The electric field of a plane electromagnetic | vedclass

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(B) The given electric field is $\vec{E} = E_0 \hat{n} e^{i k_0[(x+y+z) - ct]}$.Comparing this with the standard form $\vec{E} = E_0 \hat{n} e^{i(\vec

Vedclass · Accepted Answer

$\hat{n} = \frac{\hat{i} - \hat{k}}{\sqrt{2}}; v = \frac{c}{\sqrt{3}}$

Vedclass · Answer

(B) The given electric field is $\vec{E} = E_0 \hat{n} e^{i k_0[(x+y+z) - ct]}$.Comparing this with the standard form $\vec{E} = E_0 \hat{n} e^{i(\vec{k} \cdot \vec{r} - \omega t)}$,we identify the wave vector $\vec{k} = k_0(\hat{i} + \hat{j} + \hat{k})$.The magnitude of the wave vector is $k = |k_0(\hat{i} + \hat{j} + \hat{k})| = k_0 \sqrt{1^2 + 1^2 + 1^2} = k_0 \sqrt{3}$.The speed of the wave in the medium is $v = \frac{\omega}{k} = \frac{k_0 c}{k_0 \sqrt{3}} = \frac{c}{\sqrt{3}}$.The refractive index $n = \frac{c}{v} = \sqrt{3}$.Since the wave is transverse,$\vec{E} \cdot \vec{k} = 0$,which implies $\hat{n} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$.Given that $\vec{E}$ is polarized in the $x-z$ plane,$\hat{n}$ must be in the $x-z$ plane,so $\hat{n} = a\hat{i} + b\hat{k}$.From $\hat{n} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$,we get $a + b = 0$,so $a = -b$.Normalizing $\hat{n}$,we get $\hat{n} = \frac{\hat{i} - \hat{k}}{\sqrt{2}}$.Thus,both options $B$ and $C$ are correct.

The electric field of a plane electromagnetic wave in a medium is given by $\vec{E}(x, y, z, t) = E_0 \hat{n} e^{i k_0[(x+y+z)-ct]}$,where $c$ is the speed of light in free space. The $\vec{E}$ field is polarized in the $x-z$ plane. If the speed of the wave in the medium is $v$,then:

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