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# Help me please, A plane electromagnetic wave of wavelength λ has an intensity I. It ispropagating along the positive Y-direction. The allowed expressions for the electricand magnetic fields are given by :

A plane electromagnetic wave of wavelength λ has an intensity I. It is
propagating along the positive Y-direction. The allowed expressions for the electric
and magnetic fields are given by :

• Option 1)

$\vec{E}= \sqrt{2I/\epsilon _{0}c} \cos [\frac{2\pi }{\lambda }(y-ct)]\hat{k}; \vec{B} =+\frac{1}{c}E\hat{i}$

• Option 2)

$\vec{E}= \sqrt{2I/\epsilon _{0}c} \cos [\frac{2\pi }{\lambda }(y+ct)]\hat{k}; \vec{B} =+\frac{1}{c}E\hat{i}$

• Option 3)

$\vec{E}= \sqrt{I/\epsilon _{0}c} \cos [\frac{2\pi }{\lambda }(y-ct)]\hat{k}; \vec{B} =\frac{1}{c}E\hat{i}$

• Option 4)

$\vec{E}= \sqrt{I/\epsilon _{0}c} \cos [\frac{2\pi }{\lambda }(y-ct)]\hat{i}; \vec{B} =+\frac{1}{c}E\hat{i}$

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AS we have learned

Intensity of EM wave -

$I = \frac{1}{2} \epsilon _{o} E_{o}^{2}c$

- wherein

$\epsilon _{o}$ = Permittivity of free space

$E _{o}$ = Electric field amplitude

c = Speed of light in vacuum

$E_{0}=\sqrt{\left ( \frac{2I}{c\varepsilon _{0}} \right )}$

$B_{0}=\frac{E_{0}}{c}$

direction of $\vec{E}\times \vec{B}$  is along $\hat{j}$

$\vec{B}$ is along $\hat{i}$  and $\vec{E}$ is along $\hat{k}$

Option 1)

$\vec{E}= \sqrt{2I/\epsilon _{0}c} \cos [\frac{2\pi }{\lambda }(y-ct)]\hat{k}; \vec{B} =+\frac{1}{c}E\hat{i}$

Option 2)

$\vec{E}= \sqrt{2I/\epsilon _{0}c} \cos [\frac{2\pi }{\lambda }(y+ct)]\hat{k}; \vec{B} =+\frac{1}{c}E\hat{i}$

Option 3)

$\vec{E}= \sqrt{I/\epsilon _{0}c} \cos [\frac{2\pi }{\lambda }(y-ct)]\hat{k}; \vec{B} =\frac{1}{c}E\hat{i}$

Option 4)

$\vec{E}= \sqrt{I/\epsilon _{0}c} \cos [\frac{2\pi }{\lambda }(y-ct)]\hat{i}; \vec{B} =+\frac{1}{c}E\hat{i}$

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