Question

A monochromatic beam of light has a frequency $v=\frac{3}{2\pi }\times {10}^{12}Hz$ 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. $\hat{k}\frac{E_0}{C}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right)\cos [{10}^4\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right).\overrightarrow{r}-(3\times {10}^{12})t]$
• B. $\frac{E_0}{C}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right)\cos [{10}^4\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right).\overrightarrow{r}-(3\times {10}^{12})t]$
• C. $\frac{E_0}{C}\hat{k}\cos [{10}^4\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right).\overrightarrow{r}+(3\times {10}^{12})t]$
• D. $\frac{E_0}{C}\frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}}\cos [{10}^4\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right).\overrightarrow{r}+(3\times {10}^{12})t]$

Answer 1

The correct option is A $\hat{k}\frac{E_0}{C}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right)\cos [{10}^4\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right).\overrightarrow{r}-(3\times {10}^{12})t]$

Direction of B is ,

$=\hat{K}\times \hat{E}$

$=\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)\times \hat{K}$

$=\frac{\hat{i}+\hat{k}}{\sqrt{2}}+\frac{\hat{j}\times \hat{k}}{\sqrt{2}}$

$=\frac{\hat{j}}{\sqrt{2}}+\frac{\hat{i}}{\sqrt{2}}=\frac{\hat{i}-\hat{j}}{\sqrt{2}}$

So, ans is between (i) and (ii)

Propagation direction , $\hat{k}=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

$\overline{B}$ wave $e{q}^n$ will be

$\Rightarrow \frac{E_0}{c}\left(\hat{B}\right)$ $Cos[|k|\hat{k}-wt]$

$\downarrow$ $\downarrow$

we know it is we know it is

$\frac{\hat{i}-\hat{j}}{\sqrt{2}}$ $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

Only (i) satisfies

Hence , correct answer is (i)