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Please help! - Electromagnetic Waves - JEE Main

A plane polarized monochromatic EM wave is traveling in vacuum along z direction such that at t=t_{1} it is found that the electric field is zero at a spatial point z_{1}. The next zero that occurs in its neighbourhood is at z_{2}. The frequency of the electromagnetic wave is :

  • Option 1)

    \frac{3*10^{8}}{\left |z_{2}-z_{1} \right |}

  • Option 2)

    \frac{1.5*10^{8}}{\left |z_{2}-z_{1} \right |}

  • Option 3)

    \frac{6*10^{8}}{\left |z_{2}-z_{1} \right |}

  • Option 4)

    \frac{1}{t_{1+\frac{\left | z_{2}-z_{1} \right |}{3*10^{8}}}}

Answers (2)
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N neha

As we learnt

Electromagnetic Wave -

Combination of mutually perpendicular electric and magnetic field is referred to as Electromagnetic Wave.

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 Wavelength=2(z_{2}-z_{1})

\therefore frequency = \frac{c}{\lambda }=\frac{3*10^{8}}{2\left |z_{2}-z_{1} \right |}

                                         =\frac{1.5*10^{8}}{\left | z_{2}-z_{1} \right |}


Option 1)

\frac{3*10^{8}}{\left |z_{2}-z_{1} \right |}

Option 2)

\frac{1.5*10^{8}}{\left |z_{2}-z_{1} \right |}

Option 3)

\frac{6*10^{8}}{\left |z_{2}-z_{1} \right |}

Option 4)

\frac{1}{t_{1+\frac{\left | z_{2}-z_{1} \right |}{3*10^{8}}}}

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