Question

The electric field of an electromagnetic wave traveling through the vacuum is given by the equation $E=E_{0}sin(Kx-\omega t).$ The quantity that is independent of wavelength is:

• A. $k\omega$
• B. $\frac{k}{\omega }$
• C. $k^2\omega$
• D. $\omega$

Answer 1

Answer: (B)

$\frac{k}{\omega }$

To find: The quantity that is independent of the wavelength.

The angular frequency $\omega$ is given by:

$\omega =2\pi \nu$

The frequency of a wave varies with the wavelength. So, angular frequency is dependent on wavelength.

The quantity $k$ is defined as the wavenumber and it is given by:

$k=\frac{2\pi }{\lambda }$

It shows that $k$ is dependent on wavelength.

The value of $\frac{k}{\omega }$ can be obtained as:

$\frac{k}{\omega }=\frac{2\pi /\lambda }{2\pi \nu }⟹\frac{1}{\nu \lambda }=\frac{1}{c}(\because c=\nu \lambda )$
where c is the speed of electromagnetic wave in vacuum. It is a constant whose value is $3\times {10}^{8}m{s}^{-1}$.

So, option $\left(B\right)$ is correct.