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Question: Answered & Verified by Expert
A monochromatic beam of light has a frequency $v=\frac{3}{2 \pi} \times 10^{12} \mathrm{~Hz}$ 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:
PhysicsElectromagnetic WavesJEE MainJEE Main 2018 (15 Apr Shift 1 Online)
Options:
  • A
    $\frac{E_0}{C}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right) \cos \left[10^4\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right) \cdot \vec{r}-\left(3 \times 10^{12}\right) t\right]$
  • B
    $\frac{E_0}{C}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right) \cos \left[10^4\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \cdot \vec{r}-\left(3 \times 10^{12}\right) t\right]$
  • C
    $\frac{E_0}{C} \hat{k} \cos \left[10^4\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \vec{r}+\left(3 \times 10^{12}\right) t\right]$
  • D
    $\frac{E_0}{C} \frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}} \cos \left[10^4\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \vec{r}+\left(3 \times 10^{12}\right) t\right]$
Solution:
2296 Upvotes Verified Answer
The correct answer is:
$\frac{E_0}{C}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right) \cos \left[10^4\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right) \cdot \vec{r}-\left(3 \times 10^{12}\right) t\right]$
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