JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
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 \(\frac{{{E_0}}}{C}\frac{{\left( {\hat i + \hat j + \hat k} \right)}}{{\sqrt 3 }}\cos \left[ {{{10}^4}\left( {\frac{{\hat i + \hat j}}{{\sqrt 2 }}} \right)\cdot \vec r + \left( {3 \times {{10}^{12}}} \right)t} \right]\)
Answer & Solution
Correct Answer
(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
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…
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