# Fresnel Equations - Equal Entry/Exit Reflection

1. Aug 12, 2010

### SunMesa

My first (successful) post, hope this isn't silly/trivial. Also hope this is the appropriate area for optics-related questions.

While debugging some code that applies the Fresnel equations to transmission of light through multiple layers of planar media, I encountered an interesting feature that I had not previously noticed. Specifically, consider the transmission of a ray from medium 1 through a finite layer of medium 2, which re-emerges into medium 1 again (e.g., sunlight passing through a pane of glass in air).

It turns out that the reflection coefficient at the entry (1->2) interface is just equal to the reflection coefficient at the exit (2->1) interface. This regardless of the respective indices of refraction of the two media, the initial angle of incidence, or the polarization of the ray.

What surprises me about this symmetry is that the angle of incidence at the two interfaces is generally quite different... the refraction at the first interface just happens to produce the proper angle of incidence at the second such that the reflection coefficients are equal at both.

I guess my first question would be, is this simply wrong (i.e., am I using the wrong equations or made some other error), or, if not wrong, is there some broader (e.g., conservation or continuity) principle that makes more obvious why this would be the case?

Thanks for any info/insights-

2. Aug 12, 2010

### Redbelly98

Staff Emeritus
Welcome to PF. Interesting question.

While I was made aware of this fact many years ago because either I read it or was told about it, I have never tried to look at the Fresnel equations myself and work out the math. Presumably by combining Snell's Law with the Fresnel equations, it would be provable.

Has anyone else looked at this in more detail?

3. Aug 13, 2010

### Pete99

I do not remember the details of Fresnel equations, but it seems to me that it may have something to do with the Reciprocity theorem in electromagnetics.

4. Aug 13, 2010

### SunMesa

Redbelly98, Pete99,

Thanks for your replies. It's good to know someone else has heard of this result. It's fairly straightforward to derive from the Fresnel equations, although I don't want to do it here until I get more comfortable with inserting equations into these posts*. Snell's law does enter into it by providing a relation between the angles of incidence and transmission, but there is some underlying symmetry in the Fresnel equations involved... I'll have to look into how they are derived from first principles. Also will revisit the reciprocity theorem, can't even remember what it was. As you may imagine, EM was really not my strong suit.

Off topic a little, this is the second time I've been logged on, composed a somewhat lengthy reply, then went to preview it only to be presented with the log-in prompt again, with no way to get back to my reply text, i.e., I have to start over. Is there a time-out function I'm unaware of?

*I see you (Redbelly98) have a sticky post about using Latex for equations... I'm new to it but it looks fairly intuitive. Thanks-

5. Aug 13, 2010

### Staff: Mentor

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6. Aug 13, 2010

### Bob S

7. Aug 13, 2010

### SunMesa

Bob,

By a cruel twist of fate, the Google preview of your reference ends at page 118! However, the immediately preceding pages provide a fairly rigorous introduction, that should get me well on my way. Thanks-

8. Aug 13, 2010

### Bob S

Bob S

9. Aug 13, 2010

### SunMesa

Bob,

Tried that, it still cuts off at the end of page 118. I guess they're on to us!

10. Aug 14, 2010

### Bob S

See the attached pdf file from Prof. Rick Trebino (Georgia Tech)

Includes Fresnel's equations, Snell's equation, oblique angle of incidence, continuity of E and H at boundary, reflection, refraction, polarization, Brewster's angle.

Bob S

ps Does this link to Slater and Frank work?:

#### Attached Files:

• ###### OpticsI-08-FresnelEqs.pdf
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Last edited: Aug 14, 2010
11. Aug 14, 2010

### Staff: Mentor

12. Aug 15, 2010

### SunMesa

Bob S, jtbell,

Thanks for the links (Bob S, yes, your new Google link did bring up the missing Slater & Frank pages... how'd you do that?).

All of your links are solid fundamental treatments, and it is clear how the Fresnel equations follow from the boundary conditions at the interface(s).

I am nevertheless still struck by the quirk under discussion. Namely, a ray of arbitrary polarization and angle of incidence, encountering a planar slab of arbitrary material, will reflect (and transmit) exactly the same fraction of energy at each interface, even though the angles of incidence are different at each.

Trebino notes, almost in passing, that at *normal* incidence, reflection and transmission coefficients are the same regardless of which medium contains the incident ray. It's not clear that he, or the other authors, are aware of the more general result that obtains in the geometry considered here (oblique incidence on a 'sandwiched' material, with refraction at each interface).

As mentioned earlier, it is straightforward to derive this result from the Fresnel equations, but even in the context of the latter's first-principles derivation, it seems to emerge as something of a fluke, while having the 'feel' of something more fundamental.

But now I'm waxing philosophical, best call it a night. Thanks again for the links-

13. Aug 15, 2010

### Pete99

I will try to explain the reason I think this happens. The reasoning is probably not very rigorous, but I think the idea is correct.

If you only have one interface between two materials A and B, and you have a plane wave incident from side A to B at an angle theta_i, part of it will transmit with transmission coefficient T_ab and exit angle theta_o, and part of it will reflect with reflection coefficient R_ab.

I think it is reciprocity theorem that says that if you interchange the source and the receiver, so that the wave travels from B to A, with an incidence angle theta_o, it will have exit angle theta_i, with the same reflection and transmission coefficients as before (i.e., if we defined a two ports S parameter matrix, it would be symmetric).

Now, in your problem, you have two interfaces, from A to B and from B to A. Because of the geometry of the problem, the incidence angle in the second interface (B to A) will be the same as the exit angle from the firs interface (A to B). Then, the reflection and transmission coefficients should be the same.

14. Aug 15, 2010

### SunMesa

Pete99,

From what little grasp I have of the reciprocity theorem, I am unable to glean any insight from its application to this scenario. It does posit that the ratio of current to electric potential will remain constant under a swap of source and detector, but this symmetry does not hold true for transmitted (or reflected) power, which is really what the transmission and reflection coefficients provide.

I'm beginning to think that something like a least action principle might provide insight here, e.g., Fermat's principle of least (or actually, stationary) time, or more generally Huygens Principle, from which Snell's Law of refraction can be produced quite directly and elegantly.

15. Aug 15, 2010

### Pete99

Last edited by a moderator: Apr 25, 2017