- JEE Mains 2018
- Hard
- MCQ
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
Answer & Solution
(A) \(\frac{{{E_0}}}{C}\left( {\frac{{\hat i - \hat j}}{{\sqrt 2 }}} \right)\cos \left[ {{{10}^4}\left( {\frac{{\hat i - \hat j}}{{\sqrt 2 }}} \right)\cdot \vec r - \left( {3 \times {{10}^{12}}} \right)t} \right]\)
Step-by-step Solution
Detailed explanation
\(\vec{s}=\frac{\vec{E} \times \vec{B}}{\mu_{o}}\)
-wherein
It is total energy flowing perpendicularly per second per unit area into the surface in free space.
\(\vec{E} \times \vec{B}\) should give a direction of wave propagation
\(\Rightarrow \vec{E} \times \vec{B} \| \frac{\hat{i}+\hat{j}}{\sqrt{2}}\)
option \((1) \hat{k} \times\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)=\frac{\hat{j}-\hat{i}}{\sqrt{2}} \| \frac{\hat{i}-\hat{j}}{\sqrt{2}}\)
option \(( 2)\) and \(( 4)\) does not satisfy this wave propagation vector should be
\(\operatorname{along} \frac{\hat{i}+\hat{j}}{\sqrt{2}}\)
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