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.

  • $\begingroup$ hint: when $\eta_2=\eta_1$ you also have $\theta_2 = \theta_1$ $\endgroup$
    – hyportnex
    Feb 3 '17 at 23:30
  • $\begingroup$ @hyportnex, where does that come from? $\endgroup$ Feb 3 '17 at 23:33
  • $\begingroup$ 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. $\endgroup$ Feb 4 '17 at 0:38
  • $\begingroup$ But Snell says $\eta_1\epsilon_1\sin\theta_1 = \eta_2\epsilon_2\sin\theta_2$. $\endgroup$ 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 $n = (\mu_r \epsilon_r)^{1/2}$.

  • $\begingroup$ Good point. I always forget about the magnetic part of the impedance. $\endgroup$ 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:

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

That leads to:

$$ \cos θ_1 = \cos θ_2$$

By substitution in the first formula I derived, you get that:

$$ r_{TE} = r_{TM} = 0 $$

  • $\begingroup$ "Since the impedances are equal, also the refractive indices are equal". I don't think this is necessarily true. $\endgroup$
    – ProfRob
    Feb 4 '17 at 10:51
  • $\begingroup$ @RobJeffries, you are right, I have implicitly oversimplified it, I'll add a reference to your answer in mine. $\endgroup$
    – JackI
    Feb 4 '17 at 10:57

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