Question

An $EM$ wave from air enters a medium. The electric fields are $\overrightarrow{E_1}=E_{01}\hat{x}\cos \left[2\pi f(\frac{z}{c}-t)\right]$ in air and $\overrightarrow{E_2}=E_{02}\hat{x}\cos \left[k\right(2z-ct\left)\right]$ in medium, where the wave number $k$ and frequency $f$ refer to their values in air. The medium is non-magnetic. If ${\epsilon }_{r1}$ and ${\epsilon }_{r2}$ refer to relative permittivities of air and medium respectively, which of the following options is correct?

• A. $\frac{{\epsilon }_{r1}}{{\epsilon }_{r2}}=4$
• B. $\frac{{\epsilon }_{r1}}{{\epsilon }_{r2}}=2$
• C. $\frac{{\epsilon }_{r1}}{{\epsilon }_{r2}}=\frac{1}{4}$
• D. $\frac{{\epsilon }_{r1}}{{\epsilon }_{r2}}=\frac{1}{2}$

Answer 1

The correct option is C $\frac{{\epsilon }_{r1}}{{\epsilon }_{r2}}=\frac{1}{4}$

In the air, the electric field of $EM$ wave is

$\overrightarrow{E_1}=E_{01}\hat{x}\cos \left[2\pi f(\frac{z}{c}-t)\right]$

In medium, the electric field of $EM$ wave is

$\overrightarrow{E_2}=E_{02}\hat{x}\cos \left[k\right(2z-ct\left)\right]$

$=E_{02}\hat{x}\cos \left[\frac{2\pi }{{\lambda }_0}\right(2z-ct\left)\right](\because k=\frac{2\pi }{{\lambda }_0})$

$=E_{02}\hat{x}\cos \left[\frac{2\pi }{{\lambda }_0}c(\frac{2z}{c}-t)\right]$

$=E_{02}\hat{x}\cos \left[2\pi f(\frac{2z}{c}-t)\right]$

During refraction, frequency remains unchanged, whereas the wavelength gets changed.

From the above two equations, we can say that the velocity of $EM$ wave in the medium is,

$v=\frac{c}{2}\Rightarrow \frac{1}{\sqrt{{\mu }_0{\epsilon }_{r2}}}=\frac{1}{2}\times \frac{1}{\sqrt{{\mu }_0{\epsilon }_{r1}}}$

Since, the material is non-magnetic, the value of ${\mu }_0$ is the same for both air and the medium.

$\Rightarrow \frac{1}{{\epsilon }_{r2}}=\frac{1}{4{\epsilon }_{r1}}$

Or, $\frac{{\epsilon }_{r1}}{{\epsilon }_{r2}}=\frac{1}{4}$

Hence, option $\left(C\right)$ is correct.