Electromagnetic waves > Properties of emission and reflection of electromagnetic waves > Problem 12.1.3

Statement

$12.1.3.$ [Insert the problem statement]

Solution

For a plane sinusoidal wave, the general equation of the electromagnetic wave is given by:

\begin{equation}
E(z,t)=E_0\sin(At+Bz+\varphi_0)
\end{equation}

where $A$,$B$ and $\varphi_0$ are constants.

From the condition of the problem we know that:

\begin{equation}
E(z,0)=E_0\sin(\frac{2\pi}{\lambda}z)
\end{equation}

Using eq(1) we get:

\begin{equation}
E_0\sin{(Bz+\varphi_0)}=E_0\sin(\frac{2\pi}{\lambda}z)
\end{equation}

By comparing the phases we obtain that:

\begin{equation}
B=\frac{2\pi}{\lambda};\varphi_0=0
\end{equation}

As the wave travels in the positive direction:

\begin{equation}
\frac{dz}{dt}=c
\end{equation}

Setting the phase($\varphi=At+Bz+\varphi_0$)to a constant value allows us to derive the phase velocity($c$):

\begin{equation}
A+B\frac{dz}{dt}=\frac{d\varphi}{dt}=0
\end{equation}

So we get the constant $A$:

\begin{equation}
A=-\frac{2\pi c}{\lambda}
\end{equation}

Plugging all the constants into the eq(1) gives us:

\begin{equation}
E(z,t)=E_0\sin(\frac{2\pi}{\lambda}(z-ct))
\end{equation}

Answer

[Insert a concise answer or boxed result]