Question

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

• A. $\frac{E_{o}}{C} \frac{\left(\hat{i} - \hat{j}\right)}{\sqrt{2}} \cos\left[10^{4} \frac{\left(\hat{i} + \hat{j}\right)}{\sqrt{2}} . \vec{r} - \left(3 \times 10^{12}\right) t\right]$
• B. $\frac{E_{o}}{C} \frac{\left(\hat{i} - \hat{j}\right)}{\sqrt{2}} \cos\left[10^{4} \frac{\left(\hat{i} - \hat{j}\right)}{\sqrt{2}} . \vec{r} - \left(3 \times 10^{12}\right) t\right]$
• C. $\frac{E_{o}}{C} \hat{k} \, \, \cos\left[10^{4} \frac{\left(\hat{i} - \hat{j}\right)}{\sqrt{2}} . \vec{r} - \left(3 \times 10^{12}\right) t\right]$
• D. $\frac{E_{o}}{C} \frac{\left(\hat{i} + \hat{j} + \hat{k} \right)}{\sqrt{3}} \cos\left[10^{4} \frac{\left(\hat{i} - \hat{j}\right)}{\sqrt{2}} . \vec{r} - \left(3 \times 10^{12}\right) t\right]$

Answer 1

Answer: (A)

Given: Frequency $v=\frac{3}{2 \pi} \times 10^{12} Hz ;$ direction $=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$. The beam is polarised along $\hat{k}$ direction.
Direction of magnetic field $\vec{B}=\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}} \times \hat{k}=\frac{-\hat{j}}{\sqrt{2}}+\frac{\hat{i}}{\sqrt{2}}\right)$
Therefore, direction of $\vec{B}=\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right)$
Now from wave equation, we have
$B=B_{0} \cos (k r-\omega t) ; \omega=2 \pi v=2 \pi \times \frac{3}{2 \pi} \times 10^{12}\, Hz =3 \times 10^{12}\, Hz$
Now, $k=\frac{\hat{i}+\hat{j}}{\sqrt{2}} ; B_{0}=\frac{\varepsilon_{0}}{C}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right)$
Therefore, $B=\frac{\varepsilon_{0}}{C}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right) \cos \left[10^{4}\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \cdot r-\left(3 \times 10^{12}\right) t\right]$