Question

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

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

Answer 1

The correct option is D $\frac{k}{\omega }$

We know that:
Wave number, $k=\frac{2\pi }{\lambda }$
$\text{and}\omega =\frac{2\pi c}{\lambda }$

So, both $\omega \text{and}k$ depends upon the wavelength of the wave.

Now, $\frac{k}{\omega }=\frac{2\pi }{\lambda }\times \frac{\lambda }{2\pi c}$

$\Rightarrow \frac{k}{\omega }=\frac{1}{c}$

Also, $k\omega =\frac{2\pi }{\lambda }\times \frac{2\pi c}{\lambda }$

$\Rightarrow k\omega =\frac{(2\pi {)}^{2}c}{{\lambda }^2}$

$\therefore$ Only $\frac{k}{\omega }$ is independent of the wavelength of the wave.

Hence, option $\left(\text{C}\right)$ is correct.