Question

An $EM$ wave from air enters a medium. The electric fields are $\overrightarrow{E_1}=E_{01}\hat{x}\cos \left[2\pi v\left(\frac{z}{c}-t\right)\right]$ 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 $v$ refer to their values in air. The medium is non-magnetic. If ${ϵ}_{r_1}$ and ${ϵ}_{r_2}$ refer to relative permittivities of air and medium respectively, which of the following options is correct?

• A. $\frac{{ϵ}_{r_1}}{{ϵ}_{r_2}}=\frac{1}{4}$
• B. $\frac{{ϵ}_{r_1}}{{ϵ}_{r_2}}=\frac{1}{2}$
• C. $\frac{{ϵ}_{r_1}}{{ϵ}_{r_2}}=4$
• D. $\frac{{ϵ}_{r_1}}{{ϵ}_{r_2}}=2$

Answer 1

The correct option is A $\frac{{ϵ}_{r_1}}{{ϵ}_{r_2}}=\frac{1}{4}$

Velocity of wave $v=\frac{w}{K}$
Comparing the given equations with $E=E_o\cos (Kz-wt)$, we get
For air : $K=\frac{2\pi \nu }{c}$ $w=2\pi \nu$
$⟹v_{air}=\frac{w}{K}=c$
For medium : $K=2k$ $w=kc$
$⟹v_{medium}=\frac{w}{K}=\frac{kc}{2k}=\frac{c}{2}$
Using Snell's law, $\frac{n_{air}}{n_{medium}}=\frac{v_{medium}}{v_{air}}$
where $ϵ\propto n^2$
$\therefore \frac{{ϵ}_{air}}{{ϵ}_{medium}}=\frac{n_{air}^2}{n_{medium}^2}=\frac{v_{medium}^2}{v_{air}^2}$
$\therefore \frac{{ϵ}_{air}}{{ϵ}_{medium}}=\frac{(c/2{)}^2}{c}=\frac{1}{4}$