An EM wave from air enters a medium. The electric fields are $\vec E_1 = E_{01}\hat x\cos\left[2\pi\nu\left(\dfrac zc - t\right)\right]$ in air and $\vec E_2 = E_{02}\hat x\cos[k(2z - ct)]$ in medium, where the wave number $k$ and frequency $\nu$ refer to their values in the air. The medium is non-magnetic. If $\varepsilon_{r_1}$ and $\varepsilon_{r_2}$ refer to relative permittivities of air and medium respectively, which of the following options is correct?
Answer: (D) $\dfrac{\varepsilon_{r_1}}{\varepsilon_{r_2}} = \dfrac14$
In the medium the phase is $2kz - kct$: the angular frequency is unchanged ($kc = \omega$) but the wave number is doubled, so the speed is $\dfrac{\omega}{2k} = \dfrac c2$.
For a non-magnetic medium $v = \dfrac{c}{\sqrt{\varepsilon_r}}$, so $\varepsilon_{r_2} = 4\varepsilon_{r_1}$:
$$\frac{\varepsilon_{r_1}}{\varepsilon_{r_2}} = \frac14$$
Solution by Sreeraj P, M.Sc Physics