THE UNIVERSITY OF BRITISH COLUMBIA
 
Physics 401 Assignment # 7:
 
WAVES IN MEDIA
 
SOLUTIONS:
 
Wed. 22 Feb. 2006 - finish by Wed. 1 Mar.

1. (p. 395, Problem 9.18) - Practical Questions:¹
   1. Suppose you embedded some free charge in a piece of glass. About how long would it take for the charge to flow to the surface? ANSWER:  We know charge density dissipates exponentially in a conductor according to $\rho(t) = \rho_0 e^{-t/\tau}$ where $\tau = \epsilon/\sigma$. It's strange to think of glass as a conductor, but no insulator is perfect. Different glasses have a wide range of resistivities, from 10¹⁰ to 10¹⁴ $\Omega$-m (even 10¹⁶ for fused quartz, which most people would consider a type of glass). Since $\sigma \equiv 1/\rho_R$ (I'm using $\rho_R$ to denote resistivity here), that means $\sigma$ (the conductivity) has a range from 10^-14 to 10^-10 $\Omega^{-1}$m^-1 (10^-16 for fused quartz). Table 4.2 doesn't list glass or quartz, but we can assume they have $\mu \approx \muz$ and therefore $n = c/v = \sqrt{\epsilon\mu}/\sqrt{\epsz\muz} = \sqrt{\epsilon/\epsz} = 1.5$ or $\epsilon = 2.25 \epsz = 2 \times 10^{-11}$ C²/N-m². Thus $\tau = \epsilon/\sigma$ is for "glass" and as long as 200,000 seconds (2.3 days!) for fused quartz.
   2. Silver is an excellent conductor, but it's expensive. Suppose you were designing a microwave experiment to operate at a frequency of 10¹⁰ Hz. How thick would you make the silver coatings? ANSWER:  The skin depth is $\kappa^{-1}$, where
      
      $$\kappa = \omega \sqrt{\epsilon\mu \over 2} \sqrt{ \sqrt{1 + \left( \sigma \over \epsilon\omega \right)^2} - 1}$$
      
      
      is the imaginary part of the wavevector. For Ag, we have² $\epsilon \approx \epsz$, $\mu \approx \muz$ (so that $\sqrt{\epsilon\mu/2} \approx 1/c\sqrt{2}$), and $\sigma = 1/1.59\times10^{-8} = 6.29\times10^7$ $\Omega^{-1}$m^-1. Thus, for $\omega \approx 2\pi\times10^{10}$ s^-1, we expect ${\displaystyle \kappa = 2\pi\times10^{10} {1\over\sqrt{2}\times2.99\times10^8} \times }$ ${\displaystyle \sqrt{ \sqrt{1 + \left( 6.29\times10^7 \over 0.885\times10^{-11}\times2\pi\times10^{10} \right)^2} - 1} }$ $= 1.576\times10^6 \hbox{\rm ~m}^{-1} \, ,$ giving a skin depth of . Thus only 20.7% of the original electric field strength in a 10¹⁰ Hz microwave signal will penetrate a silver film 1 $\mu$m (one micron) thick. You may want a thicker film to get really thorough reflection.
   3. Find the wavelength and propagation speed in copper for radio waves at 1 MHz. Compare the corresponding values in air (or vacuum). ANSWER:  For Cu we have $\sigma = 1/1.68\times10^{-8} = 5.95\times10^7$ $\Omega^{-1}$m^-1. Again we assume $\epsilon \approx \epsz$ and $\mu \approx \muz$, so for $\omega = 2\pi\times10^6$ s-1 we have ${\displaystyle k = 2\pi\times10^6 {1\over\sqrt{2}\times3\times10^8} \times }$ ${\displaystyle \sqrt{ \sqrt{1 + \left( 5.95\times10^7 \over 0.885\times10^{-11}\times2\pi\times10^6 \right)^2} + 1} }$ = 1.53 x 10⁴  m $^{-1} = 2\pi/\lambda$, so (0.41 mm or 410 $\mu$m). The corresponding vacuum value at that frequency would be much larger: $\lambda_0 = c/\nu = 3\times10^8/10^6 = 300$ m.

2. (p. 396, Problem 9.19) - Skin Depth:
   1. Show that the skin depth in a poor conductor ( $\sigma \ll \omega \epsilon$) is $(2/\sigma) \sqrt{\epsilon/\mu}$ (independent of frequency). Find the skin depth (in meters) for (pure) water.³ANSWER:  For $\sigma/\omega\epsilon \ll 1$ we can approximate $\left[1 + (\sigma/\epsilon\omega)^2\right]^{1/2} \approx 1 + (1/2)(\sigma/\epsilon\omega)^2$, giving $\kappa \approx \omega \sqrt{\epsilon\mu/2} \sqrt{\not{1} + (1/2)(\sigma/\epsilon\omega)^2 - \not{1}}$ $= \omega \sqrt{\epsilon\mu/2} \times (\sigma/\epsilon\omega)/\sqrt{2}$ or $\kappa \approx \sigma\sqrt{\mu/\epsilon}/2$, so .    
      For pure water, $\sigma = 1/(2.5\times10^5) = 0.4\times10^{-5}$ $\Omega^{-1}$m^-1, $\epsilon = 80.1 \epsz$ and $\mu \approx \muz$, so it's a poor conductor with a skin depth of $\kappa^{-1} \approx (2/0.4\times10^{-5}) \times \sqrt{80.1\times0.885\times10^{-11}/4\pi\times10^{-7}}$ (about 12 km).⁴
   2. Show that the skin depth in a good conductor ( $\sigma \gg \omega \epsilon$) is $\lambda/2\pi$ (where $\lambda$ is the wavelength in the conductor. Find the skin depth (in nanometers) for a typical metal [ $\sigma \approx 10^7$ ($\Omega\,$m)^-1] in the visible range ( $\omega \approx 10^{15}$ s^-1), assuming $\epsilon \approx \epsz$ and $\mu \approx \muz$. Why are metals opaque? ANSWER:  In a good conductor, $\sigma/\epsilon\omega \gg 1$, so $\sqrt{1+(\sigma/\epsilon\omega)^2} \approx \sigma/\epsilon\omega$ and $\sqrt{\sigma/\epsilon\omega \pm 1} \approx \sqrt{\sigma/\epsilon\omega}$, giving
      
      $$k \approx \kappa \approx {k_0 \over \sqrt{2}} \sqrt{\sigma \over \epsilon\omega}$$
      
      
      where $k_0 \equiv \omega \sqrt{\epsilon\mu}$ is the wavevector of an insulator with the same $\epsilon$ and $\mu$. Since $k = 2\pi/\lambda$ and $\kappa \approx k$, it follows that the skin depth . For $\sigma \approx 10^7$ $\Omega^{-1}$m^-1 and $\omega \approx 10^{15}$ s^-1, $\kappa \approx (\omega/\sqrt{2}c) \sqrt{\sigma/\epsilon\omega}$ = (10¹⁵/1.414 x 3 x 10⁸) x $\sqrt{10^7/0.885\times10^{-11}\times10^{15}}$ = 7.93 x 10⁷ m⁽-1) and so the skin depth is . This is only a few atomic layers. How opaque can you get?
   3. Show that in a good conductor the magnetic field lags the electric field by 45$^\circ$, and find the ratio of their amplitudes. For a numerical example, use the "typical metal" in the previous question. ANSWER:  From Faraday's law we have $\tilde{B}_0/\tilde{E}_0 = \tilde{k}/\omega$, where $\tilde{k} \equiv k + i \kappa \equiv K e^{i\phi}$, ${\displaystyle K = \vert\tilde{k}\vert = \sqrt{k^2 + \kappa^2} = {\omega \over c} \left[1 + \left(\sigma\over\epsilon\omega\right)^2\right]^{1/4} }$ and $\tan \phi = \kappa/k$. Since $k \approx \kappa$ (see previous part), . All we need is $\sigma/\epsilon\omega = 10^7/0.885\times10^{-11}\times10^{15} = 1129$ to get $\vert\tilde{B}_0\vert/\vert\tilde{E}_0\vert = K/\omega$ $= \left[1 + (1129)^2\right]^{1/4}/3\times10^8$ or .

3. (p. 398, Problem 9.21) - Silver Mirror: Calculate the reflection coefficient for light at an air-to-silver interface [ $\mu_1 = \mu_2 = \muz, \; \epsilon_1 = \epsz, \; \sigma = 6 \times 10^7$ ($\Omega\,$m)^-1], at optical frequencies ( $\omega = 4 \times 10^{15}$ s^-1). ANSWER: 
   
   $$R \equiv {I_R \over I_I} \equiv {\vert\tilde{E}_R\vert^2 \o . . . . . . \tilde{\beta}^*) \over (1+\tilde{\beta})(1+\tilde{\beta}^*)}$$
   
   
   where ${\displaystyle \tilde{\beta} \equiv {\mu_1 v_1 \over \mu_2 \omega} \tilde{k}_ . . . . . . \over \omega} \tilde{k} = {\tilde{k} \over k_0} = {k + i \kappa \over k_0} }$. It's tempting to use the approximation in the previous problem to set ${\displaystyle \tilde{\beta} \approx (1 + i) \sqrt{\sigma \over 2\epsilon\omega} }$ for a good conductor like silver. However, this leads to a result $R \approx 1/5$ (independent of $\sigma, \epsilon$ or $\omega$), which is incorrect; we must not use such a crude approximation. We have $(1\pm\tilde{\beta})(1\pm\tilde{\beta}^*)$ $= (k_0 \pm k \pm i\kappa)(k_0 \pm k \mp i\kappa)/k_0^2$ $= (k_0^2 \pm 2k_0k + k^2 + \kappa^2)/k_0^2$ $= [(k_0 \pm k)^2 + \kappa^2]/k_0^2$ so ${\displaystyle R = {(k_0 - k)^2 + \kappa^2 \over (k_0 + k)^2 + \kappa^2} }$.
   If we define ${\displaystyle x \equiv \sqrt{1 + \left(\sigma\over\epsilon\omega\right)^2} }$, then ${\displaystyle \left\{k \atop \kappa\right\} = {k_0 \over \sqrt{2}} \sqrt{x \pm 1} }$, giving ${\displaystyle R = { \left(\sqrt{2} - \sqrt{x + 1}\right)^2 + x - 1 \over \left(\sqrt{2} + \sqrt{x + 1}\right)^2 + x - 1 } }$ ${\displaystyle = { 2 - 2\sqrt{2}\sqrt{x + 1} + 2x \over 2 + 2\sqrt{2}\sqrt{x + 1} + 2x } }$ ${\displaystyle = { 1 - \sqrt{2(x + 1)} \over 1 + \sqrt{2(x + 1)} } }$, at which point we can plug in numbers: assuming $\epsilon \approx \epsz$, $x = \sqrt{1 + \left(6\times10^7/0.885\times10^{-11}\times4\times10^{15}\right)^2} = 1694$ and so $R = (1-\sqrt{2\times1695})/(1+\sqrt{2\times1695})$ or . I worry about that - sign.

4.

(p. 413, Problem 9.37) - TIR:

According to SNELL'S LAW, when light passes from an optically dense medium into a less dense one (n₁ > n₂) the propagation vector $\Vec{k}$ bends away from the normal (see Figure). In particular, if the light is incident at the critical angle   $\theta_C \equiv \sin^{-1} (n_2/n_1)$,  then $\theta_T = 90^\circ$, and the transmitted ray just grazes the surface. If $\theta_I$ exceeds $\theta_C$, there is no refracted ray at all, only a reflected one. This is the phenomenon of total internal reflection,⁵ on which light pipes and fiber optics are based. But the fields are not zero in medium 2; what we get is a so-called evanescent wave, which is rapidly attenuated and transports no energy into medium 2.⁶
A quick way to construct the evanescent wave is simply to quote the results of Sect. 9.3.3, with $k_T = \omega n_2/c$ and  $\Vec{k}_T = k_T(\sin \theta_T \; \Hat{x} \; + \; \cos \theta_T \; \Hat{z})\,$;  the only change is that  $\sin \theta_T = (n_1/n_2) \sin \theta_I$  is now greater than 1, and so   $\cos \theta_T = \sqrt{1 - \sin^2 \theta_T}$  is imaginary. (Obviously, $\theta_T$ can no longer be interpreted as an angle!)

1. Show that   ${\displaystyle \tVec{E}_T(\Vec{r},t) = \tVec{E}_{0_T} \, e^{-\kappa z} e^{i(kx - \omega t)} }$,  where  ${\displaystyle \kappa \equiv {\omega \over c} \sqrt{(n_1 \sin \theta_I)^2 - n_2^2} }$  and   ${\displaystyle k \equiv {\omega n_1 \over c} \sin \theta_I \, }$. This describes a wave propagating in the x direction (parallel to the interface!) and attenuated in the z direction. ANSWER:  When $\sin\theta_T = (n_1/n_2)\sin \theta_I > 1$, $\cos\theta_T = \sqrt{1-\sin^2\theta_T} = i \sqrt{(n_1/n_2)^2\sin^2\theta_I-1}$, giving $\Vec{k}_T = k_T(\sin\theta_T \Hat{x}+\cos\theta_T\Hat{z})$ and $\Vec{k}_T\cdot\Vec{r} = k_T x (n_1/n_2)\sin\theta_I + i k_T z \sqrt{(n_1/n_2)^2\sin^2\theta_I-1} = k x + i \kappa z$ where $k = k_T \sin\theta_T = (\omega n_2/c) (n_1/n_2)\sin\theta_I = (\omega n_1/c) \sin\theta_I$ and $\kappa = k_T \sqrt{\sin^2\theta_T-1} = (\omega n_2/c) \sqrt{(n_1/n_2)^2\sin^2\theta_I-1} = (\omega/c) \sqrt{n_1^2\sin^2\theta_I-n_2^2}$.
2. Noting that ${\displaystyle \alpha \equiv {\cos \theta_T \over \cos \theta_I} }$ is now imaginary, use Eqs. (9.109),
   
   $$\hbox{\bf TM\/: } \quad { \tilde{E}_{0_R} = \left( \alph . . . . . . \left( 2 \over \alpha + \beta \right) \tilde{E}_{0_I} \, , }$$
   
   
   to calculate the reflection coefficient for polarization parallel to the plane of incidence. [Notice that you get 100% reflection, which is better than at a conducting surface (see for example Problem 9.21).] ANSWER:  $R \equiv \left\vert E_{0_R} \over E_{0_I}\right\vert^2 = \left\vert\alpha-\beta \over \alpha+\beta\right\vert^2$ where $\alpha \equiv \cos\theta_T/\cos\theta_I$ $= i \sqrt{\sin^2\theta_T - 1}/\cos\theta_I$ and $\beta = {\mu_1 n_2 \over \mu_2 n_1}$. Define $\alpha \equiv i a$ so that $R = \left\vert ia - \beta \over ia + \beta\right\vert^2$ where a and $\beta$ are both real. $R = \left\vert ia - \beta \over ia + \beta\right\vert^2 = \left(ia - \beta \o . . . . . . ia - \beta \over -ia + \beta\right) = {a^2 + \beta^2 \over a^2 + \beta^2} = 1$.
3. Do the same for polarization perpendicular to the plane of incidence (use the results of Problem 9.16):
   
   $$\hbox{\bf TE\/: } \quad { \tilde{E}_{0_R} = \left( 1 - \ . . . . . . \left( 2 \over 1 + \alpha \beta \right) \tilde{E}_{0_I} \; . }$$
   
   
   ANSWER:  Similarly, ${\displaystyle R = \left\vert 1 - ia\beta \over 1 + ia\beta\right\vert^2 = \ . . . . . . a\beta \over 1 - ia\beta\right) = {1 + a^2\beta^2 \over 1 + a^2\beta^2} = 1 }$.
4. In the case of polarization perpendicular to the plane of incidence, show that the (real) evanescent fields are
   
   $$\begin{eqnarray*} \Vec{E}(\Vec{r},t) &=& E_0 \; e^{-\kappa z} \cos (kx - \ome . . . . . . t) \; \Hat{x} + \; k \cos (kx - \omega t) \; \Hat{z} \, ] \; . \end{eqnarray*}$$
   
   ANSWER:  Refer to part (a) of this problem. For the TE mode, $\tVec{E}_T = \tilde{E}_{0_T} \Hat{y}$, so if we define $\Vec{E} \equiv \Re \tVec{E}_T$ and $E_0 \equiv \Re \tilde{E}_{0_T}$, we have the first equation.   As usual, $\tVec{B}_T \; \omega/k_T = \Hat{k}_T \times \tVec{E}_T$ $= \tilde{E}_{0_T} e^{-\kappa z} e^{i(kx-\omega t)} (\sin\theta_T \Hat{z} - \cos\theta_T \Hat{x})$. But from part (a), $\sin\theta_T = k/k_T$ and $\cos\theta_T = i \kappa/k_T$, giving $\tVec{B}_T = (\tilde{E}_{0_T}/\omega) e^{-\kappa z} e^{i(kx-\omega t)} (k \Hat{z} - i \kappa \Hat{x})$. Taking the real parts, $\Vec{B} \equiv \Re \tVec{B}_T$ and (as before) $E_0 \equiv \Re \tilde{E}_{0_T}$, we have $\Vec{B}_T = (E_0/\omega) e^{-\kappa z} [ (k \cos(kx-\omega t) \Hat{z} + \kappa \sin(kx-\omega t) \Hat{x} ]$ as predicted.

5. Check that the fields in the last part satisfy all of MAXWELL'S EQUATIONS (9.67).

   ANSWER:  GAUSS' LAW(S): Since $\Vec{E}$ is in the $\Hat{y}$ direction but has no y-dependence, $\Div{E} = 0$.
   For $\Vec{B}$ it is not so trivial: $\Div{B} = (E_0/\omega)\{\DbyD{}{x}[\kappa e^{-\kappa z} \sin(kx-\omega t)] + \DbyD{}{z}[k e^{-\kappa z} \cos(kx-\omega t)]\}$ $= (E_0/\omega)\{k\kappa e^{-\kappa z} \cos(kx-\omega t) - k\kappa e^{-\kappa z} \cos(kx-\omega t)\} = 0$.
   FARADAY'S LAW: $\Curl{E} = -\Hat{x}\DbyD{E_y}{z} + \Hat{z}\DbyD{E_y}{x}$ $= E_0 e^{-\kappa z} \{ \kappa \cos(kx-\omega t) \Hat{x} - k \sin(kx-\omega t) \Hat{z} \}$ $= - \DbyD{\Vec{B}}{t}$ $= - {E_0\over\omega} e^{-\kappa z} [-\omega \kappa \cos(kx-\omega t) \Hat{x} + \omega k \sin (kx-\omega t) \Hat{z}]$ $= E_0 e^{-\kappa z} \{ \kappa \cos(kx-\omega t) \Hat{x} - k \sin(kx-\omega t) \Hat{z} \}$.
   AMPÈRE'S LAW: $\Curl{B} = -\Hat{y}\DbyD{B_z}{x}$ $= -k(E_0/\omega) \Hat{y} e^{-\kappa z} \DbyD{}{x}\cos(kx-\omega t)$ $= (k^2/\omega) \Hat{y} E_0 e^{-\kappa z} \sin(kx-\omega t)$; meanwhile, $\Vec{J} = 0$ and $\mu_2\epsilon_2 \DbyD{\Vec{E}}{t} = \mu_2\epsilon_2 \Hat{y} E_0 e^{-\kappa z} \DbyD{}{t} \cos(kx-\omega t)$ $= \omega \mu_2\epsilon_2 \Hat{y} E_0 e^{-\kappa z} \sin(kx-\omega t)$. But $\omega \mu_2\epsilon_2 = \omega/v_2^2 = \omega (k^2/\omega^2) = k^2/\omega$ so AMPÈRE'S LAW is also satisfied.

6. For those same fields, construct the Poynting vector and show that, on average, no energy is transmitted in the z direction.

   ANSWER:  $\Vec{S} = \Vec{E} \times \Vec{B}/\mu_2$ $= (E_0^2/\omega\mu_2) e^{-2\kappa z} \cos(kx-\omega t) [\kappa \sin(kx-\omega t) \; \Hat{y}\times\Hat{x} + k \cos(kx-\omega t) \; \Hat{y}\times\Hat{z}]$
   or . Time averages are $\langle \cos^2(kx-\omega t) \rangle = {1\over2}$ and $\langle \sin(kx-\omega t) \cos(kx-\omega t) \rangle = 0$ so that . As predicted, there is no net energy flow in the z direction, only in the x direction parallel to the surface of the interface.

I don't know about you, but I found it very confusing to deal with an "angle" whose $\sin$ is greater than 1 and whose $\cos$ is imaginary. I think I would have had an easier time of it using a more formal approach, but done is done.