From "Fundamentals Of Photonics", Saleh: enter image description here

1) Why k1 = k3?

2) Why condition 2.4-2 holds? or "differ by constants"? How do we get it from invariance to x, y?

I have tried to write the continuity and derivative continuity equations and trying to put in different x and y, but I don't see why this works.

Thank you


The starting points are electric boundary conditions. For instance the normal components must satisfy $$ \epsilon_1( E_{1n}\cos(\omega_1 t-\vec k_1\cdot r) + E_{3n}\cos(\omega_3 t-\vec k_3\cdot r)= E_{2n}\cos(\omega_2 t-\vec k_2\cdot r)\, , \tag{1} $$ where $E_{kn}$ are the normal components of the incident, transmitted and reflected plane waves, and are constant. (There is also a condition on the parallel component but I don't need it).

Since both $k_1$ and $k_3$ are in the same medium, they must be equal in length since the wave vector is completely determined by the material properties of the dielectric $\mu,\sigma,\epsilon$.

Because the interface is infinite and the wave is a plane wave, the relation (1) cannot depend on $x$ or $y$ since a translation of the interface by a constant vector $\vec R=x\hat x+y\hat y$ does not change the physics. Parts of the plane wave will hit the interface at different times but the boundary conditions are local and must hold for all times and everywhere meaning that, if you determine the amplitude relations between incident, reflected and transmitted field at one point and at one time, these will hold at any point on the interface and at any time.

Because (1) must hold everywhere on the interface for every $t$, and because the $E_{1n}, E_{2n}$ and $E_{3n}$ are constant and independent of $x,y$ and $t$, we must have $$ \cos(\omega t-\vec k_1\cdot \vec r)= \cos(\omega t-\vec k_2\cdot \vec r) =\cos(\omega t-\vec k_3\cdot\vec r) $$ else the relations between $E_{1n}, E_{rn}$ and $E_{tn}$ will change with $x,y$ and with $t$. Since the argument of the cosines must be the same, and the times are the same, the relation between the $k$’s follow.

  • $\begingroup$ I don't understand how you arrived at the cosine equality. It is not coming from continuity. Where are the amplitudes? $\endgroup$ – The Capacitor Mar 29 '19 at 5:32
  • $\begingroup$ I added a bit about one of the boundary conditions to tie in with the amplitudes. $\endgroup$ – ZeroTheHero Mar 30 '19 at 4:01
  • $\begingroup$ Ok, I understand that the logic is as follows: if we take r = 0 then we get E1+E2 = E3. This must hold everywhere on the boundry, thus we get your equation (1). $\endgroup$ – The Capacitor Mar 31 '19 at 7:56
  • $\begingroup$ Yes. Pick now a different $\vec r$ and a different $t$. The same relations must hold true. $\endgroup$ – ZeroTheHero Mar 31 '19 at 12:34

1) As stated by ZeroTheHero, $k_1 = k_3$ (they are the magnitudes of vectors $\mathbf{k}_1$ and $\mathbf{k}_3$) because the incident wave (represented by $k_1$) and the reflected wave (represented by $k_2$) are in the same medium: therefore, the parameters related to the Electromagnetic field, in particular $\varepsilon$ and $\mu$, are the same.

enter image description here

2) This document (introduced in this page) exactly deals with this topic on pages 1-12. Using the same coordinate system of your picture, it obtains step-by-step the equation between the incident, reflected and transmitted fields, which holds for any point of the $(x,y)$ plane. Some of these points lead to significant relations, like 2.4-2. It is a first formulation of Snell's laws for reflection and refraction.

It also includes some 2D and 3D images to give a visual help.

enter image description here


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