- Home
- Standard 12
- Physics
એક પ્રકાશના કિરણની આવૃતિ $v = \frac{3}{{2\pi }} \times {10^{12}}\,Hz$ છે જે $\frac{{\hat i + \hat j}}{{\sqrt 2 }}$ દિશામાં પ્રસરે છે. જો તે $\hat k$ દિશામાં પોલારાઇઝ થતો હોય તો તેના માટે ચુંબકીય ક્ષેત્રનું કયું સ્વરૂપ સ્વીકાર્ય હશે?
$\frac{{{E_0}}}{C}\left( {\frac{{\hat i - \hat j}}{{\sqrt 2 }}} \right)\cos \left[ {{{10}^4}\left( {\frac{{\hat i - \hat j}}{{\sqrt 2 }}} \right)\cdot \vec r - \left( {3 \times {{10}^{12}}} \right)t} \right]$
$\frac{{{E_0}}}{C}\left( {\frac{{\hat i - \hat j}}{{\sqrt 2 }}} \right)\cos \left[ {{{10}^4}\left( {\frac{{\hat i + \hat j}}{{\sqrt 2 }}} \right)\cdot \vec r - \left( {3 \times {{10}^{12}}} \right)t} \right]$
$\frac{{{E_0}}}{C}\hat k\cos \left[ {{{10}^4}\left( {\frac{{\hat i + \hat j}}{{\sqrt 2 }}} \right)\cdot \vec r + \left( {3 \times {{10}^{12}}} \right)t} \right]$
$\frac{{{E_0}}}{C}\frac{{\left( {\hat i + \hat j + \hat k} \right)}}{{\sqrt 3 }}\cos \left[ {{{10}^4}\left( {\frac{{\hat i + \hat j}}{{\sqrt 2 }}} \right)\cdot \vec r + \left( {3 \times {{10}^{12}}} \right)t} \right]$
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
$\operatorname{along} \frac{\hat{i}+\hat{j}}{\sqrt{2}}$
