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**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,

${{z}_{1}}-{{z}_{2}}=\dfrac{\omega }{k}({{t}_{1}}-{{t}_{2}})$ ……… eq.3

Where $\omega =2\pi f$,

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

Now putting the value of $\omega$ and k in eq.3,

${{z}_{1}}-{{z}_{2}}=\dfrac{2\pi f\lambda }{2\pi }({{t}_{1}}-{{t}_{2}})$,

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,

$\dfrac{{{z}_{1}}-{{z}_{2}}}{{{t}_{1}}-{{t}_{2}}}=c$,

Or,

$\dfrac{1}{{{t}_{1}}-{{t}_{2}}}=\dfrac{c}{{{z}_{1}}-{{z}_{2}}}$,

As said earlier, that frequency is inversely proportional to time so,

We can write that,

$f=\dfrac{c}{{{z}_{1}}-{{z}_{2}}}$,

Speed of light is, $3\times {{10}^{8}}m/s$

$f=\dfrac{3\times {{10}^{8}}}{{{z}_{1}}-{{z}_{2}}}$(Answer).

We see that our equation does not matches with any of the equation in the options, so we can apply mod operator,

$f=\dfrac{3\times {{10}^{8}}}{|{{z}_{1}}-{{z}_{2}}|}$,

Then ,

$f=\dfrac{3\times {{10}^{8}}}{|{{z}_{2}}-{{z}_{1}}|}$,

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.

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