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

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1. Jan 19, 2017

### Latempe

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.

2. Jan 19, 2017

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 $1+\rho_m=(\frac{n_2}{n_1}) \tau_m$.

Last edited: Jan 19, 2017
3. Jan 19, 2017

### Latempe

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. Jan 19, 2017

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.

Last edited: Jan 19, 2017
5. Jan 19, 2017

### Latempe

6. Jan 19, 2017

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.

Last edited: Jan 19, 2017
7. Jan 19, 2017

### Latempe

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. Jan 19, 2017

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.