Fresnel equations at normal incidence

AI Thread Summary
The discussion centers on the Fresnel equations for reflection coefficients at normal incidence, questioning why the equations for perpendicular and parallel polarizations yield opposite signs despite the loss of the plane of incidence. It is noted that the reflection coefficients convey not only amplitude but also phase information, with a 180° phase shift for the electric component in one polarization. The sign conventions for the two polarizations are highlighted, explaining that the direction of the electric field differs for incident and reflected waves in parallel polarization. Participants agree that the apparent contradiction arises from differing definitions of positive direction in the context of phase shifts. The conversation concludes with a consensus that the discrepancies are rooted in convention rather than physical differences.
brianpile
Messages
4
Reaction score
0
Hello everybody, I have a question about the Fresnel equations at normal incidence. My textbook gives the reflection coefficients as:

(1) perpendicular polarization

rs=( n1*cos(theta1)-n2*cos(theta2) ) / ( n1*cos(theta1)+n2*cos(theta2) )

(2) parallel polarization

rp=( n2*cos(theta1)-n1*cos(theta2) ) / ( n1*cos(theta1)+n2*cos(theta2) )

Now, to me, it seems that for normal incidence on a dielectric interface, the polarization loses its meaning since neither the E- or H-fields have components normal to the surface. Put another way, the plane of incidence has lost its meaning. In this case, shouldn't the 2 Fresnel equations for reflection give the same result? If you put 0 degrees into those two equations they have opposite signs. What gives?

I'm probably spending too much time think about this, but any help making sense of this would be greatly appreciated!

Thanks,
Brian
 
Physics news on Phys.org
Wow, I didn't consider this aspect! Thank you for pointing this out! I'm studying these things right now.

Well, I think that you are right, the plane of incidence loses its meaning.
THOUGH, the thing here maybe another one - I say maybe because I've been thinking of it for 10 minutes, but as it makes sense to me I'll write it here.
The information from the reflection coefficients is not only about the amplitude of the reflected wave, but also about its phase.
Now, despite the lacking of a plane of incidence, at normal incidence the coefficients still have to tell you that the electric component has a 180° phase shift, while the magnetic one don't. This is why, I think, you get that the coefficients are opposite: as you say, there is no difference as regards the plane of incidence (the amplitudes of the reflected wave are the same), but there is still a difference in the phase of the reflected wave.

I think this is the explanation. I'm not sure, but it makes sense.Hope it is clear.
 
Welcome to PF.

It's a matter of the sign conventions used for the two polarizations.

For parallel polarization, rp is positive when E has an upward component for both the incident and reflected beams:

gsed_0001_0028_0_img8854.png

If you imagine the angle of incidence approaching zero (normal incidence case), this means that the incident and reflected electric fields are actually pointing in opposite directions.

For perpendicular polarization, rs is positive when E is in the same direction for both the incident and reflected beams. Changing the angle of incidence does not change the direction of E.
 
Ok I see, thanks RedBelly. So to be consistent between the two polarizations, I should define the p-polarization as positive when the magnetic field receives a pi phase shift (and therefore the E-field does not).
 
I don't think that's right. In the figure I posted earlier, B would point in the same direction -- up, out of the page -- for both the incident and reflected waves.
 
Right, I wasn't clear. If we defined the case when the H-field does change direction upon reflection (n1>n2) then the situation looks like this.

http://dl.dropbox.com/u/1505234/Fresnel%20Eqs%20-%20p-polariozation.bmp

and the boundary condition for the magnetic field would be Bi-Br=Bt. I think this change allows the signs of the reflection coefficients agree at normal incidence...like you said, just a convention.
 
Last edited by a moderator:
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
Back
Top