Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# 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{align} & A.\dfrac{3\times {{10}^{8}}}{|{{Z}_{2}}-{{Z}_{1}}|} \\ & B.\dfrac{6\times {{10}^{8}}}{|{{Z}_{2}}-{{Z}_{1}}|} \\ & C.\dfrac{1.5\times {{10}^{8}}}{|{{Z}_{2}}-{{Z}_{1}}|} \\ & D.\dfrac{1}{{{t}_{1}}+\dfrac{|{{Z}_{2}}-{{Z}_{1}}|}{3\times {{10}^{8}}}} \\ \end{align}

Last updated date: 18th Sep 2024
Total views: 80.7k
Views today: 1.80k
Answer
Verified
80.7k+ views
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.