### Solution

A1  ?? Substitute a plane wave into the Maxwell's equations and show that $\vec{E}_0$ and $\vec{B}_0$ are perpendicular to each other and to the $\vec{k}$. Also, write the expression for $|\vec{B}_0|$ in terms of $|\vec{E}_0|$, $\omega$ and $|\vec{k}|$.

To work confidently with plane waves $\DeclareMathOperator{\Rot}{curl} \DeclareMathOperator{\Div}{div} \DeclareMathOperator{\Grad}{grad} \vec{E}(\vec{r}, t) = \mathfrak{Re} \vec{E}_0 e^{i\vec{k}\vec{r} - i \omega t}$ и $\vec{B}(\vec{r}, t) = \mathfrak{Re} \vec{B}_0 e^{i\vec{k}\vec{r} - i \omega t}$ in Maxwell's equations we have to realize, how the $\Div$ and $\Rot$ operators act on the vector complex exponent $\vec{E}_0 e^{i\vec{k}\cdot\vec{r}}= \vec{E}_0 e^{i\left( x k_x + y k_y + z k_z \right)}$. Taking into account
$\frac{\partial}{\partial x} e^{i\left( x k_x + y k_y + z k_z \right)} = ik_xe^{i\left( x k_x + y k_y + z k_z \right)},$
we have
$\Div \vec{E}_0 e^{i\left( x k_x + y k_y + z k_z \right)} = i\left(k_x E_{0x} + k_y E_{0y} + k_z E_{0z} \right) e^{i\left( x k_x + y k_y + z k_z \right)} = i \vec{k} \cdot \vec{E}_0 e^{i\left( x k_x + y k_y + z k_z \right)}.$
Substituting this result into the Maxwell's equation we get
$\begin{split} \Div \vec{D} &= \frac{i}{\varepsilon_0 \varepsilon} \vec{k} \cdot \vec{E}_0 e^{i\vec{k} \cdot \vec{r} - i \omega t} = 0 \\ \Div \vec{B} &= i \vec{k} \cdot \vec{B}_0 e^{i\vec{k} \cdot \vec{r} - i \omega t} = 0 \end{split},$
so $\vec{k} \cdot \vec{E}_0 = 0$ and $\vec{k} \cdot \vec{B}_0 = 0$, which proves that $\vec{k} \perp \vec{E}, \vec{B}$.

A calculation of $\Rot \vec{E}_0 e^{i\left( x k_x + y k_y + z k_z \right)}$ is little bit more complicated:
$\Rot \vec{E}_0 e^{i\left( x k_x + y k_y + z k_z \right)} = \begin{vmatrix} \vec{x} & \vec{y} & \vec{z} \\ \partial/\partial x & \partial/\partial y & \partial/\partial z \\ E_{0x} e^{i\left( x k_x + y k_y + z k_z \right)} & E_{0y} e^{i\left( x k_x + y k_y + z k_z \right)} & E_{0z} e^{i\left( x k_x + y k_y + z k_z \right)} \end{vmatrix} = \begin{pmatrix} ik_yE_{0z} - i k_z E_{0y} \\ ik_zE_{0x} - i k_x E_{0z} \\ ik_xE_{0y} - i k_y E_{0x} \end{pmatrix} e^{i\left( x k_x + y k_y + z k_z \right)}.$
If you look closely at the final vector, you will see that it is proportional to the $i \vec{k} \times \vec{E}_0$. Indeed,
$i\vec{k} \times \vec{E}_0 = \begin{vmatrix} \vec{x} & \vec{y} & \vec{z} \\ ik_x & i k_y & ik_z \\ E_{0x} & E_{0y} & E_{0z} \end{vmatrix} = \begin{pmatrix} ik_yE_{0z} - ik_zE_{0y} \\ ik_zE_{0x} - ik_xE_{0z} \\ ik_xE_{0y} - ik_yE_{0x} \end{pmatrix}$
It leads us to the following result:
$\begin{split} \Rot \vec{E} &= i \vec{k} \times \vec{E}_0 e^{i\vec{k} \cdot \vec{r} - i \omega t} = -\frac{\partial}{\partial t} \vec{B}_0 e^{i\vec{k} \cdot \vec{r} - i \omega t}= i\omega \vec{B}_0 e^{i\vec{k} \cdot \vec{r} - i \omega t}\\ \Rot \vec{B} &= i \vec{k} \times \vec{B}_0 e^{i\vec{k} \cdot \vec{r} - i \omega t} = \frac{1}{\mu_0} \frac{\partial}{\partial t} \frac{\vec{E}_0}{\varepsilon_0 \varepsilon} e^{i\vec{k} \cdot \vec{r} - i \omega t}= -i\omega \frac{c^2 }{\varepsilon} \vec{E}_0 e^{i\vec{k} \cdot \vec{r} - i \omega t}\\ \end{split}.$
Finally, we obtain that $\vec{B}_0 = (\vec{k} \times \vec{E}_0)/\omega$ and also $\vec{E}_0 = - \varepsilon (\vec{k} \times \vec{B}_0)/(\omega c^2)$, i.e. $\vec{B}_0 \perp \vec{E}_0$ and
$|\vec{B}_0| = \frac{|\vec{k}|}{\omega} |\vec{E}_0|.$

Answer: $|\vec{B}_0| = \frac{|\vec{k}|}{\omega} |\vec{E}_0|.$

A2  ?? With the relation $\DeclareMathOperator{\Rot}{curl} \DeclareMathOperator{\Div}{div} \DeclareMathOperator{\Grad}{grad} \Rot \Rot \vec{A} = \Grad \Div \vec{A} - \Delta \vec{A}$ find a wave equation for the vector $\vec{E}$:
$\Delta \vec{E} - \frac{1}{v^2}\frac{\partial^2 E}{\partial t^2} = 0.$

What is the phase velocity $v$ of the electromagnetic wave in the medium with relative permittivity $\varepsilon$? What is the refraction index $n$ of the medium with relative permittivity $\varepsilon$?

Remark: partial derivatives with respect to different variables could be interchanged:
$\frac{\partial^2 f(x,y)}{\partial x \partial y} = \frac{\partial^2 f(x,y)}{\partial y \partial x} \quad \Rightarrow \quad \Rot \frac{\partial}{\partial t} \vec{A} = \frac{\partial}{\partial t} \Rot \vec{A}$

Let's apply $\DeclareMathOperator{\Rot}{curl} \DeclareMathOperator{\Div}{div} \DeclareMathOperator{\Grad}{grad} \Rot$ to both parts of $\Rot \vec{E} = - \partial \vec{B} /\partial t$:
$\Rot \Rot \vec{E} = \Grad \Div \vec{E} - \Delta \vec{E} = - \Rot \frac{\partial}{\partial t} \vec{B}.$
Also $\varepsilon \varepsilon_0 \Div \vec{E} = \Div \vec{D} = 0$ and according to the symmetry of the second partial derivatives we have $\Rot \partial\vec{B} /\partial t = \partial/\partial t \Rot \vec{B}$. The value of $\Rot \vec{B}$ could be subsituted from the other Maxwe's equation:
$-\Delta \vec{E} = - \mu_0 \frac{\partial^2 \vec{D}}{\partial t^2} = -\frac{\varepsilon}{c^2} \frac{\partial^2 \vec{E}}{\partial t^2} \quad \Rightarrow \quad \Delta \vec{E} -\frac{\varepsilon}{c^2} \frac{\partial^2 \vec{E}}{\partial t^2} = 0,$
so $v=c/\sqrt{\varepsilon}$.

Answer: $v=\frac{c}{\sqrt{\varepsilon}}$

A3  ?? Show, that $k_{1y} = k_{2y}$, which is an equivalent of the Snell's law $n_1 \sin \theta_1 = n_2 \sin \theta_2$.

Let's consider that the boundary condition is satisfied at the $y=0$. It will be satisfied at the other points if the field dependence on the coordinate $y$ is the same on both sides of the boundary. Above the boundary we have $\vec{E},\vec{B} \propto e^{ik_{1y}y}$, below it we have $e^{ik_{2y}y}$, therefore $k_{1y} = k_{2y}$. Herewith, $k_{1y} = k_1 \sin \theta_1 = \omega n_1 /c \sin \theta_1$ and $k_{2y} = k_2 \sin \theta_2 = \omega n_2 /c \sin \theta_2$. Thus we have Snell's law
$\frac{\omega n_1}{c} \sin \theta_1 = \frac{\omega n_2}{c} \sin \theta_2$
which is the consequence in the perspective of wave optics although it is the postulate in geometry optics.

A4  ?? Using boundary conditions, show that for $s$-polarization the amplitude reflection
$r_s = \frac{n_1 \cos \theta_1 - n_2 \cos \theta_2}{n_1 \cos \theta_1+ n_2 \cos \theta_2}$

If the angle of incidence of the is $\theta_1$:

1. wave vector of incident wave is $\vec{k}_1=\begin{pmatrix} k_1 \cos \theta_1 & k_1 \sin \theta_1 & 0\end{pmatrix}^T$
2. wave vector of reflected waves is $\vec{k}_r = \begin{pmatrix} -k_1 \cos \theta_1 & k_1 \sin \theta_1 & 0 \end{pmatrix}^T$
3. wave vector of transmitted wave is $\vec{k}_2 = \begin{pmatrix} k_2 \cos \theta_2 & k_2 \sin \theta_2 & 0\end{pmatrix}^T$.

For the $s$-polarization let's choose $\vec{p}_0=\vec{p}_r=\vec{p}_t=\vec{z}$. Then we can calculate the amplitudes of the magnetic field oscillations:

$\vec{B}_1 = \frac{1}{\omega} \vec{k}_1 \times \vec{z} E_0 = \frac{k_1 E_0}{\omega} \begin{vmatrix} \vec{x} & \vec{y} & \vec{z} \\ \cos \theta_1 & \sin \theta_1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = \frac{n_1 E_0}{c}\begin{pmatrix} \sin \theta_1 \\ - \cos \theta_1 \\ 0 \end{pmatrix}$
$\vec{B}_r = \frac{1}{\omega} \vec{k}_r \times \vec{z} rE_0 = \frac{k_1r E_0}{\omega} \begin{vmatrix} \vec{x} & \vec{y} & \vec{z} \\ -\cos \theta_1 & \sin \theta_1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = \frac{n_1 r E_0}{c}\begin{pmatrix} \sin \theta_1 \\ \cos \theta_1 \\ 0 \end{pmatrix}$
$\vec{B}_t = \frac{1}{\omega} \vec{k}_2 \times \vec{z} tE_0 = \frac{k_2t E_0}{\omega} \begin{vmatrix} \vec{x} & \vec{y} & \vec{z} \\ -\cos \theta_2 & \sin \theta_2 & 0 \\ 0 & 0 & 1 \end{vmatrix} = \frac{n_2 t E_0}{c}\begin{pmatrix} \sin \theta_2 \\ -\cos \theta_2 \\ 0 \end{pmatrix}$

To satisfy the boundary condition $E_{1\parallel}=E_{2\parallel}$ we require the relation
$1+r = t.$
The boundary condition $B_{1\parallel}=B_{2\parallel}$ for the magnetic field leads us to
$-n_1 \cos \theta_1 + n_1 r \cos \theta_1 = -n_2 t \cos \theta_2.$

To find $r$, we exclude $t$:
$1+r = \frac{n_1 \cos \theta_1}{n_2 \cos \theta_2} - \frac{n_1 \cos \theta_1}{n_2 \cos \theta_2} r$
and obtain the assumed equation for $r$:
$r = \frac{n_1 \cos \theta_1 - n_2 \cos \theta_2}{n_1 \cos \theta_1 + n_2 \cos \theta_2}$

A5  ?? Using boundary conditions, show that for $p$-polarization the amplitude reflection
$r_p = \frac{n_2 \cos \theta_1 - n_1\cos \theta_2}{n_2 \cos \theta_1+ n_1 \cos \theta_2}$

The analysis of the $p$-polarized wave is the similar to the previous one. In this case the magnetic field is in the interface plane, so we choose its positive direction along the $z$-axis. Thus, the boundary condition $B_{1\parallel}=B_{2\parallel}$ for the magnetic field could be written immediately:
$n_1 (1 + r) = n_2 t.$

To write the expression for $r$ and $t$ following from the boundary conditions for electric field we have to find its polariation $\vec{p}$. As we discussed earlier the direction of the magnetic field could be found from the $\vec{p}$ with $\vec{B} = \vec{k} \times \vec{p} |\tilde{E}|/\omega$, so $\vec{k} \times \vec{p} = |\vec{k}|\vec{z}$. On the one hand
$\vec{k} \times [\vec{k} \times \vec{p} ] = |\vec{k}| \vec{k} \times \vec{z}$
on the other hand
$\vec{k} \times [\vec{k} \times \vec{p} ] = \vec{k} \cdot 0 - \vec{p} \cdot |\vec{k}|^2,$
where we have used the vector triple product formula:
$\vec{a} \times [\vec{b} \times \vec{c} ] = \vec{b} (\vec{a} \cdot \vec{c}) - \vec{c} (\vec{a} \cdot \vec{b} ).$

Thus, $\vec{p}$ could be written as $\vec{z} \times \vec{k}/|\vec{k}|$ and furthermore the electric field is
$\vec{E}_0 = \frac{E_0}{|\vec{k}_1|} \vec{z} \times \vec{k}_1 = E_0\begin{pmatrix} -\sin \theta_1 \\ \cos \theta_1 \\ 0 \end{pmatrix}$
$\vec{E}_r = \frac{rE_0}{|\vec{k}_r|} \vec{z} \times \vec{k}_r = rE_0 \begin{pmatrix} -\sin \theta_1 \\ -\cos \theta_1 \\ 0 \end{pmatrix}$
$\vec{E}_t = \frac{tE_0}{|\vec{k}_2|} \vec{z} \times \vec{k}_2 = tE_0\begin{pmatrix} -\sin \theta_2 \\ \cos \theta_2 \\ 0 \end{pmatrix}$

The boundary condition $E_{1\parallel}=E_{2\parallel}$ gives us
$(1-r)\cos \theta_1 = t \cos \theta_2.$

Let's us exclude $t$ again:
$\frac{n_1}{n_2}\left(1+r \right) = \frac{\cos \theta_1}{ \cos \theta_2}\left( 1 -r \right)$
and find $r$:
$r = \frac{n_2 \cos \theta_1 - n_1 \cos \theta_2}{n_2 \cos \theta_1 + n_1 \cos \theta_2}.$

A6  ?? Show that the intensity $I$ is related to the complex amplitude $\tilde{E}$ by the following equation:
$I = \frac{c \varepsilon_0 n}{2}|\tilde{E}|^2,$
where $n$ is the refractive index of the medium.

Let's express the Poynting vector in terms of the complex amplitude:
$\vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B} = \frac{1}{\omega \mu_0} \mathfrak{Re} \left\lbrace \tilde{E} e^{-i\omega t} \right\rbrace \times \mathfrak{Re} \left\lbrace \vec{k} \times \tilde{E} e^{-i\omega t} \right\rbrace.$
Take into account that
$\mathfrak{Re} \left\lbrace \tilde{E} e^{-i\omega t} \right\rbrace = \mathfrak{Re} \left\lbrace \tilde{E} \right\rbrace \cos \omega t + \mathfrak{Im} \left\lbrace \tilde{E} \right\rbrace \sin \omega t.$
Substituting this into the above equation gives us
$\begin{split}\omega \mu_0\vec{S} =& \mathfrak{Re} \left\lbrace \tilde{E} \right\rbrace \times \left[ \vec{k} \times \mathfrak{Re} \left\lbrace \tilde{E} \right\rbrace \right] \cos^2 \omega t +\\& \mathfrak{Re} \left\lbrace \tilde{E} \right\rbrace \times \left[ \vec{k} \times \mathfrak{Im} \left\lbrace \tilde{E} \right\rbrace \right] \cos \omega t \sin \omega t +\\& \mathfrak{Im} \left\lbrace \tilde{E} \right\rbrace \times \left[ \vec{k} \times \mathfrak{Re} \left\lbrace \tilde{E} \right\rbrace \right] \cos \omega t \sin \omega t +\\& \mathfrak{Im} \left\lbrace \tilde{E} \right\rbrace \times \left[ \vec{k} \times \mathfrak{Im} \left\lbrace \tilde{E} \right\rbrace \right] \sin^2 \omega t \end{split}.$

With the formula for the vector triple formula $\vec{a} \times \left[ \vec{b} \times \vec{c} \right]$ and the equation $\vec{k} \cdot \tilde{E}=0$ we have:
$\vec{S} = \frac{\vec{k}}{\omega \mu_0 } \left[ \mathfrak{Re}^2 \left\lbrace \tilde{E} \right\rbrace \cos^2 \omega t + 2 \mathfrak{Re} \left\lbrace \tilde{E} \right\rbrace \cdot \mathfrak{Im} \left\lbrace \tilde{E} \right\rbrace \cos \omega t \sin \omega t + \mathfrak{Im}^2 \left\lbrace \tilde{E} \right\rbrace \sin^2 \omega t \right]$

The next step is to average $\vec{S}$ over the period $T=2\pi/\omega$. We have to calculate three integrals:
$\int\limits_0^{T} \cos^2 \omega t \, dt = \frac{1}{2} \int\limits_0^{T} \left( \cos 2 \omega t + 1 \right) dt = \frac{T}{2}$
$\int\limits_0^{T} \sin^2 \omega t \, dt = \frac{1}{2} \int\limits_0^{T} \left( 1 - \cos 2 \omega t \right) dt = \frac{T}{2}$
$\int\limits_0^{T} \cos \omega t \sin \omega t \, dt = \frac{1}{2} \int\limits_0^{T} \sin 2 \omega t \, dt = 0.$
Substituting the values we get
$\langle \vec{S} \rangle = \frac{\vec{k}}{\omega \mu_0 T} \left( \mathfrak{Re}^2 \left\lbrace \tilde{E} \right\rbrace \frac{T}{2} + \mathfrak{Im}^2 \left\lbrace \tilde{E} \right\rbrace \frac{T}{2} \right) = \frac{\vec{k}}{2 \omega \mu_0} |\tilde{E}|^2 \quad \Rightarrow \quad I = \frac{cn \varepsilon_0}{2} |\tilde{E}|^2$

A7  ?? What are the values of $R_p$ and $R_s$? Express it in terms of $n_1$, $n_2$, $\theta_1$ and $\theta_2$.

The intensity is proportional to the squre of the absolute value of the complex amplitude, so $R_p = |r_p|^2$ and $R_s = |r_s|^2$.
$R_p = \left( \frac{n_2 \cos \theta_1 - n_1 \cos \theta_2}{n_2 \cos \theta_1 + n_1 \cos \theta_2} \right)^2, \quad R_s = \left( \frac{n_1 \cos \theta_1 - n_2 \cos \theta_2}{n_1 \cos \theta_1 + n_2 \cos \theta_2} \right)^2$

Answer: $R_p = \left( \frac{n_2 \cos \theta_1 - n_1 \cos \theta_2}{n_2 \cos \theta_1 + n_1 \cos \theta_2} \right)^2, \quad R_s = \left( \frac{n_1 \cos \theta_1 - n_2 \cos \theta_2}{n_1 \cos \theta_1 + n_2 \cos \theta_2} \right)^2$

A8  ?? What is the value of the Brewster's angle $\theta_1 = \theta_\text{B}$ when $R_p=0$ i.e. the $p$-polarized light is not reflected at all.

The value of $R_p$ is zero when $n_2 \cos \theta_1 = n_1 \cos \theta_2$, so
$\cos^2 \theta_2 = \frac{n_2^2}{n_1^2} \cos ^2 \theta_1.$
From the Snell's law:
$\sin^2 \theta_2= \frac{n_1^2}{n_2^2} \sin^2 \theta_1.$
With the Pythagorean trigonometric identity $\sin^2 \alpha + \cos^2 \alpha = 1$ we obtain
$1 =\frac{n_2^2}{n_1^2} (1 - \sin^2 \theta_1) + \frac{n_1^2}{n_2^2} \sin^2 \theta_1 \quad \Rightarrow \quad \sin^2 \theta_1 = \frac{1 - \frac{n_2^2}{n_1^2}}{\frac{n_1^2}{n_2^2} - \frac{n_2^2}{n_1^2}} = n_2^2 \frac{n_1 ^2 - n_2^2}{n_1^4 - n_2^4} = \frac{n_2^2}{n_1^2 + n_2^2}.$
Finally
$\tan \theta_1 = \frac{n_2}{n_1}.$

Answer: $\tan \theta_1 = \frac{n_2}{n_1}$

A9  ?? What are the values $n_A$ and $n_B$?

For the material $A$ the Brewster's angle
$\theta_\text{B} =60^\circ + 20^\circ \frac{17~\text{px}}{101~\text{px}} = 63.3^\circ \quad \Rightarrow \quad n_A = \tan \theta_B = 2.00$
For the material $B$ the Brewster's angle
$\theta_\text{B} = 20^\circ + 20^\circ \frac{50~\text{px}}{101~\text{px}}=29.9^\circ \quad \Rightarrow \quad n_B = \tan \theta_B = 0.58.$
Also for the matherial $B$ we see that $R_p/R_s=1$ for $\theta > \theta_1$, where
$\theta_1 = 20^\circ + 20^\circ \frac{78~\text{px}}{101~\text{px}} = 35.4^\circ.$
This region corresponds to the total internal reflection and $n_B = \sin \theta_1 = 0.58$.

Answer: $n_A = 2.00, \quad n_B = 0.58$