Reflection and Refraction at Dielectric Interfaces

Reflection and refraction at an interface of a dielectric is an extremely prevalent scenario in any lab work in optics, as well as any time you want to do something interesting since any movement of light other than in a straight line will involve reflection or refraction of some sort. Here, we describe the scenario for a dielectric, but this will hold for the majority of real, physical cases.

Table of Contents:

  1. Boundary Conditions
  2. Kinematic Conditions
  3. Power Reflection and Transmission
  4. Power Reflectance
  5. Phase Shift in Total Internal Reflection
  6. Wavevector Diagrams and TIR


Described in the last section (Section 2 of “Classical Optics”) was the propagation of light in a medium of a uniform dielectric constant and refractive index. We now consider the case when light approaches a boundary between two materials of different $\epsilon$ and $\mu$. We can derive how the light will propagate, described as the “boundary condition”, using Maxwell’s equations. Typically, this derivation is first carried out by considering the special case where there are no sources for Maxwell’s equations i.e. there are no free charges or currents near the boundary. Maxwell’s equations in this case are given as:

\begin{equation} \oint_S D\cdot{dS} = \rho_{free} = 0 \label{gauss}\end{equation}

\begin{equation} \oint D\cdot{l} = -\frac{\partial}{\partial{t}}\iint_s{B}\cdot{dS} \end{equation}

\begin{equation} \oint_S B\cdot{dS} = 0 \label{ampere}\end{equation}

\begin{equation} \oint H \cdot{dl} = \frac{\partial}{\partial{t}}\iint_s{D}\cdot{dS} \end{equation}

Now, typically the derivation to find the boundary conditions is done using arbitrarily small/thin shapes placed across a boundary, so we will do the same because why be unique am I right?! Anyhow, our first scenario is a Gaussian pillbox that extends just barely over a boundary in either direction (Gauss must have a lot of pillboxes for the number of times this derivation has been made!). We can then use our handy Gauss’s Law, corresponding to Equation \eqref{gauss}, in order to state that:

\begin{equation} \oint_S D\cdot{dS} = \frac{1}{\epsilon_0} Q_{enc} = 0 \label{gauss_pill}\end{equation}

where we were able to state that the above relations equal to zero because we do not have any free charges or current. If we then consider that the sides would not contribute anything to the integral of $D$, Equation \eqref{gauss_pill} would just be $\hat{n}D_2-\hat{n}D_2 = 0$, where the $\hat{n}$ refers to the normal to the surface of the dielectric and the pillbox. Therefore, $\hat{n}D_2 = \hat{n}D_1$, meaning that the normal components of the electric field are continuous across the boundary. We can also show this is true for the magnetic field $B$.

In the other case, we have an Amperian loop across the interface. This time, the unit vector of interest will be $\hat{t}$, which is a unit vector parallel to the dielectric surface and the Amperian loop. Again, if we take the limit where the infinitesimal height approaches zero, the surface flux integrals in Equations \eqref{gauss} and \eqref{ampere}. We therefore just have:

\begin{equation} -E_2\cdot\hat{t} + E_1\cdot\hat{t} = 0 \end{equation}

\begin{equation} -H_2\cdot\hat{t} + H_t\cdot\hat{t} = 0 \end{equation}

Thus, the tangential components of $E$ and $H$ are continuous across the boundary.

I believe these slides from the University of Liverpool sum all of this up nicely.


From these above expressions for the continuity conditions, it is clear that in order for these conditions to hold true, we must have that the phase as well as the magnitude of each of the waves is continuous across a boundary. To make our following analysis simpler, we will take our sample wave to be monochromatic such that the phase is given by $cos(\omega{t}-k\cdot{r})$. In the non-monochromatic case, we would just need to prove the continuity holds for each reflected/transmitted wavelength. With a monochromatic wave, the discussion thus simplifies to needing to find the kinematic conditions for the $k$-vectors.

Since the $k$-vector represents the momentum of the wave, we can think of this as a conservation of momentum problem (this is not exactly accurate, as a complete discussion of this problem as a conservation of momentum matter would require a dive into the photon picture, but it will work heuristically). Thus, with this, we must have that each of the $k_i$, $k_r$, and $k_t$ waves must lie in the same plane (with “$i$” for the incident wave, “$r$” for the reflected wave, and “$t$” for the transmitted wave). We can see this intuitively through our physical understanding that there will be no directional component of a wave added through reflection/refraction at a boundary if the momentum is to be conserved. The continuity of the wave vectors can be neatly pictured in the figure below (photo from source in the caption).

The figure “ruins” the ending in some ways by stating some of the results we will soon be deriving, but also may give you something to look forward to at the end of this derivation! The spacing of the k-vectors in this figure (given by $\lambda_1$ and $\lambda_2$) corresponds to the case where the refractive index of the incident medium is greater than the refractive index of the transmitted medium, i.e. $n_t < n_i$. In our following discussion, we will be exploring the opposite case, i.e. $n_t > n_i$. Hopefully this won’t be too confusing!

Here is the part of the discussion where my graduate teacher would say it is useful to start thinking in $k$-space. Now, I don’t know about you, but I think we all looked at him with a little drool coming out of our mouths and our heads steaming because none of us was exactly sure how to “get into the $k$-space frame of mind”. Unfortunately, though, I now have to muster my best professor voice and say it will be at least useful to develop some intuition about how wave vectors operate, if for no reason other than to be able to successfully follow along in an Optics class or even some Solid State classes. Anyhow, we can see here that the possible $k$-vectors before the boundary would lie on the $k$-vector sphere of radius:

\begin{equation} |k_i| = \frac{n_i\omega}{c} = |k_r| \end{equation}

If we arbitrarily choose that the wave-vector lies in the x-z plane, we have that the electric field is given as:

\begin{equation} E = E_{l0}e^{\omega{t} – h_lx-\beta_l{z}} \end{equation}

where the subscript “$l$” denotes that this representation of the electric field can be generalized with perhaps different exact magnitudes for the incident, reflected, and transmitted waves. With the plane of incidence then in the x-y direction, the following figures give a physical sense of what the possible reflected/transmitted $k$-vectors in the x-z plane would be.

We will now consider the case where the refractive index in the medium of transmission is greater than the refractive index in the incident medium, i.e. $n_t > n_i$. This then gives:

\begin{equation} k_t = n_t\omega/c > n_i\omega/c = k_i = k_r \end{equation}

which would correspond to circles of different radii on the wavevector diagram. However, they would have the same extent or magnitude in the x-direction, which we knew anyways because of the translational invariance of the k-vector in the x-direction. This then means that the continuous phase condition yields $h_i = h_r = h_t$. We can then write out the equivalence for the magnitude of h using the diagrams from above to find the x-components of each of the wave-vectors as:

\begin{align} h & = k_isin(\theta_i) = k_rsin(\theta_r) = k_tsin(\theta_t) \\ & = \frac{n_i\omega}{c}sin(\theta_i) = \frac{n_i\omega}{c}sin(\theta_r) = \frac{n_t\omega}{c}sin(\theta_t)\end{align}

where we can note in this equation that the incident and refracted indices are of course the same since they are in the same medium. You have probably seen this from the beginning, but we now have our exciting kinematic boundary conditions! These are given as:

\begin{equation} \theta_i = \theta_r \end{equation}

\begin{equation} n_isin(\theta_i) = n_tsin(\theta_t) \end{equation}

These give us the directions of the reflected and transmitted waves, but what about the amplitudes? To obtain the amplitudes for the reflected/transmitted waves, we need to consider two scenarios corresponding to two different polarization directions of the incident electric field. Here, we are defining the incident field to be polarized in the x-z plane and the plane of incidence to be the x-y plane. If the incident electric field is polarized in the direction perpendicular to the plane of incidence, then this is known as “s” polarization, where the “s” stands for the German word for perpendicular “senkrecht”. If the incident electric field is polarized in the direction parallel to the plane of incidence, then this is known as “p” polarization, where the “p” stands for German word for parallel…”parallel” :-). Something to note here is that this designation of “s” and “p” is analogous to linear and circular polarized light in the xyz – coordinate plane, with the difference being that s- and p-polarization characterizations are made with respect to the plane of incidence. Thus, like in circularly polarized light, an incident electric field with arbitrary polarization in the frame of reference of an incident plane can be decomposed into its corresponding s- and p- components. The following two figures should hopefully clear up the difference between s- and p-polarization and some of these seemingly arbitrary distinctions:

Taken from:

Now, using the designations of s- and p-polarization, we can find the amplitudes for the reflected and transmitted waves:

S-polarization: In this case, we will choose our plane of incidence to be the xz-plane. Then since the electric field is perpendicular to plane of incidence for s-polarization, the electric field would be purely in the $\hat{y}$-direction. The incident electric and magnetic fields are then given as:

\begin{equation} E_i = \hat{y}E_ie^{i(\omega{t} – k_i\cdot{r})} \end{equation}

\begin{align} B_i & = \frac{1}{\omega}k_i \times E_i \\ & = \frac{n_i}{c}(-cos\theta_i\hat{x}+sin\theta_i\hat{z})E_ie^{i(\omega{t}-k_i\cdot{r})} \end{align}

where we found the k-vector from the orthogonal triad relations as:

\begin{equation} k_i = (sin\theta_i\hat{x}+cos\theta_i\hat{z})\frac{n_i\omega}{c} \end{equation}

Now we can find the reflected and transmitted fields and directions using geometrical arguments and also keeping in mind that if a field has components in one direction, it will stay in that plane on reflection and transmission. So for our example, the electric field will stay in the $\hat{y}$ direction and the magnetic field will be broken into components for the $\hat{x}$ and $\hat{z}$ direction via the corresponding trigonometric relations. For the reflected field:

\begin{equation} E_r = \hat{y}E_re^{i(\omega{t} – k_r\cdot{r})} \end{equation}

\begin{align} B_r = \frac{n_i}{c}(cos\theta_i\hat{x}+sin\theta_i\hat{z})E_re^{i(\omega{t}-k_r\cdot{r})} \end{align}

where the remaining “i” denotations in the magnetic field definition indicate that the reflected field is totally internally reflected such that $\theta_i=\theta_r$ and $n_i = n_r$. The transmitted field is then given as:

\begin{equation} E_t = \hat{y}E_te^{i(\omega{t}-k_t\cdot{r})} \end{equation}

\begin{equation} B_t = \frac{n_t}{c}(-cos\theta_t\hat{x}+sin\theta_t\hat{z})E_te^{i(\omega{t}-k_t\cdot{r})} \end{equation}

Now, we derived the phase continuity conditions using conservation of momentum arguments, but if we can to find out how much of the electric/magnetic field amplitude is reflected/refracted at a boundary, we need the electromagnetic boundary conditions that we derived in the last subsection. Our first condition is that the normal component of the electric displacement vector $D$ is continuous across the boundary. Since $D = \epsilon{E}$, this boundary condition is trivial because there is no normal component of $E$ in s-polarization (look at the above figure to see this more clearly). Thus, the electric field must be purely tangential, which means we have the continuity requirement from the tangential electric field:

\begin{equation} E_i + E_r = E_t \label{e-cont}\end{equation}

Now we can apply our magnetic field boundary conditions. We recall that the normal component of $B$ and the tangential component of $H$ must be continuous. Since the normal direction in our case would be the z-direction and the tangential component would be the x-direction, we have continuity of these components specifically. Therefore:

\begin{equation} \frac{n_i}{c}sin\theta_i(E_i+E_r) = \frac{n_t}{c}sin\theta_tE_t \end{equation}

\begin{equation} \frac{n_i}{\mu_ic}cos\theta_i(E_i-E_r) = \frac{n_t}{\mu_tc}cos\theta_tE_t \end{equation}

We can note here that the normal magnetic field relation gives us a re-derivation of Snell’s Law, which is quite nice of it! We can also see that the tangential magnetic field relation can be rewritten in terms of the wave vectors as:

\begin{equation} \frac{k_{iz}}{\mu_i}(E_i-E_r) = \frac{\k_{tz}}{\mu_t}E_t \label{k-vec}\end{equation}

where we have included the “z” distinction in the k-vector definition to make it clear that $k_{iz} = (n_i\omega/c)cos\theta_i$ and $k_{tz} = (n_t\omega/c)cos\theta_t$. Then, by substituting in Equation \eqref{e-cont} for $E_t$, we can rewrite Equation \eqref{k-vec} once more as:

\begin{equation} \frac{k_{iz}}{\mu_i}(E_i-E_r) = \frac{k_{tz}}{c}E_t \end{equation}

and rearrange this to isolate $E_r$ and $E_i$ on either side:

\begin{equation} E_i(\frac{k_{iz}}{\mu_i} – \frac{k_{tz}}{\mu_t}) = E_r(\frac{k_{iz}}{\mu_i} + \frac{k_{tz}}{\mu_t}) \end{equation}

Here we can find two very important relations, which are known as the Fresnel reflection and transmission coefficients, respectively. To motivate this, if we would want to find what the amplitudes of the reflected and transmitted electric fields are, we would know that they must be some proportion of the incident electric field. Thus, with this proportion in mind, we can define the reflection coefficient as:

\begin{equation} r_s = \frac{E_r}{E_i} = \frac{\mu_tk_{iz}-\mu_ik_{tz}}{\mu_tk_{iz}+\mu_ik_{tz}} \end{equation}

We can then find the transmission coefficient as:

\begin{equation} t_s = \frac{E_t}{E_i} = \frac{E_i + E_r}{E_i} = 1 + \frac{E_r}{E_i} = 1 + r_s \end{equation}

If we substitute in our definition for $r_s$:

\begin{align} t_s & = \frac{\mu_tk_{iz}+\mu_ik_{tz}+\mu_tk_{iz}-\mu_ik_{tz}}{\mu_tk_{iz}+\mu_ik_{tz}} \\ & = \frac{2\mu_tk_{iz}}{\mu_tk_{iz}+\mu_ik_{tz}} \end{align}

Using simplifications by rewriting with Snell’s Law and taking $\mu_i = \mu_t$, then we can find the usual definitions of the transmission and reflection coefficients for s-polarization:

\begin{equation} r_s = -\frac{sin(\theta_i-\theta_t)}{sin(\theta_i+\theta_t)} \end{equation}


\begin{equation} t_s = \frac{2cos\theta_isin\theta_t}{sin(\theta_i+\theta_t)} \end{equation}

Woo! Now we are ready to move on to p-polarization conditions.

P-Polarization– In the case of p-polarization, it is now the magnetic field that is polarized in the $\hat{y}$ direction and the electric field which lies in the $\hat{x}$ and $\hat{z}$ plane of incidence. We thus have the following definitions for the electric and magnetic fields:

\begin{equation} E_i = (cos\theta_i\hat{x}-sin\theta_i\hat{z})E_ie^{i(\omega{t}-k_i\cdot{r})} \end{equation}

\begin{equation} B_i = \hat{y}\frac{n_i}{c}E_ie^{i(\omega{t}-k_i\cdot{r})} \end{equation}

Since we went more thoroughly how to get the reflected and transmitted fields for s-polarization, we will just state the results here for p-polarization:

For the reflected fields:

\begin{equation} E_r = (-cos\theta_i\hat{x}-sin\theta_i\hat{z})E_re^{i(\omega{t}-k_r\cdot{r})} \end{equation}

\begin{equation} B_r = \hat{y}\frac{n_i}{c}E_re^{i(\omega{t}-k_r\cdot{r})} \end{equation}

and the transmitted fields:

\begin{equation} E_t = (cos\theta_t\hat{x}-sin\theta_t\hat{z})E_te^{i(\omega{t}-\k_t\cdot{r})} \end{equation}

\begin{equation} B_t = \hat{y}\frac{n_t}{c}E_te^{i(\omega{t}-k_t\cdot{r})} \end{equation}

For the p-polarization case, it is now the magnetic field $B$ that has trivially satisfied boundary conditions. The other boundary conditions are thus given as:

transverse E: \begin{equation} cos\theta_i(E_i-E_r) = cos\theta_tE_t \end{equation}

normal D: \begin{equation} \epsilon_isin\theta_i(E_i+E_r) = \epsilon_tsin\theta_tE_t \end{equation}

transverse H: \begin{equation} \frac{n_i}{\mu_i{c}}(E_i+E_r) = \frac{n_t}{\mu_tc}E_t \end{equation}

Then, using the same algebra as before, we can obtain the reflection and transmission coefficients:

\begin{equation} r_p = \frac{\epsilon_tk_{iz}-\epsilon_ik_{tz}}{\epsilon_tk_{iz}+\epsilon_ik_{tz}} \end{equation}

\begin{equation} t_p = \frac{2\epsilon_tk_{iz}}{\epsilon_tk_{iz}+\epsilon_ik_{tz}} \end{equation}


When considering the power flow through and reflected by the interface, we may think that the power coefficients are just the squares of the corresponding electric field, i.e. $R = I = |\langle{S}\rangle| = \frac{1}{2}\sqrt{\frac{\epsilon}{\mu}}|E_r|^2$. However, unfortunately this is not true; we must consider the area that the electric field makes on the interface plane. This means that we have the requirement that it is the normal component of the Poynting vector that we must consider.


The reflectance is:

\begin{equation} R_{s,p} = \frac{|\langle{S_r}\rangle|cos\theta_i}{|\langle{S_i}\rangle|cos\theta_i} = \frac{|E_r|^2}{|E_i|^2} = |r_{s,p}|^2 \end{equation}

and the transmittance is:

\begin{equation} T_{s,p} = \frac{|\langle{S_t}\rangle|cos\theta_t}{|\langle{S_i}\rangle|cos\theta_i} = \frac{n_tcos\theta_t|E_t|^2}{n_icos\theta_i|E_i|^2} = \frac{n_tcos\theta_t}{n_icos\theta_i}|t_{s,p}|^2 \end{equation}

Plots for the reflection coefficients for different values of the transmitted/incident indices of refraction are shown below for varying angles of incidence. Note that the transmission coefficients would just be $1-R$, or the difference between the below plots and 1.

silent “s”

We have several interesting cases, then, for different angles of incidence:

(a) Normal Incidence ($\theta \approx 0$): At normal incidence, the action of s- and p-polarization at the boundary is the same, meaning that it is no longer possible to differentiate between the two. We can find this mathematically once we substitute in $\theta_i = 0$ into both the transmitted and reflected coefficients for s- and p-polarization. For p-polarization:

\begin{equation} r_{\theta \rightarrow 0} = \frac{n_t-n_i}{n_t+n_i} = \frac{E_r}{E_i} \end{equation}

\begin{equation} t_{\theta \rightarrow 0} = \frac{2n_i}{n_t + n_i} = \frac{E_t}{E_i} \end{equation}

and for s-polarization:

\begin{equation} r_{\theta \rightarrow 0} = \frac{n_i-n_t}{n_t+n_i} = \frac{E_r}{E_i} \end{equation}

\begin{equation} t_{\theta \rightarrow 0} = \frac{2n_i}{n_t+n_i} = \frac{E_t}{E_i} \end{equation}

Now we can note here that there is a sign flip for the reflection coefficient of s- vs p-polarization. The usual convention is that there is a sign flip for s-polarization when the transmitted index of refraction is greater than the incident, and vice versa for p-polarization.

(b) Brewster’s angle: Now looking at the expression for P-polarization reflection coefficient, we can explore the case when the denominator approaches one of its limits. Since the p-polarization reflection coefficient has the expression $tan(\theta_i+\theta_t)$ in the denominator, the limit as the addition of these two angles approaches 90$^\circ$ is that $tan(\theta_i+\theta_t) \rightarrow \infty$. This then means the whole expression will evaluate to zero. We can write this out explicitly as follows:

\begin{equation} r_p = \frac{tan(\theta_i-\theta_t)}{tan(\theta_i+\theta_t)} \rightarrow 0 \end{equation}

We can further investigate this condition by explicitly finding the incident angle that causes this effect. To do this, we can use Snell’s Law and rewrite in terms of $\theta_i$:

\begin{equation} \frac{n_t}{n_i} = \frac{sin\theta_i}{sin\theta_t} = \frac{sin\theta_i}{sin(90^{\circ}-\theta_i} = \frac{sin\theta_i}{cos\theta_i} = tan(\theta_i)\end{equation}

In the last relation, we have our condition for what angle of incidence will give the zero-reflection point for P-polarization! We call this angle “Brewster’s Angle”, denoted $\theta_B$. The explicit expression for this angle is then:

\begin{equation} tan\theta_B = \frac{n_t}{n_i} \end{equation}

What Brewster’s Angle physically means is that all P-polarized light is transmitted through the interface at this point; there is zero reflection. We don’t have a corresponding condition for S-polarized light because the reflection (nor transmission for that matter) ever goes to zero.

(c) Total internal reflection: Now we can consider the case where we have light propagation from a dense medium to a less dense medium, which, since we know that the density is related to atomic spacing and that the atomic spacing gives us the dielectric index and thus the index of refraction, means that $n_i > n_t$. Using Snell’s Law, we have:

\begin{equation} n_isin\theta_i = n_tsin\theta_t \rightarrow \theta_t = sin^{-1}(\frac{n_i}{n_t}sin\theta_i) \end{equation}

Since the term $sin^-1(\theta)$ is undefined for $\theta > 1$, this means that we have a condition for a critical angle via $(\frac{n_i}{n_t})sin\theta_c =1$, which means that $\theta_t$ must equal $90^\circ$. To state this explicitly:

\begin{equation} \theta_c = sin^{-1}(\frac{n_t}{n_i}) \end{equation}

This holds for both s- and p-polarization, meaning that the reflection coefficient equals to 1 at the critical angle. You may be wondering, doesn’t there need to be some kind of continuity? Didn’t we spend a lot of time deriving those boundary conditions to find which components exactly are continuous? Please don’t tell me that physics breaks sometimes?! Don’t panic, this is exactly the case, and I am happy to hear that you are paying attention! We do indeed know that there has to be continuity of fields along the boundary by our boundary conditions. We can find out what exactly happens by writing down the general expression for the phase of the transmitted wave:

\begin{equation} E_t \approx e^{-ik_t\cdot{r}} = e^{-ik_t(xsin\theta_t + zcos\theta_t)} \end{equation}

To find $sin\theta_t$ in terms of the incident and critical angle, we can use Snell’s Law as follows:

\begin{equation} sin\theta_t = \frac{n_i}{n_t}sin\theta_i = \frac{sin\theta_i}{sin\theta_c} \end{equation}

and, using the trigonometric identity for cosine and sine, we can find:

\begin{equation} cos\theta_t = \sqrt{1-sin^2\theta_t} = \sqrt{1-(\frac{sin\theta_i}{sin\theta_c})^2} \end{equation}

Now we can note that when $\theta_i > \theta_c$, the cosine term becomes imaginary:

\begin{equation} cos\theta_t = i\sqrt{(\frac{sin\theta_i}{sin\theta_c})^2-1} = ia \end{equation}

So then the transmitted field is then:

\begin{align} E_t & = e^{ik_t[x(sin\theta_i/sin\theta_c)-iaz]} \\ & = e^{-ak_tz}e^{-ik_txsin\theta_i/sin\theta_c} \end{align}

which is the form of a wave that runs along the interface due to the $sin\theta_i/sin\theta_c$ term. The amplitude would then decay exponentially away from the surface. This is called an evanescent wave. It solves the boundary condition but it is not able to propagate in the same way as an electric field.

Methods of evanescent-wave excitation. (a) Total internal reflection... |  Download Scientific Diagram
Taken from:


The evanescent field can be measured by coupling into another medium brought in close proximity to the first medium where TIR is taking place; this is actually a popular measurement method called frustrated TIR (FTIR). The figure below demonstrates how, in the second image, the evanescent wave is coupled through the slight air gap:

Interestingly, there is also something known as touch FTIR:

Finally, since a fiber waveguide confines light by TIR, the two methods to couple light in are through incidence on the end or through a prism coupler:

See also C.K. Drexhage, J. Opt. Soc. Am., 62, 479 (1972)


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