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 where . 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 -m (even 1016 for fused quartz, which most people would consider a type of glass). Since (I'm using to denote resistivity here), that means (the conductivity) has a range from 10-14 to 10-10 m-1 (10-16 for fused quartz). Table 4.2 doesn't list glass or quartz, but we can assume they have and therefore or  C2/N-m2. Thus 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 , where

is the imaginary part of the wavevector. For Ag, we have2 , (so that ), and  m-1. Thus, for  s-1, we expect 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 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  m-1. Again we assume and , so for  s-1 we have = 1.53 x 104  m , so (0.41 mm or 410 m). The corresponding vacuum value at that frequency would be much larger:  m.

2. (p. 396, Problem 9.19) - Skin Depth:
1. Show that the skin depth in a poor conductor ( ) is (independent of frequency). Find the skin depth (in meters) for (pure) water.3ANSWER For we can approximate , giving or , so .
For pure water,  m-1, and , so it's a poor conductor with a skin depth of (about 12 km).4
2. Show that the skin depth in a good conductor ( ) is (where is the wavelength in the conductor. Find the skin depth (in nanometers) for a typical metal [  (m)-1] in the visible range (  s-1), assuming and . Why are metals opaque? ANSWER In a good conductor, , so and , giving

where is the wavevector of an insulator with the same and . Since and , it follows that the skin depth . For  m-1 and  s-1, = (1015/1.414 x 3 x 108) x = 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, 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 , where , and . Since (see previous part), . All we need is to get or .

3. (p. 398, Problem 9.21) - Silver Mirror: Calculate the reflection coefficient for light at an air-to-silver interface [  (m)-1], at optical frequencies (  s-1). ANSWER

where . It's tempting to use the approximation in the previous problem to set for a good conductor like silver. However, this leads to a result (independent of or ), which is incorrect; we must not use such a crude approximation. We have so .
If we define , then , giving , at which point we can plug in numbers: assuming , and so 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 bends away from the normal (see Figure). In particular, if the light is incident at the critical angle   ,  then , and the transmitted ray just grazes the surface. If exceeds , 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 and  ;  the only change is that    is now greater than 1, and so     is imaginary. (Obviously, can no longer be interpreted as an angle!)
1. Show that   ,  where    and   . This describes a wave propagating in the x direction (parallel to the interface!) and attenuated in the z direction. ANSWER When , , giving and where and .
2. Noting that is now imaginary, use Eqs. (9.109),

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 where and . Define so that where a and are both real. .
3. Do the same for polarization perpendicular to the plane of incidence (use the results of Problem 9.16):

4. In the case of polarization perpendicular to the plane of incidence, show that the (real) evanescent fields are

ANSWER Refer to part (a) of this problem. For the TE mode, , so if we define and , we have the first equation.   As usual, . But from part (a), and , giving . Taking the real parts, and (as before) , we have as predicted.

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

ANSWER GAUSS' LAW(S): Since is in the direction but has no y-dependence, .
For it is not so trivial: .