An electromagnetic wave,going through vacuum, | vedclass

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Question

(C) The angular wave number is given by $k = \frac{2\pi}{\lambda}$,where $\lambda$ is the wavelength.The angular frequency is given by $\omega = 2\pi\

Vedclass · Accepted Answer

$k/\omega$

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(C) The angular wave number is given by $k = \frac{2\pi}{\lambda}$,where $\lambda$ is the wavelength.The angular frequency is given by $\omega = 2\pi
u$,where $
u$ is the frequency.The ratio $\frac{k}{\omega} = \frac{2\pi/\lambda}{2\pi
u} = \frac{1}{\lambda
u}$.Since the speed of light in vacuum is $c = \lambda
u$,we have $\frac{k}{\omega} = \frac{1}{c}$.Since $c$ is a constant,the ratio $\frac{k}{\omega}$ is independent of the wavelength $\lambda$.

An electromagnetic wave,going through vacuum,is described by $E = E_0 \sin(kx - \omega t)$. Which of the following is independent of wavelength?

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