For the Fresnel Equations for TM light why is 1 + r not t?

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The discussion centers on the Fresnel equations for TM (transverse magnetic) light, specifically questioning why the relationship 1 + r is not equal to t, as it is for TE (transverse electric) light. The participants clarify that TM waves do not have a magnetic field in the direction of propagation and explore the implications of energy conservation and the behavior of electric and magnetic components at the interface. They note that while the Fresnel coefficients for TE light follow a simple relationship, TM light exhibits more complexity due to the nature of the polarization and the angle of incidence. The conversation emphasizes the need for a conceptual understanding of these relationships and suggests consulting additional sources for clarity. The intricacies of TM and TE polarization are highlighted as essential for accurately applying the Fresnel equations.
Latempe
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The complex amplitude ratios for light are defined as:

rTM = ErTM / EiTM
tTM = EtTM / EtTM

I've done the derivation from Wikipedia and see that

(n2/n1) * tTM = rTM + 1.

But I don't understand what is going on physically. I understand that these values are not power or intensity so I can't really invoke conservation of energy (or can I?). Why is

r + 1 = t

for TE light, but not TM?

Thanks.
 
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If I understand your question, you are using reflection coefficients for the magnetic component of the TEM wave. For the electric component, it has ## \nabla \times E =- \frac{\partial{ B}}{\partial{t}} ##. Using Stokes'theorem, you can make the rectangle around the area over which the line integral is performed arbitrarily thin, which makes ## E_1=E_2 ## so that ## \\ ## (1) ## E_i+E_r=E_t ## for the components of ## E ## parallel to the interface. ## \\ ## Also, energy (intensity) ## I=n E^2 ##. (assuming normal incidence and leaving out constants of proportionality.) With energy conservation ## \\ ## (2) ## I_i=n_1 E_i^2=n_1 E_r^2+n_2 E_t^2 ##. ## \\ ## The Fresnel coefficients for the electric field ## \rho=\frac{E_r}{E_i}=\frac{n_1-n_2}{n_1+n_2} ## and ## \tau=\frac{E_t}{E_i}=\frac{2n_1}{n_1+n_2} ## can be computed from the two equations above. ## \\ ## The magnetic components of these waves, other than proportionality constants, are basically ## B=\hat{n} \times E ##, where ## \hat{n} ## is a unit vector and points in the direction of propagation. I think this cross product puts a minus sign on the ## B_r ## term so that ## B_i-B_r =B_t ##. (The energy equation would be left unchanged by this cross product.) This would make any Fresnel relations different for the magnetic components. Hopefully this answers your question. ## \\ ## Note: For the electric field coefficients the result is ## 1-\rho=(\frac{n_2}{n_1}) \tau ## . For the magnetic field coefficients ## 1+\rho_m=(\frac{n_2}{n_1}) \tau_m ##.
 
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Charles Link said:
If I understand your question, you are using reflection coefficients for the magnetic component of the TEM wave. For the electric component, it has ## \nabla \times E =- \frac{\partial{ B}}{\partial{t}} ##. Using Stokes'theorem, you can make the rectangle around the area over which the line integral is performed arbitrarily thin, which makes ## E_1=E_2 ## so that ## \\ ## (1) ## E_i+E_r=E_t ## for the components of ## E ## parallel to the interface. ## \\ ## Also, energy (intensity) ## I=n E^2 ##. (assuming normal incidence and leaving out constants of proportionality.) With energy conservation ## \\ ## (2) ## I_i=n_1 E_i^2=n_1 E_r^2+n_2 E_t^2 ##. ## \\ ## The Fresnel coefficients for the electric field ## \rho=\frac{E_r}{E_i}=\frac{n_1-n_2}{n_1+n_2} ## and ## \tau=\frac{E_t}{E_i}=\frac{2n_1}{n_1+n_2} ## can be computed from the two equations above. ## \\ ## The magnetic components of these waves, other than proportionality constants, are basically ## B=\hat{n} \times E ##, where ## \hat{n} ## is a unit vector and points in the direction of propagation. I think this cross product puts a minus sign on the ## B_r ## term so that ## B_i-B_r =B_t ##. (The energy equation would be left unchanged by this cross product.) This would make any Fresnel relations different for the magnetic components. Hopefully this answers your question. ## \\ ## Note: For the electric field coefficients the result is ## 1-\rho=(\frac{n_2}{n_1}) \tau ## . For the magnetic field coefficients ## 1+\rho_m=(\frac{n_2}{n_1}) \tau_m ##.

Thank you Charles, but I don't mean the reflection coefficients for the magnetic component of a TEM wave. I mean a TM polarized wave, that is to say, my wave doesn't have a magnetic field in the direction of propagation. It is also called a parallel polarized wave or a p polarized wave.

I'm also looking for a more conceptual understanding.

Thank for your help so far.
Latempe
 
Latempe said:
Thank you Charles, but I don't mean the reflection coefficients for the magnetic component of a TEM wave. I mean a TM polarized wave, that is to say, my wave doesn't have a magnetic field in the direction of propagation. It is also called a parallel polarized wave or a p polarized wave.

I'm also looking for a more conceptual understanding.

Thank for your help so far.
Latempe
I think what I did might be what you are looking for. The ## B ## is perpendicular to the direction of propagation. (along with the ## E ##). At normal incidence, the direction of polarization does not enter into the picture. The p polarization is polarized with the E field in the plane of incidence and reflection, and the perpendicular polarization has ## E ## perpendicular to this plane. At normal incidence, the Fresnel coefficients are independent of polarization. A google of the term TM polarized wave showed that it is just another name for what I have always known as parallel polarization of a transverse electromagnetic wave.
 
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At normal incidence said:
That makes sense. But what about a more general case like TM light at an arbitrary angle θ?
 
Those are called the Kirchhoff relations, and they are presented in many E&M textbooks including J.D. Jackson's E&M textbook. They are a complicated function of the angle of incidence, (along with ## n_1 ## and ## n_2 ##), and are different for parallel and perpendicular polarization. ## \\ ## To describe them qualitatively, the energy reflection coefficient ## R=\rho^2 ## increases monotonically from ## \theta_i =0 ## to ## \theta_i=\pi/2 ##, for perpendicular polarization, from the normal incidence value (at ## \theta_i=0 ## ) to ## R= 1.0 ## at ## \theta_i=\pi/2 ##, (grazing incidence). The parallel polarization case takes a dip to ## R=0 ## at the Brewster angle before going to ## R=1.0 ## at the grazing angle of ## \theta_i=\pi/2 ##. ## \\ ## I think Halliday-Resnick's E&M textbook shows these two graphs. ## \\ ## Note: I edited this a couple of times, and now it hopefully reads reasonably well.
 
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Charles Link said:
Those are called the Kirchhoff relations, and they are presented in many E&M textbooks including J.D. Jackson's E&M textbook. They are a complicated function of the angle of incidence and are different for parallel and perpendicular polarization. To describe them qualitatively, the energy reflection coefficient ## R=\rho^2 ## increases monotonically from ## \theta_i =0 ## to ## \theta_i=\pi/2 ## from the normal incidence value (at \theta_i=0 ) to ## R= 1.0 ## at ## \theta_i=\pi/2 ##, (grazing incidence), while the parallel case takes a dip to ## R=0 ## at the Brewster angle before going to ## R=1.0 ## at the grazing angle of ## \theta_i=\pi/2 ##.
The text I am referencing is Fundamentals of Photonics 2nd Ed. by Saleh and Tiech. They give the Fresnel equations as:

$$ r_{TE} = \frac{ η_2*secθ_2 - η_1*secθ_1 } {η_2*secθ_2 + η_1*secθ_1 }, t_{TE} = 1 + r_{TE} $$

$$ r_{TM} = \frac{ η_2*cosθ_2 - η_1*cosθ_1 } {η_2*cosθ_2 + η_1*cosθ_1 }, t_{TM} = (1+r_{TM}) \frac{cosθ_2} {cosθ_1} $$

I have a fairly good understand of the derivation and know math mathematically why ## t_{TM} = (1+r_{TM}) \frac{cosθ_2} {cosθ_1} ##, but I don't grasp this conceptually.
 
I believe the ## t_{TE}=1+r_{TE} ## equation is in error. At normal incidence it is incorrect, and thereby is incorrect in general. I would suggest you google another source on the subject. Once you understand the derivation for normal incidence, the derivations for other angles of incidence can be figured out with a little effort. Most textbooks also show these formulas with ## cos(\theta) ## instead of ## sec(\theta) ##. They could be converted to expressions with ## cos(\theta) ## for comparison to see if they are correct. ## \\ ## Editing: Unless they are doing something very different from what I think they are trying to do, the ## r_{TE} ## expression is completely wrong. In fact, other than getting a minus sign on the whole expression, it agrees with another google article for ## \rho_{||} ##. Meanwhile, the ## \rho_{TM} ## agrees with ##\rho_{perpendicular} ## ,other than again an overall minus sign. ## \\ ## The TM , according to the google,should be parallel polarization. I don't know that TM and TE are in widespread usage. They were always referred to as parallel and perpendicular in the courses that I took. ## \\ ## I highly recommend you google a couple of other sources and compare. I think Saleh and Tiech have it incorrect. ## \\ ## Additional comment: I see one source I googled uses the boundary conditions on ## B ## , instead of an energy conservation equation, as the second equation which it works with. In any case, the derivations are still similar.
 
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