Question

An EM wave from the air enters a medium. The electric field are
$\overrightarrow{E_1}=E_{01}\hat{x}\cos \left[2\pi f(\frac{z}{c}-t)\right]$
in air and $\overrightarrow{E_2}=E_{02}\hat{x}\cos \left[k(2z-ct)\right]$

in medium, where the wave number $k$ and frequency $f$ refer to the values in air. The medium is non-magnetic. If ${ϵ}_{r1}\text{and}{ϵ}_{r2}$ refer to relative permittivities of air and the medium, which of the following options is correct?

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

Answer 1

The correct option is A $\frac{{ϵ}_{r1}}{{ϵ}_{r2}}=4$

In air, the equation of EM wave is,
$\overrightarrow{E_1}=E_{01}\hat{x}\cos \left[2\pi f(\frac{z}{c}-t)\right]$

$\overrightarrow{E_1}=E_{01}\hat{x}\cos \left[k(z-ct)\right](\because k=\frac{2\pi }{{\lambda }_0}=\frac{2\pi f}{c})$

In the medium, the equation of EM wave is,
$\overrightarrow{E_2}=E_{02}\hat{x}\cos \left[k(2z-ct)\right]$

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

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

From the above two equations we can say that velocity of EM wave in the two media is,
$c_a=c\text{and}{c}_m=\frac{c}{2}$

$\Rightarrow \frac{c_a}{c_m}=\frac{1}{2}$

$\frac{1}{\sqrt{{\mu }_0{ϵ}_{r1}}}=\frac{1}{2}\times \frac{1}{\sqrt{{\mu }_0{ϵ}_{r2}}}$

Since, the material is non-magnetic the value of ${\mu }_0$ is same for both air and the medium.
$\Rightarrow \frac{1}{{ϵ}_{r1}}=\frac{1}{4}\times \frac{1}{{ϵ}_{r2}}$

$\therefore \frac{{ϵ}_{r1}}{{ϵ}_{r2}}=4$

Hence, $\left(A\right)$ is the correct answer.