Question: An electromagnetic wave with frequency $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{14}}\mathbf{}\mathbf{Hz}$travels through vacuum in the positive direction of an xaxis. The wave has its electric field oscillating parallel to the y axis, with an amplitude ${\mathbf{E}}_{\mathbf{m}}$. At time $\mathbf{t}\mathbf{=}\mathbf{0}$, the electric field at point Pon the xaxis has a value of $\mathbf{+}\frac{{\mathbf{E}}_{\mathbf{m}}}{\mathbf{4}}$and is decreasing with time. What is the distance along the xaxis from point Pto the first point with$\mathbf{E}\mathbf{=}\mathbf{0}$if we search in (a) the negative direction and (b) the positive direction of the x axis?

The frequency of the electromagnetic wave is$\mathrm{f}=4.00\times {10}^{14}\mathrm{Hz}$.

The amplitude of the electric field along they-axis is${\mathrm{E}}_{\mathrm{m}}$.

The value of the electric field at the point P on the x-axis at the time $\mathrm{t}=0$is ${\mathrm{E}}_{\mathrm{y}}=+\frac{{\mathrm{E}}_{\mathrm{m}}}{4}$.

We can use the concept of the electric field as sinusoidal functions of position x and time t.

Formula:

${\mathrm{E}}_{\mathrm{y}}={\mathrm{E}}_{\mathrm{m}}\sin \left(\mathrm{kx}-\mathrm{\omega t}\right)$

$\mathrm{k}=\frac{2\mathrm{\pi f}}{\mathrm{c}}$

$c=f\lambda$

The expression of the electric field as sinusoidal functions of position x and time t is:

${\mathrm{E}}_{\mathrm{y}}={\mathrm{E}}_{\mathrm{m}}\sin \left(\mathrm{kx}-\mathrm{\omega t}\right)$

The value of the electric field at a point P on the x-axis at the time $\mathrm{t}=0$is ${\mathrm{E}}_{\mathrm{y}}=+\frac{{\mathrm{E}}_{\mathrm{m}}}{4}$, then the above equation becomes as:

$+\frac{{\mathrm{E}}_{\mathrm{m}}}{4}={\mathrm{E}}_{\mathrm{m}}sin\left(\mathrm{kx}\right)$

$\frac{1}{4}=sin\left(\mathrm{kx}\right)$

The distance along x axis from the point P to the first point with$\mathrm{E}=0$if we search in the negative direction of thex-axis is:

${\mathrm{x}}_{\mathrm{P}}=\frac{1}{\mathrm{k}}\left(\frac{1}{4}\right).........\left(1\right)$

The expression of the velocity and the wave number is:

$\mathrm{k}=\frac{2\mathrm{\pi f}}{\mathrm{c}}$

The equation (1) becomes as:

$x_P=\frac{c}{2\mathrm{\pi f}}\left(\frac{1}{4}\right)$

$\begin{array}{rcl}{\mathrm{x}}_{\mathrm{P}}& =& \frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{2\times 3.14\times 4.00\times {10}^{14}\mathrm{Hz}}\times \frac{1}{4}\\ & =& 30.1\times {10}^{-9}\mathrm{m}\times \frac{1\mathrm{nm}}{{10}^{-9}\mathrm{m}}\\ & =& 30.1nm\end{array}$

Along the right side of thex-axis, we can find another point where${\mathrm{E}}_{\mathrm{y}}=0$at a distance of one-half wavelengths from the last point where ${\mathrm{E}}_{\mathrm{y}}=0$.

$x=\frac{1}{2}\lambda$

The expression of the relationship between velocity, frequency, and wavelength of the electromagnetic wave is

role="math" localid="1663004119856" $$\begin{gathered} c=f\lambda \\ \lambda =\frac{c}{f} \\ x=\frac{1}{2}\times \frac{c}{f} \end{gathered}$$

Hence, the distance is

$$\begin{gathered} x\text{'}=x-x_P \\ x\text{'}=\frac{c}{2f}-x_P \end{gathered}$$

$\begin{array}{rcl}\mathrm{x}\text{'}& =& \left(\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{2\times 4.00\times {10}^{14}\mathrm{Hz}}\right)-30.1\times {10}^{-9}\mathrm{m}\\ & =& 345\times {10}^{-9}\mathrm{m}\times \frac{1\mathrm{nm}}{{10}^{-9}\mathrm{m}}\\ & =& 345\mathrm{nm}\end{array}$