Question

An electromagnetic wave of frequency $v=3MHz$ passes from a vacuum into a dielectric medium with permittivity 4. What happens to the wavelength?

Answer 1

Step1: Given data

1. The frequency of the wave is $v=3MHz$.
2. the dielectric constant of the medium is $k=4$.

Step2: Velocity of light

1. From Maxwell's electromagnetic theory we know that the velocity of light in a medium, $c=\frac{1}{\sqrt{{\mu }_o{\epsilon }_o}}$, where, ${\mu }_o$ is the magnetic permeability and ${\epsilon }_o$ permittivity of the medium.
2. Also, velocity, $c=\nu \lambda$, where, $\nu$ is the frequency of light and $\lambda$ is the wavelength of light.

Step3: Finding the wavelength

As we know, the velocity of light in a medium is $v_{med}=\frac{1}{\sqrt{{\mu }_o\epsilon }}$.

So,

$$\begin{gathered} v_{med}=\frac{1}{\sqrt{{\mu }_{o}k{\epsilon }_o}}=\frac{1}{\sqrt{\left({\mu }_o\times 4\times {\epsilon }_o\right)}} \\ or{v}_{med}=\frac{1}{2}\times \frac{1}{\sqrt{{\mu }_o{\epsilon }_o}}\left(Since,\epsilon =k{\epsilon }_o\right) \\ or{v}_{med}=\frac{c}{2}\left(\sin ce,c=\frac{1}{\sqrt{{\mu }_o{\epsilon }_o}}\right) \end{gathered}$$

Again, we know the velocity, $c=\nu \lambda$

We know that frequency of light is constant in the medium.

Now,

$$\begin{gathered} \frac{{\lambda }_{med}}{{\lambda }_{air}}=\frac{{\displaystyle \frac{c}{2}}}{c}=\frac{1}{2} \\ or\frac{{\lambda }_{med}}{{\lambda }_{air}}=\frac{1}{2}. \end{gathered}$$

Therefore, the wavelength of the wave will be halved.