MCQ
An electromagnetic wave going through vacuum is described by $\text{E}=\text{E}_0\sin(\text{kx}-\omega\text{t})$ which of the following is/are independent of the wavelength?
- A$\text{k}$
- B$\omega$
- ✓$\frac{\text{k}}{\omega}$
- D$\text{k}\omega$
✓
Answer
Correct option: C.
$\frac{\text{k}}{\omega}$
The given quantities can be expressed as,
k is given by,
$\omega$ is given by,
$\omega=2\pi\text{v}$
$\text{c}=\text{v}\lambda$
$\omega=2\pi\frac{\text{c}}{\lambda}$
$\frac{\text{k}}{\omega}$ is given by,
$\frac{\text{k}}{\omega}=\frac{\frac{2\pi}{\lambda}}{\frac{2\pi\text{c}}{\lambda}}=\frac{1}{\text{c}}$
$\text{k}\omega$ is given by,
$\text{k}\times\omega=\frac{2\pi}{\lambda}\times\frac{2\pi\text{c}}{\lambda}=\frac{4\pi^2\text{c}}{\lambda^2}$
Thus, $\frac{\text{k}}{\omega}$ is independent of the wavelength.
k is given by,
$\omega$ is given by,
$\omega=2\pi\text{v}$
$\text{c}=\text{v}\lambda$
$\omega=2\pi\frac{\text{c}}{\lambda}$
$\frac{\text{k}}{\omega}$ is given by,
$\frac{\text{k}}{\omega}=\frac{\frac{2\pi}{\lambda}}{\frac{2\pi\text{c}}{\lambda}}=\frac{1}{\text{c}}$
$\text{k}\omega$ is given by,
$\text{k}\times\omega=\frac{2\pi}{\lambda}\times\frac{2\pi\text{c}}{\lambda}=\frac{4\pi^2\text{c}}{\lambda^2}$
Thus, $\frac{\text{k}}{\omega}$ is independent of the wavelength.
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