Question

A plane polarized monochromatic EM wave is traveling a vacuum along Z direction such that at t=t${}_1$ it is found that at the electric field is zero at a spatial point z${}_1$The next zero that occurs in its neighbourhood is at z${}_2$The frequency of the electromagnetic wave is :


$\begin{array}{rl}& A.{\displaystyle \frac{3\times {10}^8}{|Z_2-Z_1|}}\\ & B.{\displaystyle \frac{6\times {10}^8}{|Z_2-Z_1|}}\\ & C.{\displaystyle \frac{1.5\times {10}^8}{|Z_2-Z_1|}}\\ & D.{\displaystyle \frac{1}{t_1+{\displaystyle \frac{|Z_2-Z_1|}{3\times {10}^8}}}}\end{array}$

Answer 1

Hint: We need to figure out the electric field at both time T, as both the values of E=0 at time t${}_1$ and t${}_2$, hence we can compare them to each other, now we represent the equation in such a way that it shows time t in terms of point z and speed of light.

Complete step-by-step answer:
As of the question we know that,
When time t= t${}_1$and at point z${}_1$ we are getting electric field E=0,
So, let us assume that, when time t= t${}_2$, the wave is at point z${}_2$, and E=0.
Now if we consider the equation of electric field,
$E=E_{\circ }{e}^{-(kz-\omega t)}$ .$E_{\circ }$is a constant,$\omega$ is the angular frequency, k is the wave number.
Now placing the value of point Z${}_1$ at time t${}_1$,
$E=0=E_{\circ }{e}^{-(k{z}_1-\omega t_1)}$………. Eq.1
Now placing the value of point Z${}_2$ at time t${}_2$,
$E=0=E_{\circ }{e}^{-(k{z}_2-\omega t_2)}$……….. Eq.2
On comparing eq.1 and eq.2 we get,
$E_{\circ }{e}^{-(k{z}_1-\omega t_1)}=E_{\circ }{e}^{-(k{z}_2-\omega t_2)}$
Here $E_{\circ }$ cancels out each other,
And as in both the sides base value is same hence we can write,
$$-k{z}_1-\omega t_1=-k{z}_2-\omega t_2$$

,

On further solving we get,
$z_1-z_2={\displaystyle \frac{\omega }{k}}(t_1-t_2)$ ……… eq.3
Where $\omega =2\pi f$,
And $k={\displaystyle \frac{2\pi }{\lambda }}$ ,
Now putting the value of $\omega$ and k in eq.3,
$z_1-z_2={\displaystyle \frac{2\pi f\lambda }{2\pi }}(t_1-t_2)$,
We know that $c=f\lambda$ where c is the speed of light,
$z_1-z_2=c(t_1-t_2)$,
We also know that frequency is inversely proportional to time,
Hence, we can represent the following equation in,
${\displaystyle \frac{z_1-z_2}{t_1-t_2}}=c$,
Or,
${\displaystyle \frac{1}{t_1-t_2}}={\displaystyle \frac{c}{z_1-z_2}}$,
As said earlier, that frequency is inversely proportional to time so,
We can write that,
$f={\displaystyle \frac{c}{z_1-z_2}}$,
Speed of light is, $3\times {10}^{8}m/s$
$f={\displaystyle \frac{3\times {10}^8}{z_1-z_2}}$(Answer).
We see that our equation does not matches with any of the equation in the options, so we can apply mod operator,
$f={\displaystyle \frac{3\times {10}^8}{|z_1-z_2|}}$,
Then ,
$f={\displaystyle \frac{3\times {10}^8}{|z_2-z_1|}}$,
Therefore option A is the correct option.
Note: In the equation $E=E_{\circ }{e}^{-(kz-\omega t)}$, z is the direction of propagation, frequency is inversely proportional to time, and $\omega$ is the angular frequency, and k is the wave number, c is the speed of light.