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Snell's Law - complex angle of refraction (need complement )

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einfall
#1
Nov16-12, 07:12 AM
P: 11
I'm tying myself in knots trying to calculate something I think should be very simple. I'm writing/debugging some code at the moment, and I simply don't know if this silly problem is the source of the major errors I'm currently seeing. Some help would be hugely appreciated.

BACKGROUND:

For a standard Snell's law calculation we can write [itex] n_{1}sinθ_{i} = n_{2}sinθ_{t} [/itex]. All angles are defined relative to the normal.

But when the incident angle [itex]θ_{i}[/itex] exceeds the critical angle [itex]θ_{crit}[/itex], we get total internal reflection (TIR) and the transmitted angle [itex]θ_{t}[/itex] is complex (but still calculable!) and so in general no energy is propagated through the interface. For my application, the evanescent wave which is generated at a TIR interface is important and so I often need to be able to calculate complex [itex]θ_{t}[/itex].

MY PROBLEM:

In the usual case where [itex]θ_{i}[/itex] < [itex]θ_{crit}[/itex], we can obviously define the complement of the transmitted angle [itex]θ_{t}[/itex] as Pi/2 - [itex]θ_{t}[/itex]. This is simply the angle of the refracted ray relative to the interface rather than relative to the normal.

But when [itex]θ_{i}[/itex] > [itex]θ_{crit}[/itex] and [itex]θ_{t}[/itex] is explicitly complex, how can I calculate the "complement" of this? I definitely need a numerical value for this "complement" for subsequent calculations, but I'm not sure how to go about calculating it and I'm driving myself slowly insane because it seems like it should be very simple. The attached sketch might clarify my question.

POSSIBLE SOLUTIONS:

1) Simply taking Pi/2 - [itex]θ_{t}[/itex]. Since this completely ignores the complex component, I simply don't know if this is conceptually sensible but it is what I have been doing.

2) Let [itex]z_{1} = θ_{t}[/itex]. Take the principal argument of [itex]z_{1}[/itex], [itex]Arg|z_{1}|[/itex], to get the [itex]θ[/itex] defined in [itex] z_{1} = r(cos θ + i sin θ)[/itex] since [itex]θ = Arg|z_{1}|[/itex] and then calculate [itex]r[/itex] by rearranging. Note that [itex]r[/itex] shouldn't change under rotation. Calculate [itex]θ_{new} = Pi/2 - θ[/itex] and then substitute this back into [itex] z_{2} = r(cos θ_{new} + i sin θ_{new})[/itex]...... so then the angle I want (the "complement") is assumed to be [itex]z_{2}[/itex]...... I really don't think this approach makes sense, but it's where my mind has wandered to.
Attached Thumbnails
sketch.jpg  
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mfb
#2
Nov16-12, 08:21 AM
Mentor
P: 11,815
The period of sin remains 2pi (real) even in the complex numbers. pi/2-θ looks like a reasonable value for a "complement", but it really depends on the way you plan to use that complement.
einfall
#3
Nov16-12, 08:36 AM
P: 11
Thanks for your tentative confirmation.

In what way would it depend on the intended use of that "complement"? Surely it's a geometrically well-defined complex number?

n.b. In every use of this "complement" θ (let's call it θ'), it's called as cos(θ') so the implicit infinity of periodic solutions is irrelevant.

mfb
#4
Nov16-12, 08:54 AM
Mentor
P: 11,815
Snell's Law - complex angle of refraction (need complement )

All possible complements are well-defined numbers.

In what way would it depend on the intended use of that "complement"?
The word "complement" itself does not include any physical use. The correct number depends on the physical application of your complement.
einfall
#5
Nov16-12, 09:05 AM
P: 11
I certainly take your point. I should clarify - my question relates specifically to the "complement" of an complex angle of refraction, and its physical interpretation in the context of total internal reflection.

In this case, the complement I mean to obtain is the one with physical significance in the context of Snell's law (or the equivalent Fresnel expressions) when the incident angle exceeds the critical angle. If Snell's law was defined with angles relative to the interface, at TIR this refracted angle would be the complex number I am looking for.

I still confused about whether simply taking [itex]Pi/2 - θ_{t}[/itex] makes physical sense in this context.
DrDu
#6
Nov16-12, 09:38 AM
Sci Advisor
P: 3,593
It is still not clear to me what you are after, but the Snell angle is the angle between the wavevector k and the normal vector n, i.e. [itex] \cos(\theta)=\vec{k}/|k|\cdot \hat{n}[/itex].
You are interested in the angle [itex] \cos(\theta')=\vec{k}/|k|\cdot \hat{n'}[/itex]. Where n' is the vector tangential to the surface. But this projection onto the surface has to be equal to the projection of the incident wavevector k_i onto n' as the incident and the transmitted (or evanescent) wave have to be in phase everywhere on the surface.
einfall
#7
Nov16-12, 10:01 AM
P: 11
Quote Quote by DrDu View Post
You are interested in the angle [itex] \cos(\theta')=\vec{k}/|k|\cdot \hat{n'}[/itex]. Where n' is the vector tangential to the surface.

Yes, I'm pretty sure this is the [itex]θ'[/itex] I'm talking about and I think we're on the same wavelength(/vector... ohh, puns).

But when you say:
Quote Quote by DrDu View Post
But this projection onto the surface has to be equal to the projection of the incident wavevector k_i onto n' as the incident and the transmitted (or evanescent) wave have to be in phase everywhere on the surface.
Please tell me if I'm interpreting this incorrectly, but is this a more physically rigorous way of implying that the real component of the angle [itex]θ'[/itex] must be zero? i.e. [itex]θ'[/itex] is pure-imaginary? This is the result when I take Pi/2 - [itex]θ_{t}[/itex], and it does seem to make some intuitive sense.
DrDu
#8
Nov16-12, 10:10 AM
Sci Advisor
P: 3,593
No, the vector theta' is real: [itex]\theta'=90^\deg-\theta_i=\theta'_i[/itex].
The vector k' has a purely imaginary component parallel the vector n and a real component parallel the vector n' which is equal to the component of the n' component of the vector k_i.
einfall
#9
Nov16-12, 10:49 AM
P: 11
Thanks so much for trying to bludgeon this into my head... I'm afraid I'm still quite confused. Please forgive me in advance...

The incident [itex]k_i[/itex] is at a real angle, say 70 degrees, to a TIR interface. If it's a glass prism ([itex]n_1 ≈ 1.52[/itex]) and an aqueous medium ([itex]n_1 ≈ 1.33[/itex]) this gives [itex]θ_t = Pi/2 - 0.367 i[/itex]. My first reaction was just to say [itex]θ' = Pi/2 - θ_t = 0.367 i [/itex], but this doesn't sit well with me... it feels like I'm being nave.

I get that the imaginary component is the only bit that can propagate in the direction n, because otherwise this would carry energy away from the TIR interface.

I'm not sure if this is what you were suggesting - but just because k' has to contain a real component parallel to n', why would this imply that the angle it forms with the interface has to have a non-zero real component?

I fear I have the wrong end of some stick. Please can you have a look at the attached figure as it may (or may not) clarify the nature of my problem. You can ignore the sphere...

Thanks again.
Attached Thumbnails
sketch of k vectors.jpg  
DrDu
#10
Nov18-12, 03:42 PM
Sci Advisor
P: 3,593
Ok, maybe I have been too quick in post #8. The question is: Why don't you work with the vector k instead of the angle theta?
einfall
#11
Nov20-12, 04:56 AM
P: 11
The k-vector for the transmitted wave is the entity I need to quantify, yes but I specifically need the angle that k-vector forms with the interface (rather than the normal, which is the angle that falls out of standard Snell/Fresnel expressions).

Despite being irked by ignoring the imaginary component, since complex numbers are still periodic on 2 Pi and taking θ' = Pi/2 - θt gives a pure-imaginary result (necessary for energy conservation) I am given some confidence. The clincher, though, is that using this θ' and after debugging a few lines of code, my outputs have started to make sense.

I'm now satisfied on the basis of physical reasoning that simply taking θ' = Pi/2 - θt gives the correct θ' even when the transmitted k-vector is complex.

Thanks mfb & DrDu for your responses, they were helpful.
DrDu
#12
Nov20-12, 05:57 AM
Sci Advisor
P: 3,593
I don't think that's correct. I still don't see why you want that angle. It's definition may depend crucially on what you intend to do with it.


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