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Physics Question on Electromagnetic waves

A plane polarized monochromatic EMEM wave is traveling in vacuum along zz direction such that at t=t1t = t_1 it is found that the electric field is zero at a spatial point z1z_1. The next zero that occurs in its neighbourhood is at z2z_2. The frequency of the electromagnetic wave is :

A

3×108∣z2−z1∣\frac{3 \times 10^8}{|z_2 - z_1|}

B

1.5×108∣z2−z1∣\frac{1.5 \times 10^8}{|z_2 - z_1|}

C

6×108∣z2−z1∣\frac{6 \times 10^8}{|z_2 - z_1|}

D

1t1+∣z2−z1∣3×108\frac{1}{t_1 + \frac{|z_2 - z_1|}{3 \times 10^8}}

Answer

1.5×108∣z2−z1∣\frac{1.5 \times 10^8}{|z_2 - z_1|}

Explanation

Solution

The correct option is(B): 1.5×108∣z2−z1∣\frac{1.5 \times 10^8}{|z_2 - z_1|}

Since c=fλ⇒f=cλc=f \lambda \Rightarrow f=\frac{c}{\lambda}.
Here, λ=2∣z2−z1∣\lambda=2\left|z_{2}-z_{1}\right| and c=3×108.c=3 \times 10^{8} . Therefore,
f=3×1082∣z2−z1∣=1.5×108∣z2−z1∣f=\frac{3 \times 10^{8}}{2\left|z_{2}-z_{1}\right|}=\frac{1.5 \times 10^{8}}{\left|z_{2}-z_{1}\right|}