THE UNIVERSITY OF BRITISH COLUMBIA

*Physics 401 *
Assignment #
**7: **

** WAVES IN MEDIA **

*SOLUTIONS:*

Wed. 22 Feb. 2006 - finish by Wed. 1 Mar.

Wed. 22 Feb. 2006 - finish by Wed. 1 Mar.

- (p. 395, Problem
**9.18**) -**Practical Questions:**^{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 10*^{10}to 10^{14}-m (even 10^{16}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 C^{2}/N-m^{2}. Thus is for "glass" and as long as 200,000 seconds (2.3 days!) for fused quartz. - Silver is an excellent conductor, but it's expensive.
Suppose you were designing a microwave experiment to operate
at a frequency of 10
^{10}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 have*^{2}, (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 10^{10}Hz microwave signal will penetrate a silver film 1 m (one micron) thick. You may want a thicker film to get really thorough reflection. - 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 10^{4}m , so (0.41 mm or 410 m). The corresponding vacuum value at that frequency would be much larger: m.

- 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?
- (p. 396, Problem
**9.19**) -**Skin Depth:**- Show that the skin depth in a poor conductor
(
) is
(independent of frequency).
Find the skin depth (in meters) for (pure) water.
^{3}**ANSWER**:*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} - 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}, = (10^{15}/1.414 x 3 x 10^{8}) x = 7.93 x 10^{7}m^{(}-1) and so the skin depth is . This is only a few atomic layers. How opaque can you get? - 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 .*

- Show that the skin depth in a poor conductor
(
) is
(independent of frequency).
Find the skin depth (in meters) for (pure) water.
- (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:***n*_{1}>*n*_{2}) 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*!)- 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 .* - 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. . - Do the same for polarization perpendicular to the plane of incidence
(use the results of Problem 9.16):

**ANSWER**:*Similarly, .* - 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. - 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: .

FARADAY'S LAW: .

AMPÈRE'S LAW: ; meanwhile, and . But so AMPÈRE'S LAW is also satisfied. - For those same fields, construct the Poynting vector
and show that, on average, no energy is transmitted
in the
*z*direction.**ANSWER**:

or . Time averages are and 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 is greater than 1 and whose is imaginary. I think I would have had an easier time of it using a more formal approach, but done is done.* - Show that
,
where
and
.
This describes a wave propagating in the

2006-02-28