Question

A plane polarized monochromatic $EM $ wave is traveling in vacuum along $z$ direction such that at $t = t_1$ it is found that 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 :

• A. $\frac{3 \times 10^8}{|z_2 - z_1|}$
• B. \(\frac{1.5 \times 10^8}{|z_2 - z_1|}\)
• C. $\frac{6 \times 10^8}{|z_2 - z_1|}$
• D. $\frac{1}{t_1 + \frac{|z_2 - z_1|}{3 \times 10^8}} $

Answer 1

The Correct Option is B

To determine the frequency of a plane polarized monochromatic electromagnetic (EM) wave, we need to consider the relationship between the spatial zero points of the electric field. The electric field of a monochromatic plane wave can be expressed as:

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

where:

• \(E_0\) is the amplitude of the wave,
• \(k\) is the wave number,
• \(\omega\) is the angular frequency, and
• \(\phi\) is the phase constant.

Given that the electric field \(E(z, t)\) is zero at two points \(z_1\) and \(z_2\), the equation becomes zero for these positions:

\(\sin(kz_1 - \omega t_1 + \phi) = 0\), \(\sin(kz_2 - \omega t_1 + \phi) = 0\).

The condition for the sine function to be zero is when its argument is an integer multiple of \(\pi\), i.e.,

\(kz_1 - \omega t_1 + \phi = n\pi\), \(kz_2 - \omega t_1 + \phi = (n+1)\pi\).

Subtracting these equations gives:

\(k(z_2 - z_1) = \pi\).

Solving for \(k\):

\(k = \frac{\pi}{z_2 - z_1}\).

Since the wave number \(k\) is related to the wavelength \(\lambda\) by \(k = \frac{2\pi}{\lambda}\), we have:

\(\frac{2\pi}{\lambda} = \frac{\pi}{z_2 - z_1}\).

This implies the wavelength is:

\(\lambda = 2(z_2 - z_1)\).

Finally, the frequency \(f\) of the wave is given by the speed of light \(c = 3 \times 10^8 \text{ m/s}\) over the wavelength:

\(f = \frac{c}{\lambda} = \frac{3 \times 10^8}{2(z_2 - z_1)}\).

This simplifies to:

\(f = \frac{1.5 \times 10^8}{|z_2 - z_1|}\).

Thus, the correct answer is:

\(f = \frac{1.5 \times 10^8}{|z_2 - z_1|}\).