# Electromagnetic wave in glass

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1. Apr 18, 2016

### TheSodesa

1. The problem statement, all variables and given/known data

A harmonic EM-wave is propagating in glass in the +x-direction. The refractive index of the glass $n = 1.4$. The wave number of the wave $k = 30 \ rad/m$. The magnetic portion of the wave is parallel to the y-axis and its amplitude $H_0 = 0.10A/m$. At $t=0$ and $x = 0$ the magnetic field vector $\vec{H} = H_0 \hat{j} = +0.10A/m \ \hat{j}$.

a) Approximate the values for the permittivity ($\epsilon$) and permeability ($\mu$) of the type of glass in question.
b) Calculate the propagation speed, wavelength, angular frequency and frequency of the wave.
c) Calculate the amplitude of the electric field component of the wave at x=0, t=0 (remember to show the direction of the vector).
d) Write the expressions for the electric and magnetic field components in vector form.

2. Relevant equations

Basic wave equation:

v = \lambda f

The wave number:

k = \frac{\omega}{v} = \frac{2\pi f}{v}

Relation between the magnetic and electric components of the wave resulting from them being in phase:

E = vB

The definition of the refractive index:

n:= \frac{c}{v} \stackrel{Maxwell}{=} \frac{\sqrt{\epsilon \mu}}{\sqrt{\epsilon_0 \mu_0}}

Total energy density of an EM-wave:

w = w_E + w_M = \frac{\epsilon E^2}{2} + \frac{\mu H^2}{2}

As a result of $E = vB$, the ratio of the energy densities of the electric and magnetic field components:

\frac{w_E}{w_M} = \mu \epsilon v^2 = 1

3. The attempt at a solution

a) I'm having trouble with this part. Looking at equations (4) and (6):
\begin{cases}
\epsilon \mu &= n^2 \epsilon_0 \mu_0\\
\epsilon \mu &= \frac{1}{v^2}
\end{cases}
This should be easily solvable, but my math head is not working(again). If I solve for either $\epsilon$ or $\mu$ in one equation and plug it into the other equation, it will cancel out and I'm only left with things that I already have values for. It seems I need more information. Not sure what.

b) Looking at equation (3) the velocity:
$$v = \frac{c}{n} = \frac{2.9979 \times 10^{8}m/s}{1.4} = 214 \ 135 \ 714.3 \ m/s$$
Now from equation (1) the angular frequency:
\begin{array}{lll}
\omega
&= kv \\
&= (30rad/m)(214 \ 135 \ 714.3 \ m/s)\\
&= 6 \ 424 \ 071 \ 429 \ rad/s\\
&\approx 6.4 \ G \ rad/s
\end{array}
and frequency:
\begin{array}{lll}
f
&= \frac{\omega}{2 \pi}\\
&= \frac{6 \ 424 \ 071 \ 429 \ rad/s}{2\pi}\\
&= 1 \ 022 \ 422 \ 723 \ Hz\\
&\approx 1.0 \ GHz
\end{array}
Equation (1) gives us the wavelength:
\begin{array}{ll}
\lambda
&= \frac{c}{f}\\
&= \frac{214 \ 135 \ 714.3 \ m/s}{1 \ 022 \ 422 \ 723 \ Hz}\\
&= 0.2 \ 094 \ 395 \ 102 \ m\\
&\approx 0.21 \ m
\end{array}

c) From equation (3):

\begin{array}{ll}
E_0
&= vB_0\\
&= v \mu H_0
\end{array}
In need the value of $\mu$ for this (from part a). The direction can be found out by using the right hand rule. If $\hat{H}$ is in the y-direction and the wave is propagating in the x-direction, the electric component must be in the z- direction. Since the waves are in phase, $\hat{E}$ must be in the +z-direction.(Right?)

d) Since the wave is harmonic, its components are sinusoidal. From c we know the directions of the components, so we can write:
\begin{array}
\vec{E} = E_0 sin(\omega t - kx) \hat{k}\\
\vec{H} = H_0 sin(\omega t - kx) \hat{j}
\end{array}

And there you have it. All I really need help with is part a. After that I should be able to solve for the electric field in part c. I assume the rest of my answer is correct.

Last edited: Apr 18, 2016
2. Apr 19, 2016

### TheSodesa

To actually show some work in part a):

\begin{cases}
\epsilon \mu &= n^2 \epsilon_0 \mu_0\\
\epsilon \mu &= \frac{1}{v^2}
\end{cases}

\begin{cases}
\mu &=\frac{n^2 \epsilon_0 \mu_0}{\epsilon }\\
\epsilon &= \frac{1}{ \mu v^2}
\end{cases}

\begin{cases}
\mu &=\frac{n^2 \epsilon_0 \mu_0}{\frac{1}{ \mu v^2}} = n^2 \epsilon_0 \mu_0 \mu v^2|| \ \text{Divide by } \mu\\
\epsilon &= \frac{1}{ \mu v^2}
\end{cases}

\begin{cases}
1 &= n^2 \epsilon_0 \mu_0 v^2\\
\epsilon &= \frac{1}{ \mu v^2}
\end{cases}
This would allow me to solve for $v$, but I've already done that by other means... It's as if equations (4) and (6) were the same equation.

3. Apr 20, 2016

### rude man

Why don't you just solve for μ from this?
Everything else looks OK to me.

4. Apr 22, 2016

### TheSodesa

I'm a bit late in answering this, but it turned out I was supposed to assume that the permeability of glass, $\mu$, is equal to $\mu _0$. From there on the question solved itself, except part d. In part d I forgot to add the phase angle of $\frac{\pi}{2}$ inside of the sines as follows:
\begin{cases}
E = E_0 sin(\omega t - kx + \frac{\pi}}{2}) \hat{k}\\
H = H_0 sin(\omega t - kx + \frac{\pi}}{2}) \hat{j}
\end{cases}
since at $t=0$ and $x=0$, $H=H_0$. Also the sign of E was brought into question by the assistants in part c. If I had used the traditional right handed coordinate system, $E$ would have been negative at $t=0$ and $x=0$.

In any case, I will mark this as solved.