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
$B=B_0\cos (kr-\omega t);\omega =2\pi v=2\pi \times \frac{3}{2\pi }\times 10^{12}Hz=3\times 10^{12}Hz$
Now, $k=\frac{\hat{i}+\hat{j}}{\sqrt{2}};B_0=\frac{{\epsilon }_0}{C}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right)$
Therefore, $B=\frac{{\epsilon }_0}{C}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right)\cos \left[10^4\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)\cdot r-\left(3\times 10^{12}\right)t\right]$