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Physics Snell's Law and complex refractive indices

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
OK, I've looked around the net but there just doesn't seem to be a straightforward answer that doesn't require you to be an electromagnetics guru, so I'm hoping someone on here can help me out.

I have a light wave incident to a planar interface, coming from a medium with complex refractive index $n_1 + i k_1$, the other side of the interface is a medium with complex refractive index $n_2 + i k_2$. The light wave is incident with the interface normal at an angle $\theta_\text{i}$. You can assume optical wavelengths.

1. What is the (real) angle of refraction $\theta_\text{t}$? Snell's Law here gives a complex angle, which doesn't really make much sense to me. What direction does the refracted wave go in?

2. Is there a phase change upon reflection or transmission? I know about the $\pi$ phase change upon reflection from a medium with higher refractive index, but what does "higher" mean when the indices are complex? Higher real part, or modulus, or something else? And is there any other phase change?

Basically, I'm confused big time about reflection/refraction from absorbing media.

For question 1, I found the following expression in "Descartes-Snell law of refraction with absorption":

$$\frac{\sin \theta_\text{i}}{\sin \theta_\text{t}} = \frac{n_2 \left ( 1 - \frac{k_2^2}{n_2^2} \right )}{n_1}$$

Which looked promising, but it does not work at all when $k_2 \geq n_2$, and for instance for copper, $k = 2.558$ and $n = 1.133$ for 500nm wavelength, so either I am failing to understand the equation, or my data is wrong.

I'm not looking for a "physics student" answer, but actual equations for $\sin \theta_\text{t}$, I'd just like to figure this out so I can move on to the rest of my project, since I need to model absorbing media. Thanks for your insight!
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
OK, I've looked around the net but there just doesn't seem to be a straightforward answer that doesn't require you to be an electromagnetics guru, so I'm hoping someone on here can help me out.

I have a light wave incident to a planar interface, coming from a medium with complex refractive index $n_1 + i k_1$, the other side of the interface is a medium with complex refractive index $n_2 + i k_2$. The light wave is incident with the interface normal at an angle $\theta_\text{i}$. You can assume optical wavelengths.

1. What is the (real) angle of refraction $\theta_\text{t}$? Snell's Law here gives a complex angle, which doesn't really make much sense to me. What direction does the refracted wave go in?

2. Is there a phase change upon reflection or transmission? I know about the $\pi$ phase change upon reflection from a medium with higher refractive index, but what does "higher" mean when the indices are complex? Higher real part, or modulus, or something else? And is there any other phase change?

Basically, I'm confused big time about reflection/refraction from absorbing media.

For question 1, I found the following expression in "Descartes-Snell law of refraction with absorption":

$$\frac{\sin \theta_\text{i}}{\sin \theta_\text{t}} = \frac{n_2 \left ( 1 - \frac{k_2^2}{n_2^2} \right )}{n_1}$$

Which looked promising, but it does not work at all when $k_2 \geq n_2$, and for instance for copper, $k = 2.558$ and $n = 1.133$ for 500nm wavelength, so either I am failing to understand the equation, or my data is wrong.

I'm not looking for a "physics student" answer, but actual equations for $\sin \theta_\text{t}$, I'd just like to figure this out so I can move on to the rest of my project, since I need to model absorbing media. Thanks for your insight!
Hi Bacterius!

Sorry for playing the role of "physics student". ;)

As you can see in wiki:
When light passes through a medium, some part of it will always be absorbed. This can be conveniently taken into account by defining a complex index of refraction,
Here, the real part of the refractive index
indicates the phase speed, while the imaginary part
indicates the amount of absorption loss when the electromagnetic wave propagates through the material.

In other words, you should do your calculations with just the real part and you're good to go.

I'm not aware of this formula from "
Descartes-Snell law of refraction with absorption" you quote... it doesn't seem right to me.
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Thanks for the reply ILikeSerena - but I still have a few questions.

1. If I could just chop off the imaginary part and keep the real part for the angle calculations, why make it a complex number? Why not two different variables? There must be some way this complex number, as a whole, is advantageous in some calculations?

2. I found this link online: Snells Law going into metal *

I quote:

Okay, so I have a complex refractive index for a metal. I want to
calculate the direction of the ray going into the metal (to be quickly
absorbed of course). Do I just use Snell's Law with the real part of
the refractive index?

...

Use the complex refractive index and get a complex angle of refraction.
Then you just have to figure out what the complex angle of refraction
means.

...

No. Use Snell's law with the full blown complex index. Calculate the complex
angle of refraction. Then you can interpret what that means.

Using the methods in Ramo and Whinnery, you can calculate how an oblique
plane wave refracts into and propagates within the metal.
They could all be wrong, of course but I am still confused. I consulted the links but.. I don't understand a thing (Nerd)

I know the absorption coefficient is also used in the Beer-Lambert law to attenuate light waves exponentially with distance travelled, but there seems to be *something* additional happening at the interface with complex refractive angles.

But Fresnel's been kicking my butt all night with his equations... I will probably just ignore the imaginary part until this is resolved, but apparently that's not good enough for metals :(

Besides, anything that looks like an imaginary number was tacked on it just feels like magic to me. How do physicists come up with this stuff?! (Smoking)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Just as a side note: using complex numbers in the middle of a calculation, and then taking the real part in the end is a standard procedure in a number of disciplines such as circuit theory. The difference is that complex arithmetic essentially gives you vector arithmetic in a convenient form. That's why it is sometimes incorrect to use real arithmetic in the middle of the calculation.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Thanks for the reply ILikeSerena - but I still have a few questions.

1. If I could just chop off the imaginary part and keep the real part for the angle calculations, why make it a complex number? Why not two different variables? There must be some way this complex number, as a whole, is advantageous in some calculations?

2. I found this link online: Snells Law going into metal *

I quote:



They could all be wrong, of course but I am still confused. I consulted the links but.. I don't understand a thing (Nerd)

I know the absorption coefficient is also used in the Beer-Lambert law to attenuate light waves exponentially with distance travelled, but there seems to be *something* additional happening at the interface with complex refractive angles.

But Fresnel's been kicking my butt all night with his equations... I will probably just ignore the imaginary part until this is resolved, but apparently that's not good enough for metals :(

Besides, anything that looks like an imaginary number was tacked on it just feels like magic to me. How do physicists come up with this stuff?! (Smoking)
It fits conveniently.

Electromagnetic waves are usually modelled with an equation of the form $e^{i(\mathbf k \cdot \mathbf x - \omega t)}$.
Put in an imaginary constant and you'll get something which contains $e^{i(n + i \kappa)} = e^{-\kappa}e^{in}$.
In other words, the real part still behaves like a "regular" refraction, but you get an extra component for free that describes how the wave is reduced in intensity.
Somehow it feels right: it fits so beautifully in a formula, just as if nature intended it that way.

As for your quote: I guess it's fine to find a nice complex refraction angle - just be aware that it is the real part that identifies the angle, whereas the imaginary part identifies how much its intensity diminishes (the quote does suggest that you need to know how to interpret it).
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
It fits conveniently.

Electromagnetic waves are usually modelled with an equation of the form $e^{i(\mathbf k \cdot \mathbf x - \omega t)}$.
Put in an imaginary constant and you'll get something which contains $e^{i(n + i \kappa)} = e^{-\kappa}e^{in}$.
In other words, the real part still behaves like a "regular" refraction, but you get an extra component for free that describes how the wave is reduced in intensity.
Somehow it feels right: it fits so beautifully in a formula, just as if nature intended it that way.

As for your quote: I guess it's fine to find a nice complex refraction angle - just be aware that it is the real part that identifies the angle, whereas the imaginary part identifies how much its intensity diminishes (the quote does suggest that you need to know how to interpret it).
Thanks for the answer. So what I should do is compute the sine using Snell's with the real part of the refractive index and just use the imaginary part with e^(-kx) to model absorption (that I know how to do :D)?

Reduced in intensity as in, as it travels within the medium? Or a sharp intensity loss at the interface? Is there *ever* any non-negligible energy loss at an interface, at least at optical wavelengths? Thought I'd ask :)

It's starting to make a lot more sense now, electromagnetics can be very confusing for a poor computer graphics developer just trying to get things right! (Tmi) Thanks to everyone who answered by the way!
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Thanks for the answer. So what I should do is compute the sine using Snell's with the real part of the refractive index and just use the imaginary part with e^(-kx) to model absorption (that I know how to do :D)?

Reduced in intensity as in, as it travels within the medium? Or a sharp intensity loss at the interface? Is there *ever* any non-negligible energy loss at an interface, at least at optical wavelengths? Thought I'd ask :)

It's starting to make a lot more sense now, electromagnetics can be very confusing for a poor computer graphics developer just trying to get things right! (Tmi) Thanks to everyone who answered by the way!
Yep.

In glass the decay is not noticeable and $\kappa \approx 0$.
In metal the intensity decays exponentially.
The word skin depth is used to identify the distance where the intensity has decreased by $e^{-1}$.

As far as I know, there is no noticeable loss at the interface.
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Thank you again :)

One last question: when you say in glass the absorption coefficient is almost zero, it still decays exponentially, right? (just really slowly?). It doesn't "become linear" at some critical coefficient or something like that :confused: just checking I've got everything
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Thank you again :)

One last question: when you say in glass the absorption coefficient is almost zero, it still decays exponentially, right? (just really slowly?). It doesn't "become linear" at some critical coefficient or something like that :confused: just checking I've got everything
Yes, it always decays exponentially.

And it's almost zero... for visible wave lengths.
Infrared now is an entirely different matter - glass looks black where infrared is concerned (try it with an infrared camera).

I used to work for a company working with infrared leds.
At some point in time I complained that I couldn't see whether they were on or not.
I had a colleague that was surprised - he said he could simply see it.
Then I learned that when I took my glasses off, I could see it! :D
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Suppose the intensity would decay linearly in glass at some point.
Then at a distance very very far into the glass, it would reach zero... and then become negative...
That's a contradiction.
Therefore a wave does not decay linearly. $\qquad \blacksquare$