Electromagnetic Waves

JEE Main level, single correct, medium

117 of 165

Question 117 of 165

A monochromatic beam of light has a frequency v=32π×1012v=\dfrac{3}{2\pi }\times {10}^{12} HzHz and is propagating along the direction i^+j^2\dfrac{\hat{i}+\hat{j}}{\sqrt{2}} . It is polarized along the k^\hat{k} direction. The acceptance form the magnetic field is:

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Answer: (C) E0C(i^−j^)2Cos⁡[104(i^+j^)2.r⃗+(3×1012)t]\dfrac{{E}_{0}}{C}\dfrac{\left(\hat{i}-\hat{j}\right)}{\sqrt{2}}Cos⁡\left[{10}^{4}\dfrac{\left(\hat{i}+\hat{j}\right)}{\sqrt{2}}.\vec{r}+\left(3\times {10}^{12}\right)t\right]

E^×B^\hat{E}\times \hat{B} should have given direction of wave propagation,⇒K^×B^∥i^+j^2\Rightarrow \hat{K}\times \hat{B}\parallel \dfrac{\hat{i}+\hat{j}}{\sqrt{2}}
K^×(i^+j^2)=j^−(−i^)2=i^+j^2∥i^+j^2\hat{K}\times \left(\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}\right)=\dfrac{\hat{j}-\left(-\hat{i}\right)}{\sqrt{2}}=\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}\parallel \dfrac{\hat{i}+\hat{j}}{\sqrt{2}}.

Wave propagation vector K^\hat{K} should be along i^+j^2\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}.

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