Question

the poynting vector of an electromagnetic wave in vacuum is s+{(120 w /m^2)sin^2[(8.0 rad/m)z+(2.4*10^9rad/s )t]}k. what is the wavelength

Answer 1

Answer: $\lambda = 0.785 m$

Explanation:

Given that,

The poynting vector of an electromagnetic wave in vacuum is

$s = {(120 w /m^2)sin^2[(8.0 rad/m)z+(2.4*10^9rad/s )t]}k.$

Wave number = 8.0 rad/m

We know that,

The wave length is

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

$\lambda = \dfrac{2\pi}{8.0}$

$\lambda = 0.785 m$

Hence, the wave length is 0.785 m.

Answer 2

"Given:

From the given equation of pointing vector of an electromagnetic wave in vacuum,

$s+ {(120 \frac { w }{ m^{ 2 } }) sin^2[(8.0 \frac {rad}{m}) z+(2.4\times10^9 \frac {rad}{s}) t]} k$

Wave number $= 8.0 \frac {rad}{m}$

Solution:

The wave length is given by the following formula:

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

Where, \lambda is the wave number.

On substituting the value of wave number,

$\lambda = \frac {2\pi} {\8.0}$

Thus, the value of $\lambda$ is 0.785 m"