Chapter 30, Plane Electromagnetic Waves Iii Video Solutions, Electromagnetic Fields and Waves: Including Electric Circuits

Problem 1

$(30,2)$ The ray equation
A wave travels in a stratified medium whose index of refraction is a function only of the coordinate $y$.
(a) Show that the angle $\theta$ between a ray and the $y$-axis obeys the following law:
$$
\frac{d \theta}{d l}=-\frac{1}{n} \frac{d n}{d y} \sin \theta
$$
where the distance $l$ is measured along the ray.
(b) You can now verify the ray equation
$$
\frac{d}{d l}(n \vec{i})=\nabla n
$$
where $i$ is a unit vector tangent to the ray at a point where the index of refraction is $n$.

Sandeep Desai

Numerade Educator

01:49

Problem 2

(30 3) Reflection and refraction at the surface of a dense medium
Write down Fresnel's equations for the case where $\mu_{n}=1, \mu_{n}=1$, $n_{2} \gg n_{1}$. You will find a surprising result: if the $E$ vector of the incident wave is parallel to the plane of incidence, the amplitude of the reflected wave is independent of the angle of incidence!
For what range of $\theta_{l}$ are your formulas valid?

Sheh Lit Chang

University of Washington

01:27

Problem 3

(30.3) Fresnel's equations expressed in terms of $\theta_{t}$ and $\theta_{T}$ alone
(a) First show that
$$
\begin{aligned}
&\sin \left(\theta_{l}-\theta_{T}\right) \cos \left(\theta_{l}+\theta_{T}\right)=\sin \theta_{l} \cos \theta_{l}-\sin \theta_{\tau} \cos \theta_{T} \\
&\sin \left(\theta_{l}+\theta_{T}\right) \cos \left(\theta_{l}-\theta_{T}\right)=\sin \theta_{l} \cos \theta_{l}+\sin \theta_{\tau} \cos \theta_{T}
\end{aligned}
$$
(b) Show that, for nonmagnetic nonconductors,
(i) $\left(\frac{E_{\text {Rm }}}{E_{f m}}\right)_{+}=\frac{\sin \left(\theta_{l}-\theta_{T}\right)}{\sin \left(\theta_{l}+\theta_{T}\right)}$
(ii) $\left(\frac{E_{T m}}{E_{l \mathrm{~m}}}\right)_{+}=\frac{2 \cos \theta_{l} \sin \theta_{T}}{\sin \left(\theta_{t}+\theta_{T}\right)}$,
(iii) $\left(\frac{E_{\text {fim }}}{E_{\text {Im }}}\right)_{1}=\frac{\tan \left(\theta_{I}-\theta_{T}\right)}{\tan \left(\theta_{l}+\theta_{T}\right)}$
(iv) $\left(\frac{E_{T m}}{E_{I m}}\right)_{I}=\frac{2 \cos \theta_{l} \sin \theta_{T}}{\sin \left(\theta_{r}+\theta_{\tau}\right) \cos \left(\theta_{l}-\theta_{T}\right)}$

Julie Silva

Numerade Educator

01:36

Problem 4

(30.3) Measuring an index of refraction
Set
$$
p=\left(\frac{E_{R m}}{E_{I m}}\right), \quad s=\left(\frac{E_{R m}}{E_{l m}}\right)_{i} .
$$
Show that, with a laser beam incident at $45^{\circ}$ in air on a medium of index of refraction $n$,
$$
n^{2}=\frac{(1-p)(1-s)}{(1+p)(1+s)}
$$
Here, $s$ is negative, from Fig. 30-5. The ratio $p$ is also negative, from Fig. $30-6 .$

In practice, instruments measure a beam power. So $p$ is equal to minus the square root of the reflected to incident powers with parallel polarization, and similarly for $s$.

Ranjeet Singh

Numerade Educator

03:07

Problem 5

(30.5) The Brewster angle
Calculate the Brewster angles for the following cases:
(a) light incident on a glass whose index of refraction is $1.6$.
(b) light emerging from the same type of glass,
(c) a radio frequency wave incident on water ( $n=9$ at radio frequencies).

Joseph Fritchman

Numerade Educator

04:47

Problem 6

(30.5) The Brewster angle and the ratio $n_{2} / n_{1}$
(a) Show that, if $n_{2}>n_{1}$, then $\theta_{I B}>45^{\circ}$.
(b) Show that, if $n_{2}<n_{1}$, then $\theta_{I B}<45^{\circ}$.

Ameer Said

Numerade Educator

03:58

Problem 7

(30.5) The Brewster angle
(a) Show that Brewster's angle is also given by
$$
\sin ^{2} \theta_{I B}=\frac{1}{1+n_{1}^{2} / n_{2}^{2}}
$$
It follows that there exists a Brewster angle only if the ratio $n_{1} / n_{2}$ is real.
(b) Show that $\sin \theta_{l B}=\cos \theta_{T}$.

Gregory Higby

Numerade Educator

04:47

Problem 7

(30.5) The Brewster angle and the ratio $n_{2} / n_{1}$
(a) Show that, if $n_{2}>n_{1}$, then $\theta_{I B}>45^{\circ}$.
(b) Show that, if $n_{2}<n_{1}$, then $\theta_{I B}<45^{\circ}$.

Ameer Said

Numerade Educator

01:25

Problem 8

(30.5) Brewster windows for lasers
The mirrors of some gas lasers are outside the glass tube that contains the discharge. Then the tube is sealed at both ends with windows set at the Brewster angle.

Show that there is no reflection from such a window as long as the $E$. vector of the incident wave lies in the plane of incidence.

Zulfiqar Ali

Numerade Educator

01:25

Problem 8

(30.5) Brewster windows for lasers
The mirrors of some gas lasers are outside the glass tube that contains the discharge. Then the tube is sealed at both ends with windows set at the Brewster angle.

Show that there is no reflection from such a window as long as the $\boldsymbol{E}$ vector of the incident wave lies in the plane of incidence.

Zulfiqar Ali

Numerade Educator

06:20

Problem 9

(30.5) The valuc of $R_{\perp}$ at the faces of a dielectric plate set at the Brewster angle

A beam of light in a medium of index of refraction $n_{1}$ falls on a plate of dielectric $n_{2}$ at the Brewster angle.
(a) Show that, at the first interface,
$$
R_{\perp}=\cos ^{2} 2 \theta_{l N}=\left(\frac{1-n_{2}^{2} / n_{1}^{2}}{1+n_{2}^{2} / n_{1}^{2}}\right)^{2}
$$
(b) Show that $R_{\perp}$ has the same numerical value at the second interface.
(c) Find the value of $R_{i}$ for glass whose $n$ is $1.5$, in air.

Averell Hause

Carnegie Mellon University

06:20

Problem 9

(30.5) The value of $R_{\perp}$ at the faces of a dielectric plate set at the Brewster angle

A beam of light in a medium of index of refraction $n_{1}$ falls on a plate of dielectric $n_{2}$ at the Brewster angle.
(a) Show that, at the first interface,
$$
R_{\perp}=\cos ^{2} 2 \theta_{l B}=\left(\frac{1-n_{2}^{2} / n_{1}^{2}}{1+n_{2}^{2} / n_{1}^{2}}\right)^{2}
$$
(b) Show that $R_{\pm}$has the same numerical value at the second interface.
(c) Find the value of $R_{\perp}$ for glass whose $n$ is $1.5$, in air.

Averell Hause

Carnegie Mellon University

01:58

Problem 10

(30.5) A "pile of plates" polarizer with pellicles
A pellicle is a very thin film of cellulose nitrate that is stretched taut over a flat ring. The cellulose nitrate is transparent and can serve as a support for various types of coating. The film is so thin that multiple reflections inside it do not give rise to ghost images.

Now it has long been known that a series of parallel glass plates set at Brewster's angle filters out waves polarized with $\boldsymbol{E}$ normal to the plane of incidence. See the two preceding problems. The same can be done with pellicles in less space and without the inconvenience of ghost images. Also, pellicles are virtually lossless and can thus polarize high-power laser beams.
(a) Find $(R / T)_{1}$ for a pellicle set at Brewster's angle. Take both interfaces into account, but disregard multiple reflections.
(b) Calculate this ratio for a pellicle whose $n$ is $1.5$ in air.
(c) Find a general expression for the ratio $(R / T)_{1}$ for $N$ interfaces.
This result is grossly wrong because we have neglected multiple reflections inside the pellicles. In actual fact, the ratio is approximately equal to $N R$. With 40 interfaces, the above result is too large by 2 orders of magnitude!

Suzanne W.

Numerade Educator

07:47

Problem 11

(30.5) The Brewster angle for magnetic media
A wave is incident in air on a nonconducting magnetic medium such as ferrite.
(a) Show that the ratio $\left(E_{R m} / E_{l m}\right)_{||}$is zero for
$$
\tan ^{2} \theta_{l}=\frac{\epsilon_{r}\left(\epsilon_{r}-\mu_{r}\right)}{\epsilon_{n} \mu_{r}-1}
$$
There is a Brewster angle only if $\epsilon_{r}>\mu_{r}$.
(b) Show that $\left(E_{R m} / E_{l m}\right)_{\perp}$ is zero when
$$
\tan ^{2} \theta_{l}=\frac{\mu_{r}\left(\mu_{r}-\epsilon_{r}\right)}{\epsilon, \mu_{r}-1}
$$
Now there is a Brewster angle, but only if $\mu,>\epsilon_{r}$.

Kai Chen

Princeton University

01:48

Problem 12

(30.6) The condition that makes $R=T$ at normal incidence Find the ratio $n_{1} / n_{2}$ that makes $R=T=0.5$ at normal incidence.

Harsh Gadhiya

Numerade Educator

01:42

Problem 13

(30.6) $E, H, R$, and $T$ at normal incidence on a water surface
A 60-watt light bulb is situated in air 1 meter above a water surface.
(a) Calculate the root mean square (rms) values of $E$ and $H$ for the incident, reflected, and refracted rays at the surface of the water directly under the bulb. Assume that all the power is dissipated as electromagnetic radiation. The index of refraction of water is $1.33$.
(b) Calculate the coefficients of reflection and transmission.

Penny Riley

Numerade Educator

01:58

Problem 14

(30.6) Antireflection coatings for photographic lenses and solar cells
There are instances where the reflection coefficient of a dielectric must be close to zero. The best known examples are photographic lenses and solar cells.

Clearly, the way to eliminate the reflected wave is by interference. Coating the dielectric with a thin film of another type of dielectric provides two reflected waves that can cancel. The situation is, however, complicated by the presence of multiple reflections in the film. Also, the degree of cancellation varies with the angle of incidence and with the wave length.
(a) Show that there is no reflected wave at normal incidence in air $\left(n_{1}=1\right)$ when the dielectric of index of refraction $n_{3}$ is coated with a quarter-wavelength film of a dielectric $n_{2}=n_{3}^{1 / 2}$. Take multiple reflections into account, and use the notation of Fig. $30-12$.
(b) Calculate and sum the amplitudes of the first four reflected waves when $n_{3}=4$, to four significant figures.
(c) A silicon solar cell has an index of refraction of $3.9$ at 600 nanometers. Calculate the reflection coefficient for normal incidence at that wavelength.
(d) Calculate the thickness and the index of refraction of a coating that would eliminate reflection at normal incidence at that wavelength.

At the interface between air and glass, $R=0.04$. In complex optical systems with many interfaces, the loss is important. Moreover, stray reflections reduce contrast in the image. Good-quality lenses are coated with magnesium fluoride ( $n=1.38$ at 550 nanometers). This reduces $R$ to $0.015$, on average, over the visible spectrum.

Suzanne W.

Numerade Educator

03:41

Problem 15

(30.6) A simple and accurate method for measuring an index of refraction

Possibly the most practical and most accurate way of measuring an index of refraction is to measure the ratio $R_{\perp} / R_{\|}$for a beam incident on the material in air at $45^{\circ}$.
(a) Show that, if $\theta_{I}=45^{\circ}$,
$$
R_{1}=\frac{1-\sin 2 \theta_{T}}{1+\sin 2 \theta_{T}}, \quad R_{\|}=\left(\frac{1-\sin 2 \theta_{T}}{1+\sin 2 \theta_{T}}\right)^{2}
$$
It follows that
$$
\frac{R_{\perp}}{R_{\|}}=\frac{1+\sin 2 \theta_{T}}{1-\sin 2 \theta_{T}}
$$This ratio is much larger than unity. For example, with $\theta_{T}=30^{\circ}$ $(n=1.414)$, it is equal to $13.93$
(b) Show that
$$
n=\frac{\left[1+\left(1-f^{2}\right)^{1 / 2}\right]^{1 / 2}}{f}
$$
where
$$
f=\sin 2 \theta_{T}=\frac{R_{\perp} / R_{\|}-1}{R_{\perp} / R_{\|}+1}
$$
The signs before the square roots are positive.

Averell Hause

Carnegie Mellon University

01:42

Problem 16

(30.7) Ducting in the ionosphere
Under certain circumstances, the index of refraction of the ionosphere varies with altitude in such a way that a ray that starts out horizontally follows a path at a constant altitude above the earth's surface. The ionosphere then acts as a duct, and the phenomenon is called ducting. Of course, the required condition applies only over a certain distance. When the ray emerges from this region, it is deflected either upward or downward. Radar signals occasionally travel over large distances in this way.
(a) How must the index of refraction vary with altitude?
(b) How must the plasma frequency vary with altitude?

Mayukh Banik

Numerade Educator