Q155 / 155 · WBJEE 2024 · Physics

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

  • A
    $\hat{\mathrm{n}}=\hat{\mathrm{i}}-\hat{\mathrm{k}} ; v=\mathrm{c}$.
  • Correct answer
    B
    $\hat{\mathrm{n}}=\frac{\hat{\mathrm{i}}-\hat{\mathrm{k}}}{\sqrt{2}} ; v=\frac{\mathrm{c}}{\sqrt{3}}$
  • Correct answer
    C
    refractive index of the medium is $\sqrt{3}$
  • D
    $\hat{n}=\frac{\hat{\mathrm{i}}+\hat{\mathrm{k}}}{\sqrt{2}} ; v=\frac{\mathrm{c}}{\sqrt{2}}$

Solution

$\begin{aligned} & \text { Hint : } \overrightarrow{\mathrm{E}}=\mathrm{E}_0 \hat{n} e^{i k_0(\hat{\mathrm{k}} \cdot \hat{\mathrm{r}}-\mathrm{wt})} \\ & \Rightarrow \overrightarrow{\mathrm{k}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \Rightarrow v=\frac{\mathrm{c}}{|\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}|}=\frac{\mathrm{c}}{\sqrt{3}} \\ & \Rightarrow \text { refractive index }=\sqrt{3} \\ & \overrightarrow{\mathrm{E}} \text { is in } z-x \text { plane and } \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{V}}=0 \\ & \Rightarrow \hat{\mathrm{n}}=\frac{\hat{\mathrm{i}}-\hat{\mathrm{k}}}{\sqrt{2}}\end{aligned}$

Hence, the correct answer is options: 'B', 'C'.

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