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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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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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