The electric field associated with an electromagnetic wave travelling in vacuum is given by $\overrightarrow{E}=E_0\sin (3y+4z+\omega t)\hat{i}$ where ω is the angular frequency. All quantities are in SI units. The correct statement(s) about this wave is/are:

[Given: speed of light in vacuum c = 3 × 10⁸ m s⁻¹]

Correct Options:

1. The wave is travelling in $-\frac{1}{5}(3\hat{j}+4\hat{k})$ direction.

2. The magnitude of wave vector is 5 m⁻¹

3. The value of ω is 1.5 × 10⁹ rad s⁻¹

Step-by-Step Explanation:

1. Finding the Direction of Propagation and Wave Vector ($\overrightarrow{k}$):

The given electric field of the electromagnetic wave is:

$\overrightarrow{E}=E_0\sin (3y+4z+\omega t)\hat{i}$

A general wave equation for a plane wave propagating in space is expressed as:

$\overrightarrow{E}={\overrightarrow{E}}_0\sin (\overrightarrow{k}·\overrightarrow{r}+\omega t)$

Comparing the spatial part inside the sine function $(3y+4z)$ with $\overrightarrow{k}·\overrightarrow{r}=k_{x}x+k_{y}y+k_{z}z$, we identify the components of the wave vector $\overrightarrow{k}$:

$k_x=0$, $k_y=3$, $k_z=4$

Thus, the wave vector is:

$\overrightarrow{k}=3\hat{j}+4\hat{k}$

Since the phase term is of the form $(\overrightarrow{k}·\overrightarrow{r}+\omega t)$ with a plus sign, the direction of wave propagation is along the negative unit vector direction of $\overrightarrow{k}$:

$\hat{n}=-\frac{\overrightarrow{k}}{\left|\overrightarrow{k}\right|}=-\frac{3\hat{j}+4\hat{k}}{\sqrt{3^2+4^2}}=-\frac{1}{5}(3\hat{j}+4\hat{k})$

Hence, the statement that the wave is travelling in the $-\frac{1}{5}(3\hat{j}+4\hat{k})$ direction is correct.

2. Magnitude of Wave Vector ($k$):

The magnitude of the wave vector $\overrightarrow{k}$ is given by:

$k=\left|\overrightarrow{k}\right|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5{\text{m}}^{-1}$

Hence, the magnitude of the wave vector is indeed 5 m⁻¹.

3. Value of Angular Frequency ($\omega$):

The relationship between angular frequency $\omega$, wave number $k$, and speed of light in vacuum $c$ is:

$c=\frac{\omega }{k}\Rightarrow \omega =c·k$

Given $c=3\times {10}^8{\text{m s}}^{-1}$ and $k=5{\text{m}}^{-1}$:

$\omega =(3\times {10}^8)\times 5=15\times {10}^8=1.5\times {10}^9{\text{rad s}}^{-1}$

Hence, the value of $\omega$ is $1.5\times {10}^9{\text{rad s}}^{-1}$.