Question

An electromagnetic wave of frequency $3GHz$ enters a dielectric medium of relative electric permittivity $2.25$ from vacuum. The wavelength of this wave in that medium will be _______ $\times {10}^{-2}cm$.

Answer 1

Step 1. Given data:

${\epsilon }_r=2.25$

${\mu }_r=1$, Assuming non-magnetic material

$f=3GHz$, Frequency of electromagnetic waves

$f=3\times {10}^{9}Hz$

$c=\frac{1}{\sqrt{{\mu }_o{\epsilon }_o}}=3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$, Speed of light

$\lambda =?$, Wavelength of the wave in the medium

Step 2. Calculating wavelength of the wave in the medium:

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

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

$\Rightarrow$$v=\frac{c}{\sqrt{{\epsilon }_r}}$ $\left[c=\frac{1}{\sqrt{{\mu }_o{\epsilon }_o}}\&{\mu }_r=1\right]$

Putting all values in above equation, we get

$\begin{array}{rcl}& \Rightarrow & v=\frac{3\times {10}^8}{\sqrt{2.25}}\\ & \Rightarrow & v=\frac{3\times {10}^8}{1.5}\\ \therefore v& =& 2\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

Now, As we know $v=f\lambda$

$\begin{array}{rcl}v& =& f\lambda \\ & \Rightarrow & 2\times {10}^8=3\times {10}^9\times \lambda \\ & \Rightarrow & \lambda =\frac{2\times {10}^8}{3\times {10}^9}\\ & \Rightarrow & \lambda =0.0667m\\ \therefore \lambda & =& 667\times {10}^{-2}cm\end{array}$

Hence, the wavelength of the wave in the medium is $\begin{array}{rcl}\lambda & =& 667\times {10}^{-2}cm\end{array}$