# Impedance matching in Fresnel's equations

I have Fresnel's equations:

\begin{align} \begin{aligned} r_\text{TE} = \dfrac{\eta_2\cos\theta_1 - \eta_1\cos\theta_2}{\eta_2\cos\theta_1 + \eta_1\cos\theta_2}, && t_\text{TE} = \dfrac{2\eta_2\cos\theta_1}{\eta_2\cos\theta_1 + \eta_1\cos\theta_2},\\ r_\text{TM} = \dfrac{\eta_2\cos\theta_2 - \eta_1\cos\theta_1}{\eta_1\cos\theta_1 + \eta_2\cos\theta_2}, && t_\text{TM} = \dfrac{2\eta_2\cos\theta_1}{\eta_1\cos\theta_1 + \eta_2\cos\theta_2}, \end{aligned} \end{align}

where $\theta_1$ is the incoming angle, and $\theta_2$ is the transmitted angle from medium 1 with impedance $\eta_1$ to medium 2 with impedance $\eta_2$. These are deduced from Snell's law and the law of reflection, as well as from the continuity relations of the electric and magnetic fields on the interface of the mediums.

How can it be deduced from these equations that if $\eta_2=\eta_1$ there is no reflection? I'm not seeing it. Any help is very much appreciated.

• hint: when $\eta_2=\eta_1$ you also have $\theta_2 = \theta_1$ – hyportnex Feb 3 '17 at 23:30
• @hyportnex, where does that come from? – Vladimir Vargas Feb 3 '17 at 23:33
• from Snell. Moreover this is obvious from the physics since in this case both materials are identical in the electromagnetic response, and they might as well be the same material from a propagation perspective. – ZeroTheHero Feb 4 '17 at 0:38
• But Snell says $\eta_1\epsilon_1\sin\theta_1 = \eta_2\epsilon_2\sin\theta_2$. – Vladimir Vargas Feb 4 '17 at 1:18

The answer is that it is only true for non-magnetic media (or media with identical relative permeability). In those cases, the impedances are inversely proportional to refractive indices and Snell's law ensures that the angles of incidence and refraction are the same.

If you had materials with identical impedance but different refractive indices, then I don't think it is true there would be zero reflection, except at normal incidence.

The relevant relationships are $\eta = (\mu_r \mu_0/\epsilon_r \epsilon_0)^{1/2}$, whereas $n = (\mu_r \epsilon_r)^{1/2}$.

• Good point. I always forget about the magnetic part of the impedance. – ZeroTheHero Feb 4 '17 at 13:08

First of all, if $η_1 = η_2$, you can simplify them and get:

$$r_{TE} = r_{TM} = {\cos θ_1 - \cos θ_2 \over \cos θ_1 + \cos θ_2}$$

Then, by using Snell's law:

$$n_1 \sin θ_1 = n_2 \sin θ_2$$

with $n_i$ is the refractive index of a certain medium. Since the impedances are equal, also the refractive indexes are equal (this is true only under certain hypothesis; please make reference to @RobJeffries answer) and you can simplify them. Then you get:

$$\sin θ_1 = \sin θ_2$$

And, from trigonometry:

$$\sqrt{1 - \cos^2 θ_1} = \sqrt{1 - \cos^2 θ_2}$$

$$\cos θ_1 = \cos θ_2$$
$$r_{TE} = r_{TM} = 0$$