A monochromatic beam of light has a frequency | vedclass

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Question

(B) The direction of wave propagation is given by the unit vector $\hat{n} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.
The electric field is polarized alo

Vedclass · Accepted Answer

$\frac{E_0}{c} \left( \frac{\hat{i} - \hat{j}}{\sqrt{2}} ight) \cos \left[ 10^4 \left( \frac{\hat{i} + \hat{j}}{\sqrt{2}} ight) \cdot \vec{r} - (3 	imes 10^{12})t ight]$

Vedclass · Answer

(B) The direction of wave propagation is given by the unit vector $\hat{n} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.
The electric field is polarized along $\hat{k}$, so $\vec{E} = E_0 \hat{k} \cos(\vec{k} \cdot \vec{r} - \omega t)$.
The magnetic field $\vec{B}$ must be perpendicular to both the direction of propagation $\hat{n}$ and the electric field $\vec{E}$.
Thus, the direction of $\vec{B}$ is $\hat{n} 	imes \hat{k} = \left( \frac{\hat{i} + \hat{j}}{\sqrt{2}} ight) 	imes \hat{k} = \frac{\hat{j} - \hat{i}}{\sqrt{2}} = - \left( \frac{\hat{i} - \hat{j}}{\sqrt{2}} ight)$.
The magnitude of the magnetic field is $B_0 = \frac{E_0}{c}$.
The wave vector $\vec{k}$ has magnitude $k = \frac{2\pi v}{c} = \frac{2\pi (3/2\pi 	imes 10^{12})}{3 	imes 10^8} = 10^4 \, m^{-1}$.
Thus, $\vec{B} = \frac{E_0}{c} \left( \frac{\hat{j} - \hat{i}}{\sqrt{2}} ight) \cos \left[ 10^4 \left( \frac{\hat{i} + \hat{j}}{\sqrt{2}} ight) \cdot \vec{r} - (2\pi v)t ight]$.
Comparing this with the given options, option $B$ matches the required direction and wave vector.

$A$ monochromatic beam of light has a frequency $v = \frac{3}{2\pi} \times 10^{12} \, Hz$ and is propagating along the direction $\vec{n} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$. It is polarized along the $\hat{k}$ direction. The acceptable form for the magnetic field $\vec{B}$ is:

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