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

In summary, complex amplitude ratios for light, such as rTM and tTM, are defined as ErTM/EiTM and EtTM/EiTM, respectively. These values are not power or intensity, so conservation of energy cannot be invoked. The Fresnel coefficients for the electric and magnetic components of a transverse electromagnetic wave can be computed using Stokes' theorem and energy conservation equations. At normal incidence, the Fresnel coefficients are independent of polarization, but for a general case such as TM light at an arbitrary angle θ, the Kirchhoff relations must be used to determine the coefficients, which vary for parallel and perpendicular polarization.
  • #1
Latempe
6
0
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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  • #2
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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  • #3
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
 
  • #4
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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  • #5
At normal incidence said:
That makes sense. But what about a more general case like TM light at an arbitrary angle θ?
 
  • #6
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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  • #7
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.
 
  • #8
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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Related to For the Fresnel Equations for TM light why is 1 + r not t?

1. Why is 1 + r not t in the Fresnel Equations for TM light?

The Fresnel Equations for TM (transverse magnetic) light describe the behavior of light that is polarized parallel to the boundary between two media. In this case, 1 + r is not equal to t because of the conservation of energy. The value of 1 + r represents the ratio of the amplitude of the reflected wave to the amplitude of the incident wave, while t represents the ratio of the amplitude of the transmitted wave to the amplitude of the incident wave. Since energy is conserved, the sum of the reflected and transmitted waves cannot exceed the amplitude of the incident wave, thus 1 + r cannot equal t.

2. What is the significance of the Fresnel Equations for TM light?

The Fresnel Equations for TM light are important in understanding the behavior of light at the interface between two media. They can be used to determine the amount of light that is reflected and transmitted at the boundary, as well as the angle of reflection and refraction. These equations are also used in various applications such as optics, semiconductors, and nanotechnology.

3. How are the Fresnel Equations for TM light different from those for TE light?

The Fresnel Equations for TM light and TE (transverse electric) light differ in the polarization of the incident light. In the case of TM light, the electric field is parallel to the boundary between the two media, while in TE light, it is perpendicular. This difference in polarization leads to different equations for the reflection and transmission coefficients.

4. Can the Fresnel Equations for TM light be applied to all types of materials?

The Fresnel Equations for TM light can be applied to all types of materials as long as they have a well-defined boundary and the incident light is polarized parallel to the boundary. This includes materials such as air, water, glass, and metals. However, these equations may not be accurate for highly absorbing materials or materials with rough surfaces.

5. How are the Fresnel Equations for TM light derived?

The Fresnel Equations for TM light are derived from Maxwell's equations, which describe the behavior of electromagnetic waves. By applying boundary conditions at the interface between two media, the reflected and transmitted waves can be determined. The resulting equations, known as the Fresnel Equations, can be used to calculate the reflection and transmission coefficients for TM light.

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