A plane polarized monochromatic EM wave is traveling in vacuum along z direction such that at t = t1 it is found that the electric field is zero at a spatial point z1. The next zero that occurs in its neighbourhood is at z2. The frequency of the electroagnetic wave is :

1.5 \times 10^8 \over | z_2 - z_1 |

3 \times 10^8 \over | z_2 - z_1 |

6 \times 10^8 \over | z_2 - z_1 |

1 \over t_1 + | z_2 - z_1 | \over 3 \times 10^8

Electromagnetic Waves PYQ — JEE Main 2018 · 15 Apr · Shift 2 · Q46 with Solution

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Correct answer: B

A monochromatic plane polarized EM wave traveling along the $+z$ direction can be written as

$E(z,t)=E_0\sin(kz-\omega t+\phi)$

where:

At time $t=t_1$ , the electric field is zero at $z=z_1$ :

$E(z_1,t_1)=0$

So,

$\sin(kz_1-\omega t_1+\phi)=0$

This means

$kz_1-\omega t_1+\phi=n\pi$

for some integer $n$ .

The next zero in its neighbourhood occurs at $z=z_2$ , so:

$kz_2-\omega t_1+\phi=(n+1)\pi$

Subtracting the two equations,

$k(z_2-z_1)=\pi$

Hence,

$|z_2-z_1|=\frac{\pi}{k}$

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

$|z_2-z_1|=\frac{\pi}{2\pi/\lambda}=\frac{\lambda}{2}$

So,

$\lambda=2|z_2-z_1|$

In vacuum,

$f=\frac{c}{\lambda}$

with $c=3\times 10^8\ \text{m/s}$ . Therefore,

$f=\frac{3\times 10^8}{2|z_2-z_1|}$

$f=\frac{1.5\times 10^8}{|z_2-z_1|}$

This corresponds to:

$\boxed{\text{B: } \frac{1.5\times 10^8}{|z_2-z_1|}}$

Stored correct answer is B, which matches the derived result.