How does one show that thickness and wavelength determine the full transmission between two different dielectric media if the boundary condition equations between two dielectric media are independent of these quantities?

For example I have a plane wave normally incident on a medium with two layers. The first layer has an index of refraction $n_1$ and is of thickness $d$. The second layer has an index of refraction $n_2$ and is of infinite thickness.

Jackson provides the following boundary condition equations, see page 306

\begin{align} \dfrac{E_0(\text{reflected})}{E_0(\text{incident})} = \dfrac{2n_1}{n_2+n_1} && \dfrac{E_0(\text{transmitted})}{E_0(\text{incident})} = \pm \dfrac{n_2-n_1}{n_2+n_1} \end{align} for normal incidence, which are independent of the wavelength and thickness. What is missing from this analysis? Thanks.

  • $\begingroup$ what you mean by full transmission? There is no reflection at the interface between the two media? $\endgroup$
    – Sofia
    Commented Jan 14, 2015 at 21:01
  • $\begingroup$ Now, the two media are absorption free? The full wave-intensity is transmitted? It's a little hard to believe. The thicker a medium is, the more it absorbs. $\endgroup$
    – Sofia
    Commented Jan 14, 2015 at 21:03
  • $\begingroup$ The thickness shouldn't indeed play any role in boundary conditions. $\endgroup$
    – Sofia
    Commented Jan 14, 2015 at 21:04
  • $\begingroup$ I assumed "no observed reflection" meant full transmission, however if the media has attenuation, then full transmission would not be guaranteed. $\endgroup$ Commented Jan 14, 2015 at 21:06
  • $\begingroup$ Are you talking about a thin film where you need to consider interference effects - which is how you can get full transmission? $\endgroup$
    – ProfRob
    Commented Jan 14, 2015 at 21:07

1 Answer 1


It is possible you are thinking of the case of a thin film coating, where destructive interference between light reflected from the first and second interface can result in "full transmission".

If you look at anti-reflection coatings you will find a criterion for destructive interference that is $$2 d n_1 \cos \theta = (m-1/2)\lambda,$$ where $d$ is the thickness of the film, $\theta$ is the angle of transmission at the first interface (between air/vacuum and medium of index $n_1$, $\lambda$ is the vacuum wavelength and $m$ is an integer.

For normal incidence $\theta_2=0$ and the thinnest possible layer for full transmission would be given by $m=1$ and $d = \lambda/4n_1$.

  • $\begingroup$ Why is there a difference between thick and thin film analysis? $\endgroup$ Commented Jan 14, 2015 at 21:24
  • $\begingroup$ @linuxfreebird Sounds like another question $\endgroup$
    – ProfRob
    Commented Jan 14, 2015 at 21:32
  • $\begingroup$ I will post it as a new question if you deem it worth. Thanks. $\endgroup$ Commented Jan 14, 2015 at 21:37

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