# Fresnel equations, transmission higher than one?

Hi, I was wondering how come I can get a transmission coefficient higher than one?

In the equation at normal incidence:

t = E_t/E_0 = 2*n2/(n1+n2) > 1 if n2 > n1

How is it even possible to get a higher electric field amplitude out compared to the one that got in?

The transmission coefficient is not defined as the ratio of the amplitudes of the electric fields in the two media.

I'm quite sure it is. In any case, would you be so kind as to present any evidence to the contrary?
Please keep in mind I'm not talking about the ratio of intensities, but the electric field ratio.

The word "transmission coefficient" can mean lot's of things. The most useful meaning is the percent of incident power that is transmitted. In that case:

T = (n2 cos θt)/(n1 cos θi) (|Et|2/|Ei|2)

You need those factors out front because you have to remember that the beam refracts upon transmission, and therefore the power spreads out. When properly defined this way, the transmission coefficient will not go above one.

I'm quite sure it is. In any case, would you be so kind as to present any evidence to the contrary?
Please keep in mind I'm not talking about the ratio of intensities, but the electric field ratio.

The transversal components of the electric and magnetic fields have to be continuous. Assuming normal incidence to the $x = 0$ plane, and that the electric field vector is polarized along y-direction, then the magnetic field-vector is polarized along the z-direction, we have:

Medium 1:

$$E_1 = E_0 e^{i k_1 x} + E_r e^{-i k_1 x}$$
$$H_1 = \frac{n_1}{Z_0} \left( E_0 e^{i k_1 x} - E_r e^{-i k_1 x} \right), \ Z_0 \equiv \sqrt{\frac{\mu_0}{\varepsilon_0}}$$

Medium 2:

$$E_2 = E_t e^{i k_2 x}$$
$$H_2 = \frac{n_2}{Z_0} E_t e^{i k_2 x}$$

where:
$$k_i = \frac{n_i \, \omega}{c}, \; n_i = \frac{1}{\sqrt{\varepsilon_i \, \mu_i}}, \ (i = 1, 2)$$

The continuity conditions give:
$$\begin{array}{l} E_0 + E_r = E_t \\ \frac{n_1}{Z_0} \left(E_0 - E_r \right) = \frac{n_2}{Z_0} E_t \end{array}$$
and, solving for $E_r/E_0$ and $E_t/E_0$, gives:
$$\begin{array}{l} \frac{E_t}{E_0} - \frac{E_r}{E_0} = 1 \\ \frac{n_2}{n_1} \frac{E_t}{E_0} + \frac{E_r}{E_0} = 1 \end{array}$$
$$\begin{array}{l} \left(1 + \frac{n_2}{n_1} \right) \frac{E_t}{E_0} = 2 \\ \left(1 + \frac{n_2}{n_1} \right) \frac{E_r}{E_0} = 1 - \frac{n_2}{n_1} \end{array}$$

But, the power flux is given by the Poynting vector, which, in harmonic time dependence is given by:
$$\mathbf{S} = \frac{1}{2} \mathrm{Re}\left(\mathbf{E} \times \mathbf{H}^{\ast} \right)$$

In our case, the cross product $\hat{\mathbf{y}} \times \hat{\mathbf{z}} = \hat{\mathbf{x}}$ is directed along the x-direction and the projections are:
Medium 1:
$$S_1 = \frac{n_1}{2 \, Z_0} \mathrm{Re} \left[ \left(E_0 e^{i k_1 x} + E_r e^{-i k_1 x} \right) \left(E^{\ast}_0 e^{-i k_1 x} - E^{\ast}_r e^{i k_1 x} \right) \right]_{x = 0} = \frac{n_1}{2 Z_0} \left( |E_0|^2 - |E_r|^2 \right)$$
Medium 2:
$$S_2 = \frac{n_2}{2 \, Z_0} \mathrm{Re} \left[ E_t e^{i k_2 x} E^{\ast}_t e^{-i k_2 x} \right]_{x = 0} = \frac{n_2}{2 Z_0} |E_t|^2$$
Keeping in mind that these amplitudes are all real, and the above continuity equations, it is easy to see that:
$$S_1 = S_2$$
which means that the Poynting vector is continuous on the boundary. This is an expression of the Law of Conservation of energy. The incident power density is:
$$S_{inc} = \frac{n_1}{2 Z_0} E^2_0$$
The reflected power density is (notice the flux is in the negative x-direction):
$$S_r = \frac{n_1}{2 Z_0} E^2_r$$
and the transmitted power density is:
$$S_t = \frac{n_2}{2 Z_0} E^2_t$$
The ratios:
$$T \equiv \frac{S_t}{S_{inc}} = \frac{n_2}{n_1} \left(\frac{E_t}{E_0}\right)^2 = \frac{2 \frac{n_2}{n_1}}{\left( 1 + \frac{n_2}{n_1} \right)^2}$$
and
$$R \equiv \frac{S_r}{S_{inc}} = \frac{E^2_r}{E^2_0} = \frac{\left(1 - \frac{n_2}{n_1} \right)^2}{\left(1 + \frac{n_2}{n_1} \right)^2}$$
are defined as the transmission and reflection coefficients, respectively. Notice that each is less than 1, and that:
$$T + R = 1$$

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Hi, I was wondering how come I can get a transmission coefficient higher than one?

In the equation at normal incidence:

t = E_t/E_0 = 2*n2/(n1+n2) > 1 if n2 > n1

How is it even possible to get a higher electric field amplitude out compared to the one that got in?

If transmission coefficient is higher than one, it is all dissipated in the medium. There is no transmission.

Thanks to your explanation, Dickfore, I've discovered where I was wrong.

I was assuming that the Intensity was proportional to the amplitude of the electric field squared, and did not foresee that such proportion would change from one medium to another. So I just happily assumed T=(Et/E0)^2 and instead of checking "T" I turned to look at the "t" coefficients.

I stand corrected now, thanks.

Ok, so I'm back with more doubts about the applications of Fresnel equations and optics in general, if you don't mind helping me again.

Imagine a ray of light incident on a dielectric slab, such that the reflected light has intensity Ir=R*I_0 and the transmitted one It=T*I_0, being R and T the usual transmission and reflection coefficients, which satisfy R+T=1.

If the transmitted light is reflected on a mirror so that it crosses the path of the first reflected ray, the intensities in the crossing point sum with:

I=I_0*R+I_0*T+2*I_0*sqrt(R*T)*cos(phi)

Where phi is the phase difference between the two rays, and "I" the resulting intensity.

Which, using the relation R+T=1 becomes:

I/I_0=R+(1-R)+2*sqrt(R(1-R))*cos(phi)=1+2*sqrt(R-R^2)*cos(phi)

Which means that if the phase is right, I'll be getting more intensity than I put in, and it just seems too weird.

So, what I am doing wrong now, or not getting? I've tried to wrap my head around this for the last week, but I haven't found a suitable answer.

So the light is reflected at the interface between medium 1 and 2. The transmitted light is in medium 2. In which one of the media do you consider the crossing of the two beams?
If it's medium 1, how do you "reflect" the light from medium 2 back into 1 without having another reflection-transmission at the interface?

And when you consider interference of tow beams, you add the amplitudes and then take the square of the resulting amplitude to find the intensity. If this is what you are doing.

If I understand the new problem you are considering two interfaces now. The problem gets more complicated now because at each surface a ray splits into transmitted and reflected rays, and the light will continue to bounce back and forth between the two surfaces, splitting at every bounce. This is called a Fabry-Perot etalon. The transmission and reflection phenomena of the whole system is now more a result of the interference of multiple bounces than just single-surface reflection/refraction. The best way to approach this problem is not treat the light as rays, but as total electromagnetic fields. That way, you don't have to track a million splitting rays. You just set up total fields in each region and apply boundary conditions.