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\subsection*{Problem 1}
\ramme{ {\bfseries\sffamily Notes chapter 4, exercise 10} \\
        We will once again consider the interference from two loudspeakers. One speaker is placed on the $x$-axis at $x=-L$ and the 
        other is placed on the $y$-axis at $y=-L$, as shown in the figure below. The speaker at $x=-L$ emits a spherical wave 
        described by the complex wave
        \begin{equation}
         \tilde f_x (\vec r, t) = \frac{A}{r_x} e^{i(kr_x -\omega t)} ,
        \end{equation}
        while the speaker at $y=-L$ emits a spherical wave with a different phase
        \begin{equation}
         \tilde f_y (\vec r, t) = \frac{A}{r_y} e^{i(kr_y -\omega t + \delta)} .
        \end{equation}
        The amplitude $A$ is real. }

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  %Bølger fra de to kilder:
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  \multido{\r=0.8+0.4}{29}{ \pscircle[linewidth=0.2pt,linecolor=GrayLight](-4,0){\r} }
  %Koordinatsystem:
  \psline{->}(0,-5)(0,2)  \uput[180](0,2){$y$}
  \psline{->}(-10,0)(7,0)  \uput[-90](7,0){$x$}
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  \psline[linestyle=dotted]{c-c}(-5,-5)(2,2)
  \psline[linestyle=dashed](-1,-1)(-1,1)(1,1)(1,-1)(-1,-1)
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\subsubsection*{Question a}
\ramme{ Find the amplitude of the oscillation at all positions in the plane $y=x$, which is shown with a dotted line in the figure. }
When I am only looking on the plane where $x=y$ (I will call the new parameter $s$, so that $s=x=y$, as not to get any confusion), I get that $r_s = r_x = r_y$. Now I can add the two wavefunctions together to see how they interact. I get that the sum of the two, on the line $y=x$, is the total wave function $f_s$, given by
\begin{equation*}
 \tilde f_s(\vec r, t) = \tilde f_x(\vec r, t) + \tilde f_y(\vec r, t)  
                       = \frac{A}{r_s} e^{i(kr_s -\omega t)} + \frac{A}{r_s} e^{i(kr_s -\omega t + \delta)} 
                       = \frac{A}{r_s} e^{-i\omega t} \left( e^{ikr_s} + e^{i(kr_s + \delta)} \right) ,
\end{equation*}
which can be written as 
\begin{equation}
 \tilde f_s(\vec r, t) = |\tilde C (\vec r)| e^{-i(\omega t + \phi)} ,
 \label{p1:qa:form}
\end{equation}
where $\phi$ is some unknown (and unimportant in this problem) phase, and 
\allowdisplaybreaks{\begin{align*}
 |\tilde C (\vec r)|^2 &= \frac{A^2}{r_s^2} \left| e^{ikr_s} + e^{i(kr_s + \delta)} \right|^2 \\
                       &= \frac{A^2}{r_s^2} \left( e^{ikr_s} + e^{i(kr_s + \delta)} \right) \left( e^{-ikr_s} + e^{-i(kr_s + \delta)} \right) \\
                       &= \frac{A^2}{r_s^2} \left( e^{-i\delta} + 1 \right)^2 e^{i\delta} \\
                       &= \frac{A^2}{r_s^2} 2 \left( \cos{\delta} + 1 \right) ,
\end{align*}}%
so that the amplitude of the wave is
\begin{equation}
 |\tilde C (\vec r)| = \frac{A}{r_s} \sqrt{2 \left( \cos{\delta} + 1 \right)} .
 \label{p1:qa:amplitude}
\end{equation}
With this knowledge I can write the wavefunction in the form from \eqref{p1:qa:form} with the amplitude given in \eqref{p1:qa:amplitude}, so that
\begin{equation*}
 \tilde f_s(\vec r, t) = \frac{A}{r_s} \sqrt{2 \left( \cos{\delta} + 1 \right)} e^{-i(\omega t + \phi)} ,
\end{equation*}
where the phase $\phi$ still isn't relevant to the answer of this question (the actual answer of the question was the amplitude in \eqref{p1:qa:amplitude}). 
 


\subsubsection*{Question b}
\ramme{ We will now consider the small region close to $x=y=z=0$ (shown by a dashed rectangle in the figure). In this region $x,y,z \ll L$ 
        so that we can ignore variations in $A/r_x$ and $A/r_y$ within the region, and we will assume $x^2,y^2,z^2 \ll L/k$, but $x$, $y$ 
        and $z$ may be bigger than $1/k$.
        
        \hspace{0.5cm}Find the amplitude of the oscillation in this region, and show that there will be planes with constructive and destructive 
        interference. Find the shortest distance between two planes with destructive interference. }
First, I note that the $r_x$ and $r_y$ are the two distances from their respective sources, not from the origo of the coordinate system. So I have to write them as functions of the coordinate system coordinates, where I simply have that
\begin{equation}
 r_x = \sqrt{(x+L)^2 + y^2 + z^2}  \qquad\text{and}\qquad  r_y = \sqrt{x^2 + (y+L)^2 + z^2} .
 \label{eq:distances}
\end{equation}
So when $x,y,z << L$, I get that the two distances $r_x$ and $r_y$ in \eqref{eq:distances} are aproximately equal to
\begin{equation*}
 r_x \approx \sqrt{L^2} = L  \qquad\text{and}\qquad  r_y \approx \sqrt{L^2} = L ,
\end{equation*}
so the variations in $A/r_x$ and $A/r_y$ are really nonexisting in the region. Hence the two spherical waves summed together inside the rectangle will approximately be the total wave equation $f$, given by
\begin{equation*}
 f(\vec r, t) = \frac{A}{r_x} e^{i(kr_x -\omega t)} + \frac{A}{r_y} e^{i(kr_y -\omega t + \delta)}
              = \frac{A}{L} e^{-i \omega t} \left( e^{i k r_x} + e^{i ( k r_y + \delta ) } \right) .
\end{equation*}
I now note that the distances can be written as something simpler, by taking the $L$ outside the square root and approximate it with the formula $(1+\epsilon)^n \approx 1 + n \epsilon$, so that I have
\allowdisplaybreaks{\begin{align*}
 r_x &= \sqrt{(x+L)^2 + y^2 + z^2} \\
     &= \sqrt{x^2 + L^2 + 2Lx + y^2 + z^2} \\
     &= L \sqrt{\frac{x^2}{L^2} + 1 + \frac{2x}{L} + \frac{y^2}{L^2} + \frac{z^2}{L^2}} \\
     &\approx L \left[ 1 + \frac{1}{2} \left( \frac{x^2}{L^2} + 1 + \frac{2x}{L} + \frac{y^2}{L^2} + \frac{z^2}{L^2} \right)  \right] \\
     &= L + \frac{x^2}{2L} + x + \frac{y^2}{2L} + \frac{z^2}{2L} ,
\end{align*}}%
and, by a calculation totaly similar (I actually just have to change $x$ and $y$), 
\begin{equation*}
 r_y \approx L + \frac{y^2}{2L} + y + \frac{x^2}{2L} + \frac{z^2}{2L} .
\end{equation*}
This gives me the total wave equation to be
\begin{equation*}
 f(\vec r, t) \approx \frac{A}{L} e^{-i \omega t} \left( e^{i k \left( L + \frac{x^2}{2L} + x + \frac{y^2}{2L} + \frac{z^2}{2L} \right)} 
                                    + e^{i \left( k \left( L + \frac{y^2}{2L} + y + \frac{x^2}{2L} + \frac{z^2}{2L} \right) + \delta \right) } \right) .
\end{equation*}
Now, this is not a particularly pretty expression, but if I note that $x^2,y^2,z^2 \ll L/k$, I see that the fractions of the form $kx^2/2L$ is much smaller than 1, so that I can simply ignore them. I then have a much tidier expression of the wave function as 
\begin{equation*}
 f(\vec r, t) \approx \frac{A}{L} e^{-i \omega t} \left( e^{i k \left( L + x\right)} + e^{i \left( k \left( L + y + \right) + \delta \right) } \right) .
\end{equation*}
Note that this is actually just two plane waves added together. This expression can be written as
\begin{equation}
 f(\vec r, t) = |\tilde C (\vec r)| e^{-i(\omega t + \phi)} ,
 \label{p1:qb:form}
\end{equation}
where $\phi$ again is some unknown phase (not nessesarily the same as in question a) and 
\allowdisplaybreaks{\begin{align*}
 |\tilde C (\vec r)|^2 
    &= \frac{A^2}{L^2} \left| e^{i k \left( L + x\right)} + e^{i \left( k \left( L + y + \right) + \delta \right) } \right|^2 \\
    &= \frac{A^2}{L^2} \left( e^{i k \left( L + x\right)} + e^{i \left( k \left( L + y + \right) + \delta \right) } \right) 
                       \left( e^{-i k \left( L + x\right)} + e^{-i \left( k \left( L + y + \right) + \delta \right) } \right) \\
    &= \frac{A^2}{L^2} 2 \left( \cos{( kx - ky - \delta )} + 1 \right) \\
    &= \frac{A^2}{L^2} 2 \left( \cos{\big( k (x - y) -\delta \big)} + 1 \right) ,
\end{align*}}%
so that
\begin{equation}
 |\tilde C (\vec r)| = \frac{A}{L} \sqrt{2 \left( \cos{\big( k (x - y) -\delta \big)} + 1 \right)} .
 \label{p1:qb:amplitude}
\end{equation}
Then I can write the wave function with the form from \eqref{p1:qb:form} with the amplitude in \eqref{p1:qb:amplitude} as
\begin{equation*}
 f(\vec r, t) \approx \frac{A}{L} \sqrt{2 \left( \cos{\big( k (x - y) -\delta \big)} + 1 \right)} e^{-i(\omega t + \phi)} 
                    = \frac{A}{L} \sqrt{2 \left( \cos{\big( k (x - y) -\delta \big)} + 1 \right)} \cos{(\omega t + \phi)} .
\end{equation*}
For this wave to be zero at all times (destructive interference), the first cosine must equal $-1$, which happens when
\begin{equation}
 k (x-y) - \delta = n \pi  \quad\Rightarrow\quad  y = x + \frac{\delta - n \pi}{k} , \qquad \text{for odd } n .
 \label{p1:qb:destructive}
\end{equation}
On the other hand, for this wave to be maximal at all times (constructive interference), the first cosine must equal $1$, which happens when
\begin{equation}
 k (x-y) - \delta = n \pi  \quad\Rightarrow\quad  y = x + \frac{\delta - n \pi}{k} , \qquad \text{for even } n .
 \label{p1:qb:constructive}
\end{equation}
These two equations in \eqref{p1:qb:destructive} and \eqref{p1:qb:constructive} describes lines in the $xy$-plane that stretches over all $z$ -- so really, they are planes in the $xyz$-space (the equation of a plane is $ax+by+cz+d=0$, which these two planes certainly holds). They are all parallel, and, as expected, every second is a plane with destructive interference while the others are a planes with constructive interference.

The distance between two planes with destructive interference is simply found by looking at two of these planes, from equation \eqref{p1:qb:destructive}, with next-to-each-other $n$'s. By looking at the line where $x=0$ and $z=0$, I can find the distance in $y$ between such two planes. I choose to look at the planes for $n=1$ and $n=3$. The distance in $y$ between these two is
\begin{equation*}
 \frac{\delta - \pi}{k} - \frac{\delta - 3\pi}{k} = \frac{2\pi}{k} ,
\end{equation*}
and since the slope of the line (that is, the line which is the interception between the plane and the $xy$-plane) is 1 (which is the constant that multiplies $x$ in the expression in \eqref{p1:qb:destructive}), the angle between the planes and the $y$-axis is $45^\circ = \pi/4$. 

Then the orthogonal distance between the two planes is 
\begin{equation*}
 d = \frac{2\pi}{k} \sin{\left( \frac{\pi}{4} \right)} 
   = \frac{2\pi}{\sqrt{2} k}
   = \frac{\sqrt{2} \pi}{k} .
\end{equation*}
This distance is drawn in to a plot of the $xy$-plane with the destructive and constructive interference planes on figure \vref{fig:interference}. Here it is easy to see how I have found the distance $d$ from the $y$-distance between two planes and the angle between the planes and the $y$-axis. 



\begin{figure}[tb]
 \centering
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   \psline[linewidth=1pt,linecolor=Periwinkle]{c-c}(-5,-5.5)(5,4.5)
   \psline[linewidth=1pt,linecolor=YellowGreen]  {c-c}(-5,-4.5)(5,5.5)
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   \psline[linewidth=1pt,linecolor=Periwinkle]{c-c}(-5,-1.5)(5,8.5)
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   \psline{->}(-4,0)(4,0)  \uput[-90](3.8,0){$x$}
   %Bestemte linier der forklarer systemet:
   \psline[linewidth=1.5pt,linecolor=YellowOrange]{*-*}(0,-0.5)(0,1.5)     
   \psline[linewidth=1.5pt,linecolor=YellowOrange]{c-c}(0,1.5)(1,0.5)      \uput[0](0.5,1.2){$d$}
   \psline[linewidth=1.5pt,linecolor=YellowOrange]{c-c}(1,0.5)(0,-0.5)
   \psarc[linewidth=1.5pt,linecolor=YellowOrange]{->}(2.5,0){1cm}{0}{45}   \uput[45](3.4,0.3){$45^\circ$}
   %Tekst:
   \uput[0](1.4,-1.8){\color{Periwinkle} Destructive} 
   \uput[0](1.4,-2.15){\color{Periwinkle} $\quad$interference}
   \uput[0](1.4,-2.75){\color{YellowGreen} Constructive}
   \uput[0](1.4,-3.1){\color{YellowGreen} $\quad$interference}
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 \caption{Plot of the planes in the $xy$-plane with an explanation of the findings of the distance $d$ between two planes with destructive
          interference.}
 \label{fig:interference}
\end{figure}





















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