8.Electromagnetic waves
hard

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

$\frac{{{E_0}}}{C}\left( {\frac{{\hat i - \hat j}}{{\sqrt 2 }}} \right)\cos \left[ {{{10}^4}\left( {\frac{{\hat i - \hat j}}{{\sqrt 2 }}} \right)\cdot \vec r - \left( {3 \times {{10}^{12}}} \right)t} \right]$

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

$\frac{{{E_0}}}{C}\hat k\cos \left[ {{{10}^4}\left( {\frac{{\hat i + \hat j}}{{\sqrt 2 }}} \right)\cdot \vec r + \left( {3 \times {{10}^{12}}} \right)t} \right]$

D

$\frac{{{E_0}}}{C}\frac{{\left( {\hat i + \hat j + \hat k} \right)}}{{\sqrt 3 }}\cos \left[ {{{10}^4}\left( {\frac{{\hat i + \hat j}}{{\sqrt 2 }}} \right)\cdot \vec r + \left( {3 \times {{10}^{12}}} \right)t} \right]$

(JEE MAIN-2018)

Solution

Poynting Vector –

$\vec{s}=\frac{\vec{E} \times \vec{B}}{\mu_{o}}$

-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

$\operatorname{along} \frac{\hat{i}+\hat{j}}{\sqrt{2}}$

Standard 12
Physics

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