## Reflection at a conducting surface: the physics of mirrors

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.21.

We can analyze reflection of an electromagnetic wave at a nonconductor-conductor interface in a similar way to that used for a nonconductor-nonconductor interface. We’ll look only at the case of normal incidence here.

As before, we start with the boundary conditions in linear media derived from Maxwell’s equations:

 $\displaystyle \epsilon_{1}E_{1}^{\perp}-\epsilon_{2}E_{2}^{\perp}$ $\displaystyle =$ $\displaystyle \sigma_{f}\ \ \ \ \ (1)$ $\displaystyle B_{1}^{\perp}-B_{2}^{\perp}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle \mathbf{E}_{1}^{\parallel}-\mathbf{E}_{2}^{\parallel}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (3)$ $\displaystyle \frac{1}{\mu_{1}}\mathbf{B}_{1}^{\parallel}-\frac{1}{\mu_{2}}\mathbf{B}_{2}^{\parallel}$ $\displaystyle =$ $\displaystyle \mathbf{K}_{f}\times\hat{\mathbf{n}} \ \ \ \ \ (4)$

We’ll take medium 1 as the nonconductor (air, say) and medium 2 as the conductor. We’re allowing for the presence of free surface charge density ${\sigma_{f}}$ and free current density ${\mathbf{K}_{f}}$ at the boundary.

If we’re dealing with a conductor that obey’s Ohm’s law, the volume current density is proportional to the electric field

$\displaystyle \mathbf{J}_{f}=\sigma\mathbf{E} \ \ \ \ \ (5)$

where here ${\sigma}$ is the conductivity, not a charge density. Recall that ${\mathbf{J}_{f}}$ is the amount of current flowing through a unit area in the conductor. If we had a surface current density ${\mathbf{K}_{f}}$, this current flows along the boundary as a sheet of moving charge with infinitesimal thickness, so that the cross-sectional area occupied by ${\mathbf{K}_{f}}$ is essentially zero, making the volume charge density infinite. For a finite conductivity ${\sigma}$ it would take an infinite electric field to produce this surface current, so we can safely assume that ${\mathbf{K}_{f}=0}$ in what follows.

The incident and reflected waves are both in medium 1, so if we polarize the wave in the ${x}$ direction, we have for the incident wave:

 $\displaystyle \tilde{\mathbf{E}}_{I}$ $\displaystyle =$ $\displaystyle \tilde{E}_{0_{I}}e^{i\left(k_{1}z-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (6)$ $\displaystyle \tilde{\mathbf{B}}_{I}$ $\displaystyle =$ $\displaystyle \frac{1}{v_{1}}\tilde{E}_{0_{I}}e^{i\left(k_{1}z-\omega t\right)}\hat{\mathbf{y}} \ \ \ \ \ (7)$

where ${v_{1}}$ is the speed of the wave in medium 1.

The reflected wave is travelling in the ${-z}$ direction and has equations

 $\displaystyle \tilde{\mathbf{E}}_{R}$ $\displaystyle =$ $\displaystyle \tilde{E}_{0_{R}}e^{i\left(-k_{1}z-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (8)$ $\displaystyle \tilde{\mathbf{B}}_{R}$ $\displaystyle =$ $\displaystyle -\frac{1}{v_{1}}\tilde{E}_{0_{R}}e^{i\left(-k_{1}z-\omega t\right)}\hat{\mathbf{y}} \ \ \ \ \ (9)$

The transmitted wave is inside the conductor, so its equations can be written as

 $\displaystyle \tilde{\mathbf{E}}_{T}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{E}_{0_{T}}e^{i\left(\tilde{k}_{2}z-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (10)$ $\displaystyle \tilde{\mathbf{B}}_{T}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \frac{\tilde{k}_{2}}{\omega}\tilde{E}_{0_{T}}e^{i\left(\tilde{k}_{2}z-\omega t\right)}\hat{\mathbf{y}} \ \ \ \ \ (11)$

where the wave vector is complex:

$\displaystyle \tilde{k}=\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}+1}+i\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}-1}\equiv k+i\kappa \ \ \ \ \ (12)$

We can now apply the boundary conditions. Equation 1 tells us that ${\sigma_{f}=0}$ since there is no perpendicular component of ${\mathbf{E}}$ (remember the wave is transverse). Equation 2 tells us nothing (${0=0}$). From 3, assuming that the boundary is at ${z=0}$, we get, since all components of ${\mathbf{E}}$ are in the ${x}$ direction:

$\displaystyle \tilde{E}_{0_{I}}+\tilde{E}_{0_{R}}=\tilde{E}_{0_{T}} \ \ \ \ \ (13)$

Finally, from 4 we get, since all components of ${\mathbf{B}}$ are in the ${y}$ direction and ${\mathbf{K}_{f}=0}$:

$\displaystyle \frac{1}{\mu_{1}v_{1}}\left(\tilde{E}_{0_{I}}-\tilde{E}_{0_{R}}\right)=\frac{\tilde{k}_{2}}{\mu_{2}\omega}\tilde{E}_{0_{T}} \ \ \ \ \ (14)$

which we can rewrite as

 $\displaystyle \tilde{E}_{0_{I}}-\tilde{E}_{0_{R}}$ $\displaystyle =$ $\displaystyle \frac{\mu_{1}v_{1}}{\mu_{2}\omega}\tilde{k}_{2}\tilde{E}_{0_{T}}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \tilde{\beta}\tilde{E}_{0_{T}} \ \ \ \ \ (16)$

Adding 13 and 16 we get

 $\displaystyle \tilde{E}_{0_{I}}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(1+\tilde{\beta}\right)\tilde{E}_{0_{T}}\ \ \ \ \ (17)$ $\displaystyle \tilde{E}_{0_{T}}$ $\displaystyle =$ $\displaystyle \frac{2}{1+\tilde{\beta}}\tilde{E}_{0_{I}} \ \ \ \ \ (18)$

Subtracting 13 and 16 we get

 $\displaystyle \tilde{E}_{0_{R}}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(1-\tilde{\beta}\right)\tilde{E}_{0_{T}}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1-\tilde{\beta}}{1+\tilde{\beta}}\tilde{E}_{0_{I}} \ \ \ \ \ (20)$

These are deceptively simple equations, since everything with a tilde on it is a complex number. To get the actual amplitudes and phases we need to extract the real and imaginary parts.

Example To put some numbers into these equations, let’s consider an air-silver interface. For a good conductor such as silver, ${\sigma\gg\epsilon\omega}$ and in 12

$\displaystyle k\approx\kappa\approx\sqrt{\frac{\mu\sigma\omega}{2}} \ \ \ \ \ (21)$

In air, ${v_{1}\approx c}$ and we can take ${\mu_{1}=\mu_{2}=\mu_{0}}$ so from 16

$\displaystyle \tilde{\beta}\approx c\sqrt{\frac{\mu_{0}\sigma}{2\omega}}\left(1+i\right) \ \ \ \ \ (22)$

For silver ${\sigma=6\times10^{7}\mbox{ S m}^{-1}}$ and at an optical wavelength of ${\omega=4\times10^{15}\mbox{ s}^{-1}}$ we get

 $\displaystyle \tilde{\beta}$ $\displaystyle =$ $\displaystyle 29.1\left(1+i\right)\ \ \ \ \ (23)$ $\displaystyle \frac{1-\tilde{\beta}}{1+\tilde{\beta}}$ $\displaystyle =$ $\displaystyle \frac{-28.1-29.1i}{30.1+29.1i}\times\frac{30.1-29.1i}{30.1-29.1i}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{1753}\left(-1693-58.2i\right) \ \ \ \ \ (25)$

To get the reflection coefficient we can write the complex amplitudes in modulus-phase form as

 $\displaystyle \tilde{E}_{0_{R}}$ $\displaystyle =$ $\displaystyle E_{0_{R}}e^{i\delta_{R}}\ \ \ \ \ (26)$ $\displaystyle \tilde{E}_{0_{I}}$ $\displaystyle =$ $\displaystyle E_{0_{I}}e^{i\delta_{I}} \ \ \ \ \ (27)$

The intensity of a wave is the average over one cycle of the magnitude of the Poynting vector, so the fact that the incident and reflected waves may have different phases doesn’t matter (since they have the same frequency). This means that

 $\displaystyle R$ $\displaystyle =$ $\displaystyle \frac{\frac{1}{2}c\epsilon_{0}E_{0_{R}}^{2}}{\frac{1}{2}c\epsilon_{0}E_{0_{I}}^{2}}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\left|E_{0_{R}}\right|^{2}}{\left|E_{0_{I}}\right|^{2}}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1693^{2}+58.2^{2}}{1753^{2}}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.93 \ \ \ \ \ (31)$

Silver reflects 93% of the incident light, so it makes a good mirror.

## Fermat’s principle of least time and Snell’s law

References: Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014) – Problem 1.1

One of the guiding principles of quantum field theory is that a particle travelling between two points actually traverses all possible paths between these two points, although with varying probabilities for different paths. Although this idea is expressed mathematically using the calculus of variations, a simpler example of the same idea is that of Fermat’s principle of least time applied to the derivation of Snell’s law of refraction in optics.

The idea is that given that the speed of light in a medium with index of refraction ${n}$ is ${c/n}$, if a light beam starts at a point ${A}$ in medium 1 and hits the interface between mediums 1 and 2 at an angle ${\theta_{1}}$ to the normal, and continues through into medium 2 at an angle ${\theta_{2}}$ to the normal eventually arriving at point ${B}$, then these angles are such that the travel time from ${A}$ to ${B}$ is a minimum. There isn’t any particular reason why this assumption is made (apart from the the fact that it gives the right answer!).

To see how it works, suppose we orient the interface so that it lies in the ${xy}$ plane, so that the normal to the interface is the ${z}$ axis. We’ll take the incident beam of light starting at point ${A}$ to lie in the ${yz}$ plane, as does the refracted beam which travels from the interface to point ${B}$. We’ll let ${y_{AB}}$ be the difference in ${y}$ coordinate of the points ${A}$ and ${B}$, and let ${y_{AO}}$ be the difference in ${y}$ coordinate of the point ${O}$ where the beam hits the interface. Thus the difference in ${y}$ coordinate between ${O}$ and ${B}$ is ${y_{OB}=y_{AB}-y_{AO}}$. Similarly, let ${z_{AO}}$ and ${z_{OB}}$ be the differences in ${z}$ coordinates between the corresponding points. Finally, let ${a}$ be the distance from ${A}$ to ${O}$, and ${b}$ the distance from ${O}$ to ${B}$.

Then by Pythagoras

 $\displaystyle a$ $\displaystyle =$ $\displaystyle \sqrt{z_{AO}^{2}+y_{AO}^{2}}\ \ \ \ \ (1)$ $\displaystyle b$ $\displaystyle =$ $\displaystyle \sqrt{z_{OB}^{2}+\left(y_{AB}-y_{AO}\right)^{2}} \ \ \ \ \ (2)$

The total travel time of the light beam is

 $\displaystyle t$ $\displaystyle =$ $\displaystyle \frac{a}{v_{1}}+\frac{b}{v_{2}}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{a}{c}n_{1}+\frac{b}{c}n_{2}\ \ \ \ \ (4)$ $\displaystyle ct$ $\displaystyle =$ $\displaystyle n_{1}\sqrt{z_{AO}^{2}+y_{AO}^{2}}+n_{2}\sqrt{z_{OB}^{2}+\left(y_{AB}-y_{AO}\right)^{2}} \ \ \ \ \ (5)$

Since the points ${A}$ and ${B}$ are fixed, as is the location of the interface, the only thing we can vary is ${y}$ coordinate of the point where the light beam hits the interface, that is, ${y_{AO}}$. We can therefore minimize ${ct}$ with respect to ${y_{AO}}$:

 $\displaystyle \frac{d\left(ct\right)}{dy_{AO}}$ $\displaystyle =$ $\displaystyle \frac{y_{AO}n_{1}}{\sqrt{z_{AO}^{2}+y_{AO}^{2}}}-\frac{n_{2}\left(y_{AB}-y_{AO}\right)}{\sqrt{z_{OB}^{2}+\left(y_{AB}-y_{AO}\right)^{2}}}=0\ \ \ \ \ (6)$ $\displaystyle \frac{y_{AO}}{\sqrt{z_{AO}^{2}+y_{AO}^{2}}}n_{1}$ $\displaystyle =$ $\displaystyle \frac{\left(y_{AB}-y_{AO}\right)}{\sqrt{z_{OB}^{2}+\left(y_{AB}-y_{AO}\right)^{2}}}n_{2}\ \ \ \ \ (7)$ $\displaystyle n_{1}\sin\theta_{1}$ $\displaystyle =$ $\displaystyle n_{2}\sin\theta_{2} \ \ \ \ \ (8)$

where the last line uses the trigonometric definition of the sine from the sides of the triangles.

## Electromagnetic waves in conductors: energy density and intensity

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.20.

We can write the electromagnetic wave inside a conductor as (if we orient the axes so that ${\mathbf{E}}$ is polarized in the ${x}$ direction)

 $\displaystyle \tilde{\mathbf{E}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{E}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E_{0}e^{-\kappa z}e^{i\left(kz-\omega t+\delta_{E}\right)}\hat{\mathbf{x}}\ \ \ \ \ (2)$ $\displaystyle \tilde{\mathbf{B}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \frac{\tilde{k}}{\omega}\tilde{E}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\epsilon\mu\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}}E_{0}e^{-\kappa z}e^{i\left(kz-\omega t+\delta_{E}+\phi\right)}\hat{\mathbf{y}} \ \ \ \ \ (4)$

where

$\displaystyle \tilde{k}=\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}+1}+i\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}-1}\equiv k+i\kappa\equiv Ke^{i\phi} \ \ \ \ \ (5)$

The actual fields are the real parts of these equations, so

 $\displaystyle \mathbf{E}\left(z,t\right)$ $\displaystyle =$ $\displaystyle E_{0}e^{-\kappa z}\cos\left(kz-\omega t+\delta_{E}\right)\hat{\mathbf{x}}\ \ \ \ \ (6)$ $\displaystyle \mathbf{B}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \sqrt{\epsilon\mu\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}}E_{0}e^{-\kappa z}\cos\left(kz-\omega t+\delta_{E}+\phi\right)\hat{\mathbf{y}} \ \ \ \ \ (7)$

The energy density in the wave is

$\displaystyle u=\frac{1}{2}\left(\epsilon E^{2}+\frac{1}{\mu}B^{2}\right) \ \ \ \ \ (8)$

Taking the time average (over one cycle) of this we have (since the average of ${\cos^{2}\omega t}$ over one cycle ${\tau=2\pi/\omega}$ is ${\frac{1}{2}}$):

$\displaystyle u=\frac{E_{0}^{2}e^{-2\kappa z}}{4}\left(\epsilon+\epsilon\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}\right) \ \ \ \ \ (9)$

For a good conductor, ${\sigma\gg\epsilon\omega}$ so

 $\displaystyle u$ $\displaystyle \approx$ $\displaystyle \frac{E_{0}^{2}e^{-2\kappa z}}{4}\left(\epsilon+\frac{\sigma}{\omega}\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \frac{E_{0}^{2}e^{-2\kappa z}}{4}\frac{\sigma}{\omega} \ \ \ \ \ (11)$

From 10, we see that the magnetic contribution (${\sigma/\omega}$) is much larger than the electric contribution (${\epsilon}$) for a good conductor.

We can express this in terms of the wave vector ${k}$ by using 5 for a good conductor.

 $\displaystyle k$ $\displaystyle \approx$ $\displaystyle \frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\frac{\sigma}{\epsilon\omega}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\omega\mu\sigma}{2}}\ \ \ \ \ (13)$ $\displaystyle \sigma$ $\displaystyle =$ $\displaystyle \frac{2k^{2}}{\omega\mu}\ \ \ \ \ (14)$ $\displaystyle u$ $\displaystyle \approx$ $\displaystyle \frac{E_{0}^{2}e^{-2\kappa z}}{2}\frac{k^{2}}{\mu\omega^{2}} \ \ \ \ \ (15)$

The intensity is the energy crossing a unit area in unit time, which is the energy density times the volume crossing a unit area per unit time, which is

$\displaystyle I=uv \ \ \ \ \ (16)$

where ${v}$ is the speed of the wave, which is ${\omega/k}$ so

$\displaystyle I=\frac{E_{0}^{2}e^{-2\kappa z}}{2}\frac{k}{\mu\omega} \ \ \ \ \ (17)$

## Electromagnetic waves in conductors: phases and amplitudes

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.19c.

Electromagnetic waves in a conductor (where there is free current but no free charge) can be written as

 $\displaystyle \tilde{\mathbf{E}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{E}}_{0}e^{i\left(\tilde{k}z-\omega t\right)}\ \ \ \ \ (1)$ $\displaystyle \tilde{\mathbf{B}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{B}}_{0}e^{i\left(\tilde{k}z-\omega t\right)} \ \ \ \ \ (2)$

where the wave vector is complex:

$\displaystyle \tilde{k}=\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}+1}+i\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}-1}\equiv k+i\kappa \ \ \ \ \ (3)$

so

 $\displaystyle \tilde{\mathbf{E}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{E}}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\ \ \ \ \ (4)$ $\displaystyle \tilde{\mathbf{B}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{B}}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)} \ \ \ \ \ (5)$

By applying Maxwell’s equations in a conductor we can get a few more properties of these waves. The equations are

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (7)$ $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (8)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \mu\sigma\mathbf{E}+\mu\epsilon\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (9)$

Using the same techniques as in analyzing waves in vacuum. Both ${\nabla\cdot\mathbf{E}=0}$ and ${\nabla\cdot\mathbf{B}=0}$ from which we get

 $\displaystyle \nabla\cdot\tilde{\mathbf{E}}$ $\displaystyle =$ $\displaystyle \left(ik-\kappa\right)\tilde{E}_{0z}e^{i\left(kz-\omega t\right)}=0\ \ \ \ \ (10)$ $\displaystyle \nabla\cdot\tilde{\mathbf{B}}$ $\displaystyle =$ $\displaystyle \left(ik-\kappa\right)\tilde{B}_{0z}e^{i\left(kz-\omega t\right)}=0 \ \ \ \ \ (11)$

Since this must be true for all ${z}$, we must have

$\displaystyle \tilde{E}_{0z}=\tilde{B}_{0z}=0 \ \ \ \ \ (12)$

That is, the wave has only ${x}$ and ${y}$ components, so it must be a transverse wave: a wave that oscillates in a plane perpendicular to the direction of propagation. If we orient the axes so that ${\mathbf{E}}$ is polarized in the ${x}$ direction then

$\displaystyle \tilde{\mathbf{E}}\left(z,t\right)=\tilde{E}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\hat{\mathbf{x}} \ \ \ \ \ (13)$

Applying 8 to this gives

 $\displaystyle \nabla\times\tilde{\mathbf{E}}$ $\displaystyle =$ $\displaystyle i\left(k+i\kappa\right)\tilde{E}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\tilde{k}\tilde{E}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\omega\tilde{\mathbf{B}}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\ \ \ \ \ (17)$ $\displaystyle \tilde{\mathbf{B}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \frac{\tilde{k}}{\omega}\tilde{E}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\hat{\mathbf{y}} \ \ \ \ \ (18)$

As in vacuum, ${\mathbf{E}}$ and ${\mathbf{B}}$ are perpendicular and transverse to the direction of propagation. Unlike in the vacuum, however, the two components of the wave may not be in phase, due to the presence of the complex variable ${\tilde{k}}$ in the equation for ${\tilde{\mathbf{B}}}$. If we write ${\tilde{k}}$ in modulus-phase form we have

$\displaystyle \tilde{k}=Ke^{i\phi} \ \ \ \ \ (19)$

where

 $\displaystyle K$ $\displaystyle =$ $\displaystyle \sqrt{k^{2}+\kappa^{2}}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \omega\sqrt{\epsilon\mu\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}}\ \ \ \ \ (21)$ $\displaystyle \phi$ $\displaystyle =$ $\displaystyle \arctan\frac{\kappa}{k} \ \ \ \ \ (22)$

Then the complex amplitudes of the two components can be written as

 $\displaystyle \tilde{E}_{0}$ $\displaystyle =$ $\displaystyle E_{0}e^{i\delta_{E}}\ \ \ \ \ (23)$ $\displaystyle \tilde{B}_{0}$ $\displaystyle =$ $\displaystyle \frac{K}{\omega}E_{0}e^{i\left(\delta_{E}+\phi\right)} \ \ \ \ \ (24)$

and the ratio of the real amplitudes is

$\displaystyle \frac{B_{0}}{E_{0}}=\frac{K}{\omega}=\sqrt{\epsilon\mu\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}} \ \ \ \ \ (25)$

Example For a good conductor, ${\sigma\gg\epsilon\omega}$ so from 3 ${k\approx\kappa}$ so from 22 the phase difference between ${\mathbf{B}}$ and ${\mathbf{E}}$ is ${\pi/4}$. The ratio of amplitudes is

$\displaystyle \frac{B_{0}}{E_{0}}=\sqrt{\frac{\sigma\mu}{\omega}} \ \ \ \ \ (26)$

For a typical good conductor ${\sigma\approx10^{7}\mbox{S m}^{-1}}$ and ${\mu\approx\mu_{0}}$ and at visible frequencies ${\omega\approx10^{15}\mbox{ s}^{-1}}$ so

$\displaystyle \frac{B_{0}}{E_{0}}=1.12\times10^{-7}\mbox{ s m}^{-1} \ \ \ \ \ (27)$

## Skin depth of water and metals

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.19a-b.

Electromagnetic waves in a conductor (where there is free current but no free charge) can be written as

 $\displaystyle \tilde{\mathbf{E}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{E}}_{0}e^{i\left(\tilde{k}z-\omega t\right)}\ \ \ \ \ (1)$ $\displaystyle \tilde{\mathbf{B}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{B}}_{0}e^{i\left(\tilde{k}z-\omega t\right)} \ \ \ \ \ (2)$

where the wave vector is complex:

$\displaystyle \tilde{k}=\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}+1}+i\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}-1}\equiv k+i\kappa \ \ \ \ \ (3)$

For a poor conductor, the conductivity ${\sigma}$ is small, so for large enough frequencies ${\sigma\ll\epsilon\omega}$ and we can approximate ${\kappa}$ by

 $\displaystyle \kappa$ $\displaystyle \approx$ $\displaystyle \frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{1+\frac{1}{2}\left(\frac{\sigma}{\epsilon\omega}\right)^{2}-1}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sigma}{2}\sqrt{\frac{\mu}{\epsilon}} \ \ \ \ \ (5)$

Since the imaginary part of ${\tilde{k}}$ governs the attenuation of the wave as it penetrates the material, the skin depth for a poor conductor is

$\displaystyle d=\frac{1}{\kappa}=\frac{2}{\sigma}\sqrt{\frac{\epsilon}{\mu}} \ \ \ \ \ (6)$

For pure (deionized) water ${\sigma=5.5\times10^{-6}\mbox{S m}^{-1}}$ and ${\epsilon=80.1\epsilon_{0}}$ (at ${20^{\circ}\mbox{C}}$) (we can take ${\mu\approx\mu_{0}}$) so the skin depth of water is

$\displaystyle d=8635\mbox{ m} \ \ \ \ \ (7)$

Because the skin depth is so large, water is transparent.

For a good conductor, ${\sigma\gg\epsilon\omega}$ and we can approximate

$\displaystyle \kappa\approx\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\frac{\sigma}{\epsilon\omega}}=\sqrt{\frac{\mu\sigma\omega}{2}}\approx k \ \ \ \ \ (8)$

so the skin depth is

$\displaystyle d=\sqrt{\frac{2}{\mu\sigma\omega}}\approx\frac{1}{k}=\frac{2\pi}{\lambda} \ \ \ \ \ (9)$

where ${\lambda}$ is the wavelength within the material. For a typical metal, ${\sigma\approx10^{7}\mbox{S m}^{-1}}$ and ${\mu\approx\mu_{0}}$ so the skin depth at visible frequencies ${\omega\approx10^{15}\mbox{s}^{-1}}$ is

$\displaystyle d\approx1.26\times10^{-8}\mbox{m} \ \ \ \ \ (10)$

With a skin depth this small, even a thin film of metal is effectively impervious to any penetration by visible light.

## Skin depth of electromagnetic waves in conductors

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.18b-c.

We’ve seen that any free charge within a conductor migrates to the surface with a characteristic time that depends on the conductor’s conductance and permittivity. Once that transient effect subsides, we can take the free charge density to be zero: ${\rho_{f}=0}$. This doesn’t mean that the free current density ${\mathbf{J}_{f}}$ is zero, though. We can still have a current in an electrically neutral conductor caused by electrons moving relative to stationary atomic nuclei.

In linear media, Maxwell’s equations are

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{1}{\epsilon}\rho_{f}\ \ \ \ \ (1)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (3)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \mu\mathbf{J}_{f}+\mu\epsilon\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (4)$

Within a conductor with conductivity ${\sigma}$ (not surface charge density!) Ohm’s law can be written as

$\displaystyle \mathbf{J}_{f}=\sigma\mathbf{E} \ \ \ \ \ (5)$

so with ${\rho_{f}=0}$, Maxwell’s equations become

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (7)$ $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (8)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \mu\sigma\mathbf{E}+\mu\epsilon\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (9)$

We can take the curl of the last two equations in the same way as when we derived the wave equation in a vacuum.

 $\displaystyle \nabla\times\left(\nabla\times\mathbf{E}\right)$ $\displaystyle =$ $\displaystyle \nabla\left(\nabla\cdot\mathbf{E}\right)-\nabla^{2}\mathbf{E}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\nabla\times\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\mu\epsilon\frac{\partial^{2}\mathbf{E}}{\partial t^{2}}-\mu\sigma\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (12)$

Since ${\nabla\cdot\mathbf{E}=0}$ we get

$\displaystyle \nabla^{2}\mathbf{E}=\mu\epsilon\frac{\partial^{2}\mathbf{E}}{\partial t^{2}}+\mu\sigma\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (13)$

A similar calculation for ${\mathbf{B}}$ gives

$\displaystyle \nabla^{2}\mathbf{B}=\mu\epsilon\frac{\partial^{2}\mathbf{B}}{\partial t^{2}}+\mu\sigma\frac{\partial\mathbf{B}}{\partial t} \ \ \ \ \ (14)$

We thus get the wave equation modified by the addition of an extra first-derivative term. Conveniently, these equations have a similar solution to the ordinary wave equation. For a plane wave travelling in the ${z}$ direction

 $\displaystyle \tilde{\mathbf{E}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{E}}_{0}e^{i\left(\tilde{k}z-\omega t\right)}\ \ \ \ \ (15)$ $\displaystyle \tilde{\mathbf{B}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{B}}_{0}e^{i\left(\tilde{k}z-\omega t\right)} \ \ \ \ \ (16)$

Substituting 15 into 13 we get, after cancelling terms

$\displaystyle \tilde{k}^{2}=\omega^{2}\mu\epsilon+i\omega\mu\sigma \ \ \ \ \ (17)$

The fact that the wave vector ${\tilde{k}}$ is complex means that the resulting wave has both an oscillatory and an exponentially decaying factor. Finding the square root of this using Maple gives

 $\displaystyle \tilde{k}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\,\sqrt{2\,\sqrt{{\omega}^{4}{\mu}^{2}{\epsilon}^{2}+{\omega}^{2}{\mu}^{2}{\sigma}^{2}}+2\,{\omega}^{2}\mu\,\epsilon}+\frac{1}{2}\, i\sqrt{2\,\sqrt{{\omega}^{4}{\mu}^{2}{\epsilon}^{2}+{\omega}^{2}{\mu}^{2}{\sigma}^{2}}-2\,{\omega}^{2}\mu\,\epsilon}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}+1}+i\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}-1}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle k+i\kappa \ \ \ \ \ (20)$

The wave is then given by

 $\displaystyle \tilde{\mathbf{E}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{E}}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\ \ \ \ \ (21)$ $\displaystyle \tilde{\mathbf{B}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{B}}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)} \ \ \ \ \ (22)$

If an electromagnetic wave hits a conductor (starting in air or vacuum, say), the wave will attenuate as it penetrates the conductor with a characteristic distance, called the skin depth ${d}$ of

$\displaystyle d=\frac{1}{\kappa}=\frac{1}{\omega}\left[\frac{\mu\epsilon}{2}\left(\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}-1\right)\right]^{-1/2} \ \ \ \ \ (23)$

The skin depth depends not only on the conductivity and permittivity, but also on the frequency of the incident radiation.

Example 1 The skin depth for good conductors such as metals is very small for a wide range of frequencies. For silver, the conductivity is ${\sigma=6.30\times10^{7}\mbox{ S m}^{-1}}$. To get the permittivity of silver, we can return to its definition:

$\displaystyle \epsilon=\epsilon_{0}\left(1+\chi_{e}\right) \ \ \ \ \ (24)$

where the electric susceptibility ${\chi_{e}}$ is defined by the amount of polarization in the material that is produced by an applied electric field:

$\displaystyle \mathbf{P}=\epsilon_{0}\chi_{e}\mathbf{E} \ \ \ \ \ (25)$

For a perfect conductor, no polarization is produced by any amount of electric field, so ${\chi_{e}=0}$ and ${\epsilon=\epsilon_{0}}$. Thus it’s a reasonable approximation for a good conductor like silver to take ${\epsilon\approx\epsilon_{0}}$. Most materials also have a permeability ${\mu\approx\mu_{0}}$. At a microwave frequency of ${10^{10}\mbox{ Hz}=2\pi\times10^{10}\mbox{ s}^{-1}}$

$\displaystyle \frac{\sigma}{\epsilon_{0}\omega}=1.13\times10^{8} \ \ \ \ \ (26)$

so to a good approximation

$\displaystyle d=\frac{1}{\omega}\sqrt{\frac{2\omega}{\mu_{0}\sigma}}=\sqrt{\frac{2}{\omega\mu_{0}\sigma}}=6.34\times10^{-7}\mbox{m} \ \ \ \ \ (27)$

The real part of ${\tilde{k}}$ gives the oscillatory part of the wave, so the wavelength is

$\displaystyle \lambda=\frac{2\pi}{k} \ \ \ \ \ (28)$

and the wave speed is

$\displaystyle v=\nu\lambda=\frac{\omega}{2\pi}\lambda=\frac{\omega}{k} \ \ \ \ \ (29)$

The index of refraction is the ratio of ${c}$ to ${v}$ as usual:

$\displaystyle n=\frac{c}{v}=\frac{ck}{\omega} \ \ \ \ \ (30)$

Example 2 For copper, ${\sigma=5.96\times10^{7}\mbox{ S m}^{-1}}$. For radio waves with a frequency of ${1\mbox{ MHz}=2\pi\times10^{6}\mbox{ s}^{-1}}$ we have

 $\displaystyle \frac{\sigma}{\epsilon_{0}\omega}$ $\displaystyle =$ $\displaystyle 1.07\times10^{12}\ \ \ \ \ (31)$ $\displaystyle k$ $\displaystyle \approx$ $\displaystyle \sqrt{\frac{\omega\mu_{0}\sigma}{2}}\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.533\times10^{4}\mbox{ m}^{-1}\ \ \ \ \ (33)$ $\displaystyle \lambda$ $\displaystyle =$ $\displaystyle \frac{2\pi}{k}=4.1\times10^{-4}\mbox{m} \ \ \ \ \ (34)$

The wave speed is

$\displaystyle v=\frac{\omega}{k}=410\mbox{ m s}^{-1} \ \ \ \ \ (35)$

This is obviously very slow for electromagnetic radiation.

In air or vacuum, ${v=c=3\times10^{8}\mbox{m s}^{-1}}$ and ${\lambda=v/\nu=300\mbox{ m}}$ which is closer to what we’d expect for radio waves.

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.18a.

When we looked at conductors in electrostatics, we saw that any free charge in a conductor must reside on the surface. In electrodynamics, however, we can place some free charge inside a conductor and then watch it (figuratively) move to the surface. How long does this migration take?

 $\displaystyle \nabla\cdot\mathbf{D}$ $\displaystyle =$ $\displaystyle \rho_{f}\ \ \ \ \ (1)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (3)$ $\displaystyle \nabla\times\mathbf{H}$ $\displaystyle =$ $\displaystyle \mathbf{J}_{f}+\frac{\partial\mathbf{D}}{\partial t} \ \ \ \ \ (4)$

For linear media, ${\mathbf{D}=\epsilon\mathbf{E}}$ and ${\mathbf{H}=\mathbf{B}/\mu}$, so these equations become

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{1}{\epsilon}\rho_{f}\ \ \ \ \ (5)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (7)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \mu\mathbf{J}_{f}+\mu\epsilon\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (8)$

Within a conductor with conductivity ${\sigma}$ (not surface charge density!) Ohm’s law can be written as

$\displaystyle \mathbf{J}_{f}=\sigma\mathbf{E} \ \ \ \ \ (9)$

so Maxwell’s equations become

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{1}{\epsilon}\rho_{f}\ \ \ \ \ (10)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (11)$ $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (12)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \mu\sigma\mathbf{E}+\mu\epsilon\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (13)$

The conservation of charge (continuity condition) says

$\displaystyle \nabla\cdot\mathbf{J}_{f}=-\frac{\partial\rho_{f}}{\partial t} \ \ \ \ \ (14)$

so from 9 and 10 we get

$\displaystyle \frac{\partial\rho_{f}}{\partial t}=-\sigma\nabla\cdot\mathbf{E}=-\frac{\sigma}{\epsilon}\rho_{f} \ \ \ \ \ (15)$

For a fixed location within the conductor we can integrate this with respect to time to get

$\displaystyle \rho_{f}\left(t\right)=\rho_{f}\left(0\right)e^{-\sigma t/\epsilon} \ \ \ \ \ (16)$

That is, the free charge density decays exponentially with a characteristic time ${\tau=\epsilon/\sigma}$.

Example For glass, the conductivity is around ${10^{-11}\mbox{ S m}^{-1}}$ (S stands for Siemens, which is the SI unit of conductance, where ${1\mbox{ S}=1\mbox{ kg}^{-1}\mbox{m}^{-2}\mbox{s}^{3}\mbox{A}^{2}}$) and the permittivity is ${4.7\epsilon_{0}=4.16\times10^{-11}\mbox{m}^{-3}\mbox{kg}^{-1}\mbox{s}^{4}\mbox{A}^{2}}$ so the charge would move to the surface in a time of about

$\displaystyle \tau=4.16\mbox{ s} \ \ \ \ \ (17)$

Some forms of glass have much smaller conductivities, so the migration time would be correspondingly larger.

## Optical properties of diamond

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.17.

We’ve seen how to derive the properties of reflected and transmitted waves in the case where the wave is polarized perpendicular to the plane of incidence. The derivation in the case of parallel polarization is very similar and is given in Griffiths 9.3.3. Here we’ll have a look at some of these properties at an interface between air and diamond.

The Fresnel equations for parallel polarization, giving the reflected and transmitted amplitudes in terms of the incident amplitude, turn out to be

 $\displaystyle E_{R}$ $\displaystyle =$ $\displaystyle \frac{\alpha-\beta}{\alpha+\beta}E_{I}\ \ \ \ \ (1)$ $\displaystyle E_{T}$ $\displaystyle =$ $\displaystyle \frac{2}{\alpha+\beta}E_{I} \ \ \ \ \ (2)$

where the angle of incidence is ${\theta_{I}}$, the angle of transmission is ${\theta_{T}}$ and

 $\displaystyle \alpha$ $\displaystyle \equiv$ $\displaystyle \frac{\cos\theta_{T}}{\cos\theta_{I}}\ \ \ \ \ (3)$ $\displaystyle \beta$ $\displaystyle \equiv$ $\displaystyle \frac{\mu_{1}n_{2}}{\mu_{2}n_{1}} \ \ \ \ \ (4)$

Taking ${\mu_{1}=\mu_{2}=\mu_{0}}$ and using diamond’s index of refraction ${n_{2}=2.42}$, we can draw plots of ${E_{R}/E_{I}}$ and ${E_{T}/E_{I}}$ (red for reflected and blue for transmitted):

At normal incidence ${\theta_{I}=\theta_{T}=0}$ and

 $\displaystyle \frac{E_{R}}{E_{I}}$ $\displaystyle =$ $\displaystyle -0.415\ \ \ \ \ (5)$ $\displaystyle \frac{E_{T}}{E_{I}}$ $\displaystyle =$ $\displaystyle 0.585 \ \ \ \ \ (6)$

The negative value for the reflected amplitude indicates that the wave is ${\pi}$ out of phase with the incident wave.

The reflected and transmitted amplitudes are equal where the curves cross, which occurs at an angle obtained from solving ${E_{R}=E_{T}}$:

$\displaystyle \theta_{R=T}=1.362\mbox{ rad}=78.06^{\circ} \ \ \ \ \ (7)$

We can see that if ${\alpha=\beta}$, ${E_{R}=0}$ and there is no reflected wave. This occurs at Brewster’s angle ${\theta_{B}}$, given by

$\displaystyle \sin^{2}\theta_{B}=\frac{1-\beta^{2}}{\left(n_{1}/n_{2}\right)^{2}-\beta^{2}} \ \ \ \ \ (8)$

For the air-diamond interface, we get

$\displaystyle \theta_{B}=1.179\mbox{ rad}=67.55^{\circ} \ \ \ \ \ (9)$

## Fresnel equations for perpendicular polarization

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.16.

Continuing our study of electromagnetic waves incident on a surface at an oblique angle we’ll now use the boundary conditions to derive the reflection and transmission coefficients. The boundary conditions are

 $\displaystyle \epsilon_{1}E_{1}^{\perp}-\epsilon_{2}E_{2}^{\perp}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (1)$ $\displaystyle B_{1}^{\perp}-B_{2}^{\perp}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle \mathbf{E}_{1}^{\parallel}-\mathbf{E}_{2}^{\parallel}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (3)$ $\displaystyle \frac{1}{\mu_{1}}\mathbf{B}_{1}^{\parallel}-\frac{1}{\mu_{2}}\mathbf{B}_{2}^{\parallel}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (4)$

where the subscript 1 refers to fields in medium 1 (${z<0}$) and 2 refers to medium 2 (${z>0}$). That is

 $\displaystyle \mathbf{E}_{1}$ $\displaystyle =$ $\displaystyle \mathbf{E}_{I}+\mathbf{E}_{R}\ \ \ \ \ (5)$ $\displaystyle \mathbf{E}_{2}$ $\displaystyle =$ $\displaystyle \mathbf{E}_{T} \ \ \ \ \ (6)$

and similarly for ${\mathbf{B}}$. As we saw last time, the space-time dependence cancels out of the boundary conditions, we can replace all fields by their (complex) amplitudes so we get

 $\displaystyle \epsilon_{1}\left(\tilde{E}_{0_{I}}^{\perp}+\tilde{E}_{0_{R}}^{\perp}\right)$ $\displaystyle =$ $\displaystyle \epsilon_{2}\tilde{E}_{0_{T}}^{\perp}\ \ \ \ \ (7)$ $\displaystyle \tilde{B}_{0_{I}}^{\perp}+\tilde{B}_{0_{R}}^{\perp}$ $\displaystyle =$ $\displaystyle \tilde{B}_{0_{T}}^{\perp}\ \ \ \ \ (8)$ $\displaystyle \mathbf{\tilde{E}}_{0_{I}}^{\parallel}+\mathbf{\tilde{E}}_{0_{R}}^{\parallel}$ $\displaystyle =$ $\displaystyle \mathbf{\tilde{E}}_{0_{T}}^{\parallel}\ \ \ \ \ (9)$ $\displaystyle \frac{1}{\mu_{1}}\left(\mathbf{\tilde{B}}_{0_{I}}^{\parallel}+\mathbf{\tilde{B}}_{0_{R}}^{\parallel}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\mu_{2}}\mathbf{\tilde{B}}_{0_{T}}^{\parallel} \ \ \ \ \ (10)$

There are actually two cases to consider: polarization parallel to the incident plane (that is, ${\mathbf{E}}$ in the ${xz}$ plane in our example) or perpendicular to the incident plane (so ${\mathbf{E}}$ is polarized along the ${y}$ axis). Griffiths does the parallel case in his section 9.3.3 so we’ll look at the perpendicular case here. (It’s important to be clear about what is perpendicular or parallel to what. In the four boundary conditions above, the ${\perp}$ and ${\parallel}$ symbols mean perpendicular and parallel to the boundary (that is, the ${xy}$ plane) not the incident plane. The polarization we’re considering is perpendicular to the incident plane.)

The incident wave travels along wave vector ${\mathbf{k}_{I}}$ at an angle of ${\theta_{I}}$ to the normal to the ${xy}$ plane, the reflected wave travels along ${\mathbf{k}_{R}}$ also at an angle of ${\theta_{I}}$, and the transmitted wave travels along ${\mathbf{k}_{T}}$ at angle ${\theta_{T}}$ such that, according to Snell’s law

$\displaystyle \frac{\sin\theta_{I}}{\sin\theta_{T}}=\frac{n_{2}}{n_{1}} \ \ \ \ \ (11)$

Condition 7 tells us nothing since ${\mathbf{E}}$ is in the ${y}$ direction, so has no component perpendicular to the ${xy}$ plane. To use the other conditions, we need to work out the components of ${\mathbf{E}}$ and ${\mathbf{B}}$. We know ${\mathbf{E}}$ is in the ${y}$ direction so that’s easy. The direction of ${\mathbf{B}}$ is given by ${\mathbf{k}\times\mathbf{E}}$. Consider ${\mathbf{B}_{I}}$. Here ${\mathbf{k}_{I}}$ points towards the ${xy}$ plane (from the left) at an angle ${\theta_{I}}$ to the normal to this plane. The cross product ${\mathbf{k}\times\mathbf{E}}$ therefore lies in the ${xz}$ plane and points to the lower right at an angle ${\frac{\pi}{2}-\theta_{I}}$ to the normal, so the components of ${\mathbf{B}_{I}}$ are (to keep the notation simple in what follows we’ll drop the 0 subscript and tilde for the amplitudes):

 $\displaystyle B_{I_{z}}$ $\displaystyle =$ $\displaystyle B_{I}\cos\left(\frac{\pi}{2}-\theta_{I}\right)=\frac{1}{v_{1}}E_{I}\sin\theta_{I}\ \ \ \ \ (12)$ $\displaystyle B_{I_{x}}$ $\displaystyle =$ $\displaystyle -B_{I}\sin\left(\frac{\pi}{2}-\theta_{I}\right)=-\frac{1}{v_{1}}E_{I}\cos\theta_{I} \ \ \ \ \ (13)$

${B_{I_{x}}}$ is negative since ${\mathbf{B}_{I}}$ points towards negative ${x}$ and positive ${z}$.

Assuming the reflected wave still has polarization in the ${+y}$ direction, the direction of ${\mathbf{B}_{R}}$ is now ${\mathbf{k}_{R}\times\mathbf{E}_{R}}$ and ${\mathbf{k}_{R}}$points away (towards the left) from the ${xy}$ plane at angle ${\theta_{R}=\theta_{I}}$ so ${\mathbf{B}_{R}}$ points to the upper right and has components

 $\displaystyle B_{R_{z}}$ $\displaystyle =$ $\displaystyle B_{R}\cos\left(\frac{\pi}{2}-\theta_{I}\right)=\frac{1}{v_{1}}E_{R}\sin\theta_{I}\ \ \ \ \ (14)$ $\displaystyle B_{R_{x}}$ $\displaystyle =$ $\displaystyle B_{R}\sin\left(\frac{\pi}{2}-\theta_{I}\right)=\frac{1}{v_{1}}E_{R}\cos\theta_{I} \ \ \ \ \ (15)$

Finally, the transmitted wave has direction ${\mathbf{k}_{T}}$ which points away (towards the right) from the ${xy}$ plane, so ${\mathbf{B}_{T}=\frac{1}{v_{2}}\mathbf{k}_{T}\times\mathbf{E}_{T}}$ points to the lower right and has components

 $\displaystyle B_{T_{z}}$ $\displaystyle =$ $\displaystyle B_{T}\cos\left(\frac{\pi}{2}-\theta_{T}\right)=\frac{1}{v_{2}}E_{T}\sin\theta_{T}\ \ \ \ \ (16)$ $\displaystyle B_{T_{x}}$ $\displaystyle =$ $\displaystyle -B_{T}\sin\left(\frac{\pi}{2}-\theta_{T}\right)=-\frac{1}{v_{2}}E_{T}\cos\theta_{T} \ \ \ \ \ (17)$

We’re now ready to apply the boundary conditions. First, we use 8, which applies to the ${z}$ components of ${\mathbf{B}}$ so we have

 $\displaystyle \frac{1}{v_{1}}E_{I}\sin\theta_{I}+\frac{1}{v_{1}}E_{R}\sin\theta_{I}$ $\displaystyle =$ $\displaystyle \frac{1}{v_{2}}E_{T}\sin\theta_{T}\ \ \ \ \ (18)$ $\displaystyle E_{I}+E_{R}$ $\displaystyle =$ $\displaystyle \frac{v_{1}}{v_{2}}\frac{\sin\theta_{T}}{\sin\theta_{I}}E_{T}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{n_{1}v_{1}}{n_{2}v_{2}}E_{T}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E_{T} \ \ \ \ \ (21)$

Condition 9 just gives us the same relation, so we don’t learn anything new from it. Condition 10 applies to ${B_{x}}$ only since ${\mathbf{B}}$ has no ${y}$ component.

 $\displaystyle \frac{1}{\mu_{1}}\left(-\frac{1}{v_{1}}E_{I}\cos\theta_{I}+\frac{1}{v_{1}}E_{R}\cos\theta_{I}\right)$ $\displaystyle =$ $\displaystyle -\frac{1}{\mu_{2}v_{2}}E_{T}\cos\theta_{T}\ \ \ \ \ (22)$ $\displaystyle E_{I}-E_{R}$ $\displaystyle =$ $\displaystyle \frac{\mu_{1}v_{1}}{\mu_{2}v_{2}}E_{T}\frac{\cos\theta_{T}}{\cos\theta_{I}}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \alpha\beta E_{T} \ \ \ \ \ (24)$

where

 $\displaystyle \alpha$ $\displaystyle \equiv$ $\displaystyle \frac{\cos\theta_{T}}{\cos\theta_{I}}\ \ \ \ \ (25)$ $\displaystyle \beta$ $\displaystyle \equiv$ $\displaystyle \frac{\mu_{1}v_{1}}{\mu_{2}v_{2}}=\frac{\mu_{1}n_{2}}{\mu_{2}n_{1}} \ \ \ \ \ (26)$

Solving these two equations gives

 $\displaystyle E_{R}$ $\displaystyle =$ $\displaystyle \frac{1-\alpha\beta}{1+\alpha\beta}E_{I}\ \ \ \ \ (27)$ $\displaystyle E_{T}$ $\displaystyle =$ $\displaystyle \frac{2}{1+\alpha\beta}E_{I} \ \ \ \ \ (28)$

These are Fresnel’s equations for perpendicular polarization. For normal incidence, ${\theta_{I}=\theta_{T}=0}$, ${\alpha=1}$ and they reduce to the equations we got in that case. Plots of ${E_{R}/E_{I}}$ (red) and ${E_{T}/E_{I}}$ (blue) for ${n_{2}/n_{1}=1.5}$ are as shown:

The negative values for ${E_{R}/E_{I}}$ indicate that the reflected wave is ${\pi}$ out of phase with the incident wave. For ${\theta_{I}=0}$ (normal incidence) 80% of the amplitude is transmitted, dropping to zero when ${\theta_{I}=\pi/2}$ (incident wave is parallel to the surface).

The Fresnel equations for parallel polarization (see Griffiths) turn out to be

 $\displaystyle E_{R}$ $\displaystyle =$ $\displaystyle \frac{\alpha-\beta}{\alpha+\beta}E_{I}\ \ \ \ \ (29)$ $\displaystyle E_{T}$ $\displaystyle =$ $\displaystyle \frac{2}{\alpha+\beta}E_{I} \ \ \ \ \ (30)$

We can see that if ${\alpha=\beta}$, ${E_{R}=0}$ and there is no reflected wave. This occurs at Brewster’s angle ${\theta_{B}}$, given by

$\displaystyle \sin^{2}\theta_{B}=\frac{1-\beta^{2}}{\left(n_{1}/n_{2}\right)^{2}-\beta^{2}} \ \ \ \ \ (31)$

For perpendicular polarization, there is no reflection if we can find an angle ${\theta_{B}}$ such that ${\alpha=1/\beta}$.

 $\displaystyle \alpha^{2}$ $\displaystyle =$ $\displaystyle \frac{1-\sin^{2}\theta_{T}}{1-\sin^{2}\theta_{I}}\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1-\left(n_{1}/n_{2}\right)^{2}\sin^{2}\theta_{I}}{1-\sin^{2}\theta_{I}}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\beta^{2}}\ \ \ \ \ (34)$ $\displaystyle \sin^{2}\theta_{B}$ $\displaystyle =$ $\displaystyle \frac{1-1/\beta^{2}}{\left(n_{1}/n_{2}\right)^{2}-1/\beta^{2}}\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\beta^{2}-1}{\beta^{2}\left(n_{1}/n_{2}\right)^{2}-1} \ \ \ \ \ (36)$

In practice, the permeabilities of media are approximately equal, so that ${\mu_{1}\approx\mu_{2}}$ and ${\beta\approx n_{2}/n_{1}}$ from 26. In this case, the expression for ${\sin^{2}\theta_{B}}$ blows up so there is no solution, and thus no Brewster angle for perpendicular polarization (unless ${\beta=1}$ which occurs only if ${n_{1}=n_{2}}$ so there is effectively no boundary).

 $\displaystyle R$ $\displaystyle =$ $\displaystyle \frac{\frac{1}{2}\epsilon_{1}v_{1}E_{R}^{2}}{\frac{1}{2}\epsilon_{1}v_{1}E_{I}^{2}}\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(1-\alpha\beta\right)^{2}}{\left(1+\alpha\beta\right)^{2}}\ \ \ \ \ (38)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{\frac{1}{2}\epsilon_{2}v_{2}E_{T}^{2}}{\frac{1}{2}\epsilon_{1}v_{1}E_{I}^{2}}\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\alpha\beta}{\left(1+\alpha\beta\right)^{2}} \ \ \ \ \ (40)$

and it can be seen that ${R+T=1}$.

## Three laws of geometrical optics

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.15.

We got the equations for reflected and transmitted waves when an electromagnetic wave is incident on a boundary head-on. Using similar (although somewhat more involved) methods, we can solve the general case where a wave is incident on a boundary at any angle.

Suppose the wave vector of the incident wave is ${\mathbf{k}_{I}}$ and the boundary is located on the ${xy}$ plane. The incident wave is

 $\displaystyle \tilde{\mathbf{E}}_{I}$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{E}}_{0_{I}}e^{i\left(\mathbf{k}_{I}\cdot\mathbf{r}-\omega t\right)}\ \ \ \ \ (1)$ $\displaystyle \tilde{\mathbf{B}}_{I}$ $\displaystyle =$ $\displaystyle \frac{1}{v_{1}}\hat{\mathbf{k}}_{I}\times\tilde{\mathbf{E}}_{I} \ \ \ \ \ (2)$

where the direction of polarization is given by the direction of ${\tilde{\mathbf{E}}_{0_{I}}}$. At this stage, we don’t know the direction of either the reflected or transmitted waves, so we can just write them as

 $\displaystyle \tilde{\mathbf{E}}_{R}$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{E}}_{0_{R}}e^{i\left(\mathbf{k}_{R}\cdot\mathbf{r}-\omega t\right)}\ \ \ \ \ (3)$ $\displaystyle \tilde{\mathbf{B}}_{R}$ $\displaystyle =$ $\displaystyle \frac{1}{v_{1}}\hat{\mathbf{k}}_{R}\times\tilde{\mathbf{E}}_{R}\ \ \ \ \ (4)$ $\displaystyle \tilde{\mathbf{E}}_{T}$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{E}}_{0_{T}}e^{i\left(\mathbf{k}_{T}\cdot\mathbf{r}-\omega t\right)}\ \ \ \ \ (5)$ $\displaystyle \tilde{\mathbf{B}}_{T}$ $\displaystyle =$ $\displaystyle \frac{1}{v_{2}}\hat{\mathbf{k}}_{T}\times\tilde{\mathbf{E}}_{T} \ \ \ \ \ (6)$

The speeds ${v_{i}}$ of the waves are determined by the medium and not by the direction of the waves and since the frequency ${\omega}$ is the same for all three waves and ${\omega=kv}$, we get

 $\displaystyle k_{I}v_{1}$ $\displaystyle =$ $\displaystyle k_{R}v_{1}\ \ \ \ \ (7)$ $\displaystyle k_{I}$ $\displaystyle =$ $\displaystyle k_{R}\ \ \ \ \ (8)$ $\displaystyle k_{I}v_{1}$ $\displaystyle =$ $\displaystyle k_{T}v_{2}\ \ \ \ \ (9)$ $\displaystyle k_{T}$ $\displaystyle =$ $\displaystyle \frac{v_{1}}{v_{2}}k_{I}=\frac{n_{2}}{n_{1}}k_{I} \ \ \ \ \ (10)$

where ${n_{i}}$ is the index of refraction.

This gives us the magnitudes of the wave vectors, but not the directions. To get those, we need to apply the boundary conditions for zero surface charge:

 $\displaystyle \epsilon_{1}E_{1}^{\perp}-\epsilon_{2}E_{2}^{\perp}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (11)$ $\displaystyle B_{1}^{\perp}-B_{2}^{\perp}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (12)$ $\displaystyle \mathbf{E}_{1}^{\parallel}-\mathbf{E}_{2}^{\parallel}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (13)$ $\displaystyle \frac{1}{\mu_{1}}\mathbf{B}_{1}^{\parallel}-\frac{1}{\mu_{2}}\mathbf{B}_{2}^{\parallel}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (14)$

where the subscript 1 refers to fields in medium 1 (${z<0}$) and 2 refers to medium 2 (${z>0}$). That is

 $\displaystyle \mathbf{E}_{1}$ $\displaystyle =$ $\displaystyle \mathbf{E}_{I}+\mathbf{E}_{R}\ \ \ \ \ (15)$ $\displaystyle \mathbf{E}_{2}$ $\displaystyle =$ $\displaystyle \mathbf{E}_{T} \ \ \ \ \ (16)$

and similarly for ${\mathbf{B}}$. Thus all these boundary conditions have the form

$\displaystyle A_{I}e^{i\left(\mathbf{k}_{I}\cdot\mathbf{r}-\omega t\right)}+A_{R}e^{i\left(\mathbf{k}_{R}\cdot\mathbf{r}-\omega t\right)}=A_{T}e^{i\left(\mathbf{k}_{T}\cdot\mathbf{r}-\omega t\right)} \ \ \ \ \ (17)$

for coefficients ${A_{i}}$ that vary, depending on the boundary condition being imposed, but that don’t depend on ${\mathbf{r}}$ or ${t}$. The space and time dependencies are entirely within the complex exponentials. Since we’ve taken the boundary to be the ${xy}$ plane, ${z=0}$ in these equations, so they become

 $\displaystyle A_{I}e^{i\left(k_{Ix}x+k_{Iy}y-\omega t\right)}+A_{R}e^{i\left(k_{Rx}x+k_{Ry}y-\omega t\right)}$ $\displaystyle =$ $\displaystyle A_{T}e^{i\left(k_{Tx}x+k_{Ty}y-\omega t\right)}\ \ \ \ \ (18)$ $\displaystyle A_{I}e^{i\left(k_{Ix}x+k_{Iy}y\right)}+A_{R}e^{i\left(k_{Rx}x+k_{Ry}y\right)}$ $\displaystyle =$ $\displaystyle A_{T}e^{i\left(k_{Tx}x+k_{Ty}y\right)} \ \ \ \ \ (19)$

These equations must be valid over the entire ${xy}$ plane, so they are valid for ${x=0}$ or ${y=0}$:

$\displaystyle A_{I}e^{ik_{Ix}x}+A_{R}e^{ik_{Rx}x}=A_{T}e^{ik_{Tx}x} \ \ \ \ \ (20)$

To see what this implies, consider the general case

$\displaystyle Ae^{iax}+Be^{ibx}=Ce^{icx} \ \ \ \ \ (21)$

for all ${x}$. That is, both sides are the same function of ${x}$, so they must have the same Taylor expansion. Starting at point ${x}$ and expanding to get the function at ${x+\Delta x}$ we get up to first order in ${\Delta x}$:

$\displaystyle Ae^{iax}\left(1+ia\Delta x\right)+Be^{ibx}\left(1+ib\Delta x\right)+\mathcal{O}\left(\Delta x^{2}\right)=Ce^{icx}\left(1+ic\Delta x\right)+\mathcal{O}\left(\Delta x^{2}\right) \ \ \ \ \ (22)$

Using 21 and cancelling common factors we get

 $\displaystyle aAe^{iax}+bBe^{ibx}$ $\displaystyle =$ $\displaystyle cCe^{icx}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c\left(Ae^{iax}+Be^{ibx}\right)\ \ \ \ \ (24)$ $\displaystyle A\left(a-c\right)e^{iax}+B\left(b-c\right)e^{ibx}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (25)$

This last equation must be true for all ${x}$ so we must have

$\displaystyle a=b=c \ \ \ \ \ (26)$

Going back to 19, we therefore must have

 $\displaystyle k_{Ix}$ $\displaystyle =$ $\displaystyle k_{Rx}=k_{Tx}\ \ \ \ \ (27)$ $\displaystyle k_{Iy}$ $\displaystyle =$ $\displaystyle k_{Ry}=k_{Ty} \ \ \ \ \ (28)$

If we choose axes so that ${\mathbf{k}_{I}}$ is in the ${xz}$ plane, then ${\mathbf{k}_{R}}$ and ${\mathbf{k}_{T}}$ must also lie in the same plane. This gives the first law of geometrical optics:

The wave vectors of the incident, reflected and transmitted waves all lie in the same plane, and this plane also contains the normal to the boundary.

If the angle of incidence is ${\theta_{I}}$ (that is, the angle between ${\mathbf{k}_{I}}$ and the normal to the ${xy}$ plane) with ${\theta_{R}}$ and ${\theta_{T}}$ the corresponding angles for reflection and transmission, then from 8 we have

 $\displaystyle k_{I}\sin\theta_{I}$ $\displaystyle =$ $\displaystyle k_{R}\sin\theta_{R}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle k_{I}\sin\theta_{R}\ \ \ \ \ (30)$ $\displaystyle \sin\theta_{I}$ $\displaystyle =$ $\displaystyle \sin\theta_{R}\ \ \ \ \ (31)$ $\displaystyle \theta_{I}$ $\displaystyle =$ $\displaystyle \theta_{R} \ \ \ \ \ (32)$

This is the familiar condition for specular reflection: the angle of incidence is equal to the angle of reflection. This is the second law of geometrical optics.

Finally, for the transmitted wave, from 10 we have

 $\displaystyle k_{I}\sin\theta_{I}$ $\displaystyle =$ $\displaystyle k_{T}\sin\theta_{T}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{n_{2}}{n_{1}}k_{I}\sin\theta_{T}\ \ \ \ \ (34)$ $\displaystyle \frac{\sin\theta_{I}}{\sin\theta_{T}}$ $\displaystyle =$ $\displaystyle \frac{n_{2}}{n_{1}} \ \ \ \ \ (35)$

This is Snell’s law, or the law of refraction, which is the third law of geometrical optics.

We haven’t actually applied the boundary conditions yet, apart from 20, but we’ll look at that next to get an idea of the reflection and transmission coefficients.