction. I've read that and re-read that a bunch of times. And I don't think it does really does get into the real/atomic reasons. Throughout the book, and even in this part, he always talks about amplitudes. And in this part he is basically saying that the amplitudes for the photon scatter throughout the medium cancel each other out somehow, almost magically, except in the direction that the light winds up traveling, but never telling us why the amplitudes/scatter cancel out. It's not a physical process he is talking about .. he is still talking about amplitudes, of which he never goes further with a description (as he might well not be able to).

Thanks for your help.
 
  • #6
Thanks, you cleared one thing up for me .. by your cross product you got rid of the division by zero (Sin(È)), which was throwing me, but it still has a problem (or, a problem in my mind). That makes the constant zero, if the light goes in through the normal. That constant is not zero if the light goes into the medium at any angle less than 90 degrees and greater than 0 degrees to the normal. This has me stumped

What this means is that you will not be able to learn anything of the index of refraction by looking at the undeflected beam.
by your cross product

Cross product has a well defined meaning in Math Physics, this is not a cross product. It is a simple algebraic expression.



As for the rest of your question, it is not clear to me. What proof was presented in the book you refer to? You may want to read it over several times, very carefully. Examine closely any algebra given. I have looked briefly and did not see an obvious geometric/algegbraic proof of that claim. But then I still have a pretty fuzzy picture of situation you discribe. Your first effort is clearly wrong as you say divide time by distance to get a velocity, this cannot be.
If you take the time a ray was in the water, and divide that by the depth of the water (not the distance the light moved through the water), wouldn't you come up with the speed of light in a vacumn?

Fynmans discribtion is completely physical, why would light not have an amplitude?
 
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  • #7
Integral,

What this means is that you will not be able to learn anything of the index of refraction by looking at the undeflected beam.

I agree, that answers one of my questions. Thanks for your patients.

Cross product has a well defined meaning in Math Physics, this is not a cross product. It is a simple algebraic expression.

Well, if you take one of the relations that I posted:

sin phi/sin theta = n1/n2

And do a "cross product" multiplication, you come up with the relation you posted:

N1Sin(È)=N2Sin{ö)

I call that "cross product". That's what I learned to call it in Algebra I, 35 years ago when we were working on "relations" (ought Oh -- age telling)! You multiply the numerator of one side by the demominator of the other side, and visa versa, and you come up with the same equality. I guess I'm getting too old! And it does away with the division by 0, one of my worries.

I did re-read the part in the book. I'm sorry, I did misrepresent it, but my main question just gets rephrased. Here is an updated question, I hope this works:

---------\-----------------------|-------------------
|...\......|.....|
|...\.....|.....|
|...\......|.....|
|...\......|.....|
|....\.....|.....|
|.....\....|.....|
|...\...|.....|
|...\...|.....|
|....\.....|.....|
|.....\....|.....|
|......\...|.....|
---------------------\-----------|-------------------

I have no software to make gifs or images, so I'm hoping this will do, and will come though the posting program here OK. That is supposed to be a tank of water. The dots (.) don't mean anything. I just put them in as a place holder so that the browser would keep the object as a rectangle, rather than spaces, which it would remove. Imagine they are not there.

"\" is the path the light takes through the water, after it is refracted. I didn't show the path the light took before it went into the water as I didn't have another symbol that I could use that is at a different angle than "\". Think of that path EXTENDED THROUGH THE WATER, and call it "?".

"|" is the path of the normal, and I from what we have said, the path with no refraction, where Snel's law would give us meaningless information. I was asking about this, but forget it for now.

OK, d(\) is the distance the light travels through the water.

t(\) is the time it takes the light to travel that path through the water.

So,

d(\)/t(\) is the speed of light in that medium, water. We should be square with each other here ...

What I'm saying (asking), would d(?)/t(\) yield the speed of light in a vacumn, or in the medium above the water, or nothing? Note that I didn't use d(\) there. I used d(?). This is my main question. I think I'm basically asking: is the amount of the angle of refraction purportional someway to the amount the light is slowed down in the medium?

Fynmans discribtion is completely physical, why would light not have an amplitude?

Oh, I'm not trying to say that. Clearly light does have an amplitude for interactions with matter -- it's part of the duality of light. I was getting at this: At the beginning of the book, when he starts talking about partial reflection through plates of glass, he writes his little vectors, which I took as really amplitudes. And he says they don't know why they work in describing the interactions, just that they do work. Then we get back into the book where I had thought he was going to describe what was going on with at an interaction level in reflection/refraction, and it seemed he was talking about that, yet he still termed it in amplitudes, which he stated before he (we) didn't know why the amplitudes gave us the right answer .. just that the did give us the right answer. It seemed like a big cop-out.

His description of what was happening at the interaction level didn't help clear up anything to me. He really didn't say anything about the interactions that were occurring, and why. He just said that of all the interactions that could occur, the amplitudes of the different interactions canceled out in every direction other than the direction the light actually traveled through the medium. Looking back, I don't know what more I expected to be said about it, or even what could be said. It was a big buildup for nothing. It caught my eye because I had never ever read a detail of what was happening in the medium to cause reflection or refraction.

I hope I haven't hit a nerve here with the book. I guess it really describes what is known .. I was just looking for more I guess.
 
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  • #8
What I'm saying (asking), would d(?)/t(\) yield the speed of light in a vacumn,

Think about it, it can't be. Your t(/) must be greater then t(?)correct? In a vacumn the time for light to travel the distance d(?) is d(?)/c, this is the absolute minimum for t(?). We must have t(\) > t(?) so your quantity d(?)/t(\) is less then c.

You must remember that Fynmans book is written for the layperson so all math has been left out. What remains is a discription, not in complete detail, of what is happening.

When you are attempting to get a grasp on concepts without the depth of understanding provided by the math there is much lost.

Sorry
 
  • #9
You know it just dawned on me that there might be an experiment that would tell if refraction is accurately modeled by amplitudes (wave nature of light) or photon/electron absorbtion/re-emission (particle nature of light).

It's basically the same test for photo-electrons that Einstein used to show the particle nature of light.

I'm thinking varying the frequency of light VS varying the intensity of light in the medium should do different things based upon the correct model. And then correlating each of those back to the amount of refraction with constant frequency and intensity.

If it's "classic", there shouldn't be any difference of varying either, except at some point (for either variance) the medium would get "saturated" I guess, and fry.

If it's "particle", shouldn't there be differences in refraction in varying one of those things?
 
  • #10
Originally posted by Nacho
So, take light and shine it through a medium, but pulse it with a very short pulse. The light should be coming out of the medium over a longer period of time than the pulse, as some of the photons got delayed more than others.
To put what Hurkyl said another way, you get a very STEEP bell curve.

Think about the sun in my example. The atmosphere scatters blue light, making the sun appear red. But the sun only appears really red when near the horizon because most of the sun's light makes it to your eyes without hitting a nitrogen atom when it is high in the sky.
 
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  • #11
Nacho,
I don't think that refraction (or reflection) can be accurately modeled as electrons absorbing and re-emitting the photons.
It had better be, because that is what is happening.

It is to bad that you cannot accept what you are being told, make an effort to understand it. Go back to Fynman, read it again, then read it again. He is not lying to you, but he is presenting in the simplist form in order that people with no math (or only 25 yr old algebra) BTW it has been over 25 years since I graduated so quit pulling the old man story, I bet I got'ch beat!

If you are unable or unwilling to take our word for what is happening this conversation can go nowhere. We cannot prove anything here, only attempt to explain to the best of our understanding and ability what we know.

The answeres you seek will not be found in an E&M book, my Optics book by Rossi has a good presentation of much of this, but it is done in terms of higher math, if you think a cross product is simply cross multiplication, you will have a steep learning curve.

Good luck, keep asking quesitions but please attempt to understand what is being said. I make a strong effort to keep this forum true to the physics as I understand it.
 
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  • #12
Integral,

If you are unable or unwilling to take our word for what is happening this conversation can go nowhere. We cannot prove anything here, only attempt to explain to the best of our understanding and ability what we know.

I don't know where you come up with "unable or unwilling". What I have been posting has always been crouched in "my understanding", "I don't think", "I think", or in a question. In case you don't know, those are offers for discussion, learning.

Don't pat yourself on your back so hard -- you haven't attempted to "explain to the best of our understanding and ability what we know". You've basically said "that's the way it is, don't question it, it can't be any different than the way I said, accept it even though you don't know why you should accept it". That is dogma to me, and a lot worse than ignorance. It's a good way to get your kids to eat their Cheerios, nothing more though.

It is to bad that you cannot accept what you are being told, make an effort to understand it. Go back to Fynman, read it again, then read it again. He is not lying to you, but he is presenting in the simplist form in order that people with no math (or only 25 yr old algebra)

I definitely hit a nerve dissing the book. Too bad he's not around to ask questions, to discuss what he's said. I wonder if he'd tell me to accept it cause he told me so, or would make an attempt to clear up lingering questions I had?

BTW it has been over 25 years since I graduated so quit pulling the old man story, I bet I got'ch beat!

Projection on your part .. I wasn't pulling up "the old man" story. I was relating difficulties in learning, which can occur at any age.

I make a strong effort to keep this forum true to the physics as I understand it.

That's fine, your forum, but if it should ever get past your level of understanding, what, you cut the discussion off?

The topic took a turn .. from trying to understand a bit more of Snel's law to a discussion for reasons for refraction. Maybe that part would have been more appropriate on another forum.
 
  • #13
Lets get back to snells law. Do you follow my agrument of why the distances and time you discribed cannot be c?
Side note:
What you are referring to is simply cross multiplication. The result is a product. Cross multiplication refers to the action taken, the products are the result.

A cross product is a vector operation, the cross product of 2 vectors yields a vector normal to the plane defined by the original 2 vectors.
To optain the a 3D cross product constuct 3x3 a matrix with the vector components (i,j,k) it the top row and the components of the 2 vectors, in the second and third row, then compute the determinate.

let the first vector be xi+yj+zk and the second Xi+Yj+Zk, Then you get the matrix:

|i,j,k|
|x,y,z|
|X,Y,Z|

This yields a vector (yZ-zY)i - (Zx-zX)j + (xY-Xy)k
 
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  • #14
In reality light does not "slow down" in water. Up on entering the water a photon is adsorped by a water molecule leaving the molecule in an excied state. A short time later a photon is reemitted. Each emission is in a random direction, perhaps with some favor in the original direction due to momentum consideraions.

When dealing with 'velocity' I guess we are dealing with total displacemet/time - using that definition the velocity of light does vary as it goes through various media.

I do find what you said fascinating - am I right that in reality it only actually travels through a vacuum (at 'c') - the rest of the time it is undergoing a delay as it is absorbed and then re-emitted?[?]
 
  • #15
Originally posted by Cyberphysics
When dealing with 'velocity' I guess we are dealing with total displacemet/time - using that definition the velocity of light does vary as it goes through various media.

I do find what you said fascinating - am I right that in reality it only actually travels through a vacuum (at 'c') - the rest of the time it is undergoing a delay as it is absorbed and then re-emitted?[?]

That is correct. Due to this process, I believe that that transist time for a photon to reach the surface of sun, from the core of the sue, is on the order of 10,000yrs.
 
  • #16
I can see that as a poor 'in a loop of delivering curriculum' teacher this forum is going to be a Godsend! A means of taking my understanding just that little bit deeper without having to do all of the math - which is now VERY rusty lol!

Thank you for that insight - I have tweaked my site pages on refraction for my pupils and will be able to stretch their understanding (and interest) just that little bit more.

I have had people argue that text-books are 'wrong' in speaking of light 'slowing down' before but I have not had someone explain it so simply and clearly before. Thanks!
 

FAQ: Understanding Refraction and Snel's Law: Common Questions Answered

What is refraction?

Refraction is the bending of light as it passes through different mediums, such as air, water, or glass.

What is Snel's Law?

Snel's Law, also known as the Law of Refraction, is a formula that describes the relationship between the angle of incidence (the angle at which light enters a medium) and the angle of refraction (the angle at which light bends as it passes through the medium).

Why does refraction occur?

Refraction occurs because light travels at different speeds in different mediums. When light enters a medium at an angle, one side of the light wave slows down before the other side, causing the light to bend.

What are some real-life applications of refraction?

Refraction has many practical applications, including in the design of lenses for eyeglasses, cameras, and telescopes. It is also used in the creation of optical illusions and in the study of the human eye.

How can I calculate the angle of refraction?

The angle of refraction can be calculated using the formula n1sinθ1 = n2sinθ2, where n1 and n2 are the refractive indices of the two mediums and θ1 and θ2 are the angles of incidence and refraction, respectively.

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