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# The electric field of an electromagnetic wave travelling through the vacuum is given by the equation $E = {E_0}\sin \left( {kx - \omega t} \right)$ . The quantity that is independent of wavelength is:(A) $k\omega$(B) $\dfrac{k}{\omega }$ (C) ${k^2}\omega$ (D) $\omega$

Last updated date: 20th Jun 2024
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Hint: The angular frequency $\omega = 2\pi \nu$. The frequency varies with wavelength as $\nu = \dfrac{c}{\lambda }$. Now to find the quantity which is independent of wavelength, we will go with the other options available.

$k$ is a wave number.
$\Rightarrow k = \dfrac{{2\pi }}{\lambda }$
It shows that k is dependent on wavelength.
Now,
$\Rightarrow$ $\dfrac{k}{\omega } = \dfrac{{\dfrac{{2\pi }}{\lambda }}}{{2\pi \nu }}$ $= \dfrac{1}{{\nu \lambda }} = \dfrac{1}{c}\left( {\because c = \nu \lambda } \right)$
where $c$ is the speed of electromagnetic waves in vacuum. Its value is $3 \times {10^8}{\raise0.5ex\hbox{$\scriptstyle m$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle s$}}$ .

Hence, option B. is correct.

NOTE: Try to use the options given as a tool to find your answer. It is necessary to observe that $\omega$ is dependent on wavelength. So, anything which includes $\omega$ as a product will also be dependent on wavelength. Only, the quantity that can be independent is the one with the division(option B. here). Hence, try to derive from it and get the answer.