Question

An electromagnetic wave of frequency $f=3.0MHz$ passes from vacuum into a dielectric medium with permittivity $\epsilon =4.0$. Then

• A. Wavelength is doubled and the frequency remains unchanged
• B. Wavelength is doubled and frequency becomes half
• C. Wavelength is halved and frequency remains unchanged
• D. Wavelength and frequency both remain unchanged

Answer 1

The correct option is C

Wavelength is halved and frequency remains unchanged

Step 1. Given data:

$f=3MHz$, Frequency of electromagnetic waves

${\epsilon }_r=4$, Permittivity of the medium

${\mu }_r=1$, Permeability of the medium [Almost equal to $1$.]

$c=\frac{1}{\sqrt{{\epsilon }_o{\mu }_o}}$, Speed of an electromagnetic wave n vacuum

$v=\frac{1}{\sqrt{{\epsilon }_o{\mu }_o{\epsilon }_r{\mu }_r}}$, Speed of the electromagnetic wave in the medium

The frequency of an electromagnetic wave is decided by the source. Hence, it does not change on changing medium.

Step 2. Calculating speed of waves in the medium:

$v=\frac{1}{\sqrt{{\epsilon }_o{\mu }_o{\epsilon }_r{\mu }_r}}$

$\Rightarrow$$v=\frac{1}{\sqrt{{\epsilon }_o{\mu }_o}\times \sqrt{{\epsilon }_r{\mu }_r}}$

$$\begin{gathered} \Rightarrow v=\frac{c}{\sqrt{4\times 1}} \\ \therefore v=\frac{c}{2} \end{gathered}$$

Step 3. Calculating wavelength of wave in the medium.

We know that, $Speed\left(v\right)=frequency\left(f\right)\times wavelength\left(\lambda \right)$

$\begin{array}{rcl}& \Rightarrow & \frac{{\lambda }_{medium}}{{\lambda }_{air}}=\frac{v}{c}\\ & \Rightarrow & \frac{{\lambda }_{medium}}{{\lambda }_{air}}=\frac{{\displaystyle \raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{c}\\ & \Rightarrow & \frac{{\lambda }_{medium}}{{\lambda }_{air}}=\frac{1}{2}\\ \therefore {\lambda }_{medium}& =& \frac{1}{2}\times {\lambda }_{air}\end{array}$

The frequency of electromagnetic waves remains unchanged and wavelength is halved.

Hence, correct option is (C).