- #1
Dom Tesilbirth
- 4
- 1
- Homework Statement
- Consider the interference of two coherent electromagnetic waves whose electric field vectors are given by ##\overrightarrow{E}_{1}=\widehat{i}E_{0}\cos \omega t## and ##\overrightarrow{E}_{2}=\widehat{j}E_{0}\cos \left( \omega t+\phi \right),## where ##\phi## is the phase difference. The intensity of the resulting wave is given by ##\dfrac{\varepsilon _{0}}{2}\langle E^{2}\rangle##, where ##\langle E^{2}\rangle## in the time average of ##E^{2}.## What is the total intensity?
- Relevant Equations
- $$I=\dfrac{\varepsilon _{0}}{2}\langle E^{2}\rangle$$
My Try:
The resultant field is given by
$$\begin{aligned}\overrightarrow{E}=\overrightarrow{E}_{1}+\overrightarrow{E}_{2}=\widehat{i} E_{0}\cos \omega t+\widehat{j}E_{0}\cos \left( \omega t+\phi \right) \\
\Rightarrow E^{2}=E_{0}^{2}\cos ^{2}\omega t+E_{0}^{2}\cos ^{2}\left( \omega t+\phi \right) \\
\Rightarrow E^{2}=E_{0}^{2}\left\{ \cos ^{2}\omega t+\cos ^{2}\left( \omega t+\phi \right) \right\} \\
\Rightarrow \langle E^{2}\rangle =E_{0}^{2}\dfrac{\left[ \int ^{t}_{0}\left\{ \cos ^{2}\omega t+\cos ^{2}\left( \omega t+\phi \right) \right\} dt\right] }{\int ^{t}_{0}dt}\\
\Rightarrow \langle E^{2}\rangle =\dfrac{E_{0}^{2}}{t}\times \int ^{t}_{0}\left[ \dfrac{1}{2}\left\{ \left( 1+\cos \left( 2\omega t\right) \right) +\left( 1+\cos \left( 2\left( \omega t+\phi \right) \right) \right) \right\} \right] \\
\Rightarrow \langle E^{2}\rangle =\dfrac{E_{0}^{2}}{2t}\times \int ^{t}_{0}\left( 2+\cos 2\omega t+\cos 2\left( \omega t+\phi \right) \right) dt\\
\Rightarrow \langle E^{2}\rangle =\dfrac{E_{0}^{2}}{2t}\left[ 2t+\dfrac{\sin \omega t}{\omega }+\dfrac{\sin 2\left( \omega t+\phi \right) }{2\omega }\right] \end{aligned}$$
$$\therefore I=\dfrac{\varepsilon _{0}}{2}\langle E^{2}\rangle$$
$$\Rightarrow I=\dfrac{\varepsilon _{0}}{2}\times \dfrac{E_{0}^{2}}{2t}\left[ 2t+\dfrac{\sin \omega t}{\omega }+\dfrac{\sin ^{2}( \omega t+\phi) }{2\omega }\right]$$
But the answer is suppose to be ##\varepsilon _{0}E_{0}^{2}.## How do I get this answer?
The resultant field is given by
$$\begin{aligned}\overrightarrow{E}=\overrightarrow{E}_{1}+\overrightarrow{E}_{2}=\widehat{i} E_{0}\cos \omega t+\widehat{j}E_{0}\cos \left( \omega t+\phi \right) \\
\Rightarrow E^{2}=E_{0}^{2}\cos ^{2}\omega t+E_{0}^{2}\cos ^{2}\left( \omega t+\phi \right) \\
\Rightarrow E^{2}=E_{0}^{2}\left\{ \cos ^{2}\omega t+\cos ^{2}\left( \omega t+\phi \right) \right\} \\
\Rightarrow \langle E^{2}\rangle =E_{0}^{2}\dfrac{\left[ \int ^{t}_{0}\left\{ \cos ^{2}\omega t+\cos ^{2}\left( \omega t+\phi \right) \right\} dt\right] }{\int ^{t}_{0}dt}\\
\Rightarrow \langle E^{2}\rangle =\dfrac{E_{0}^{2}}{t}\times \int ^{t}_{0}\left[ \dfrac{1}{2}\left\{ \left( 1+\cos \left( 2\omega t\right) \right) +\left( 1+\cos \left( 2\left( \omega t+\phi \right) \right) \right) \right\} \right] \\
\Rightarrow \langle E^{2}\rangle =\dfrac{E_{0}^{2}}{2t}\times \int ^{t}_{0}\left( 2+\cos 2\omega t+\cos 2\left( \omega t+\phi \right) \right) dt\\
\Rightarrow \langle E^{2}\rangle =\dfrac{E_{0}^{2}}{2t}\left[ 2t+\dfrac{\sin \omega t}{\omega }+\dfrac{\sin 2\left( \omega t+\phi \right) }{2\omega }\right] \end{aligned}$$
$$\therefore I=\dfrac{\varepsilon _{0}}{2}\langle E^{2}\rangle$$
$$\Rightarrow I=\dfrac{\varepsilon _{0}}{2}\times \dfrac{E_{0}^{2}}{2t}\left[ 2t+\dfrac{\sin \omega t}{\omega }+\dfrac{\sin ^{2}( \omega t+\phi) }{2\omega }\right]$$
But the answer is suppose to be ##\varepsilon _{0}E_{0}^{2}.## How do I get this answer?
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