The given electric field is:

$$ \overrightarrow{\mathrm{E}}(x, y, z, t)=\mathrm{E}_0 \hat{\mathrm{n}} \mathrm{e}^{i k_o[(x+y+z)-c t]} $$

We can rewrite this as:

$$ \overrightarrow{\mathrm{E}}(x, y, z, t)=\mathrm{E}_0 \hat{\mathrm{n}} \mathrm{e}^{i k_o[(\hat{i}+\hat{j}+\hat{k})\cdot\overrightarrow{r}-c t]} $$

where $$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$$ is the position vector. This represents a plane wave propagating in the direction of $$\hat{i}+\hat{j}+\hat{k}$$.

The wave vector is given by:

$$ \overrightarrow{k}=k_o(\hat{i}+\hat{j}+\hat{k}) $$

The magnitude of the wave vector is:

$$ k=|\overrightarrow{k}|=k_o\sqrt{3} $$

The speed of the wave in the medium is given by:

$$ v=\frac{\omega}{k}=\frac{ck_o}{k_o\sqrt{3}}=\frac{c}{\sqrt{3}} $$

The refractive index of the medium is given by:

$$ n=\frac{c}{v}=\sqrt{3} $$

Since the electric field is polarized in the $$x-z$$ plane, the polarization vector $$\hat{\mathrm{n}}$$ must be a linear combination of $$\hat{\mathrm{i}}$$ and $$\hat{\mathrm{k}}$$. To ensure the polarization is in the $$x-z$$ plane, and considering the wave is propagating along $$\hat{i}+\hat{j}+\hat{k}$$, the polarization vector must be perpendicular to the direction of propagation. Thus:

$$ \hat{\mathrm{n}}=\frac{\hat{\mathrm{i}}-\hat{\mathrm{k}}}{\sqrt{2}} $$

Therefore, the correct options are:

Option B: $$\hat{\mathrm{n}}=\frac{\hat{\mathrm{i}}-\hat{\mathrm{k}}}{\sqrt{2}} ; v=\frac{\mathrm{c}}{\sqrt{3}}$$.

Option C: Refractive index of the medium is $$\sqrt{3}$$.

Mechanics

Electricity

Optics

Modern Physics

Electromagnetic Waves · Physics · WB JEE

MCQ (More than One Correct Answer)