WBJEE 2024 Physics & Chemistry Paper

The given electric field is:$\vec{E}(x, y, z, t)=E_0 \hat{n} e^{i k_o[(x+y+z)-c t]}$We can rewrite this as:$\vec{E}(x,y,z,t)=E_0 \hat{n} e^{\hat{i} k_o[(i+\hat{j}+\hat{k})\cdot\vec{r}-c t]}$ where $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$ 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:$\vec{k}=k_o(\hat{i}+\hat{j}+\hat{k})$ The magnitude of the wave vector is:$k=|\vec{k}|=k_o \sqrt{3}$The speed of the wave in the medium is given by:$v = \frac{\omega}{k} = \frac{c k_o}{k_o \sqrt{3}} = \frac{c}{\sqrt{3}}$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 $\hat{n}$ must be a linear combination of $\hat{i}$ and $\hat{k}$. 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:$\hat{n} = \frac{\widehat{i - \hat{k}}}{\sqrt{2}}$Therefore, the correct options are:Option B: $\hat{n} = \frac{\widehat{i - \hat{k}}}{\sqrt{2}} ; v = \frac{c}{\sqrt{3}}$.Option C: Refractive index of the medium is $\sqrt{3}$.