| ### Statement | | ### Statement |
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| $12.1.3.$ [Insert the problem statement] | | $12.1.3.$ [Insert the problem statement] |
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| ### Solution | | ### Solution |
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| For a plane sinusoidal wave, the general equation of the electromagnetic wave is given by: | | For a plane sinusoidal wave, the general equation of the electromagnetic wave is given by: |
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| \begin{equation} | | \begin{equation} |
| E(z,t)=E_0\sin(At+Bz+\varphi_0) | | E(z,t)=E_0\sin(At+Bz+\varphi_0) |
| \end{equation} | | \end{equation} |
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| where $A$,$B$ and $\varphi_0$ are constants. | | where $A$,$B$ and $\varphi_0$ are constants. |
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| From the condition of the problem we know that: | | From the condition of the problem we know that: |
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| \begin{equation} | | \begin{equation} |
| E(z,0)=E_0\sin(\frac{2\pi}{\lambda}z) | | E(z,0)=E_0\sin(\frac{2\pi}{\lambda}z) |
| \end{equation} | | \end{equation} |
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| Using eq(1) we get: | | Using eq(1) we get: |
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| \begin{equation} | | \begin{equation} |
| E_0\sin{(Bz+\varphi_0)}=E_0\sin(\frac{2\pi}{\lambda}z) | | E_0\sin{(Bz+\varphi_0)}=E_0\sin(\frac{2\pi}{\lambda}z) |
| \end{equation} | | \end{equation} |
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| By comparing the phases we obtain that: | | By comparing the phases we obtain that: |
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| \begin{equation} | | \begin{equation} |
| B=\frac{2\pi}{\lambda};\varphi_0=0 | | B=\frac{2\pi}{\lambda};\varphi_0=0 |
| \end{equation} | | \end{equation} |
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| As the wave travels in the positive direction: | | As the wave travels in the positive direction: |
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| \begin{equation} | | \begin{equation} |
| \frac{dz}{dt}=c | | \frac{dz}{dt}=c |
| \end{equation} | | \end{equation} |
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| Setting the phase($\varphi=At+Bz+\varphi_0$)to a constant value allows us to derive the phase velocity($c$): | | Setting the phase($\varphi=At+Bz+\varphi_0$)to a constant value allows us to derive the phase velocity($c$): |
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| \begin{equation} | | \begin{equation} |
| A+B\frac{dz}{dt}=\frac{d\varphi}{dt}=0 | | A+B\frac{dz}{dt}=\frac{d\varphi}{dt}=0 |
| \end{equation} | | \end{equation} |
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| So we get the constant $A$: | | So we get the constant $A$: |
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| \begin{equation} | | \begin{equation} |
| A=-\frac{2\pi c}{\lambda} | | A=-\frac{2\pi c}{\lambda} |
| \end{equation} | | \end{equation} |
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| Plugging all the constants into the eq(1) gives us: | | Plugging all the constants into the eq(1) gives us: |
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| \begin{equation} | | \begin{equation} |
| E(z,t)=E_0\sin(\frac{2\pi}{\lambda}(z-ct)) | | E(z,t)=E_0\sin(\frac{2\pi}{\lambda}(z-ct)) |
| \end{equation} | | \end{equation} |
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