Question

The electric field of an electromagnetic wave travelling through 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. $\omega$
• E. $k^2\omega$

Answer 1

Answer: (B)

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