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

For an electromagnetic plane wave travelling in the +z-direction the standard form of the electric field is

$$\vec E = E_0\hat x\cos\!\bigl(kz-\omega t\bigr),$$

where the angular frequency $$\omega$$ and the wave-number $$k$$ are related to the phase velocity $$v$$ through the well-known formula

$$k=\frac{\omega}{v}\qquad\bigl(\text{since }v=\frac{\omega}{k}\bigr).$$

In a non-magnetic medium (that is, $$\mu_r=1$$) the phase velocity is connected with the relative permittivity by

$$v=\frac{c}{n},\qquad\; n=\sqrt{\epsilon_r},$$

so that

$$k=\frac{\omega}{v}= \frac{\omega n}{c}=\frac{\omega}{c}\sqrt{\epsilon_r}.$$

Now let us compare the two fields given in the problem.

Wave in air

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

Writing $$\omega=2\pi\nu$$ we get

$$\vec{E_1}=E_{01}\hat x\cos\!\left(\frac{\omega z}{c}-\omega t\right) =E_{01}\hat x\cos\!\bigl(kz-\omega t\bigr),$$

where

$$k=\frac{\omega}{c}.$$

This is exactly the expected form with

$$k_1=k,\qquad\omega_1=\omega.$$

Wave in the dielectric medium

$$\vec{E_2}=E_{02}\hat x\cos\!\bigl[k(2z-ct)\bigr].$$

Expanding the phase term:

$$k(2z-ct)=2kz-kct.$$

Because $$k=\dfrac{\omega}{c}$$, the time-dependent part becomes

$$-kct=-\bigl(\tfrac{\omega}{c}\bigr)ct=-\omega t.$$

Hence

$$\vec{E_2}=E_{02}\hat x\cos\!\bigl(2kz-\omega t\bigr).$$

Therefore the effective wave-number inside the medium is

$$k_2=2k.$$

Relating the two wave-numbers

The frequency does not change on refraction, so $$\omega_2=\omega_1=\omega.$$ From the relation $$k=\dfrac{\omega}{c}\sqrt{\epsilon_r},$$ we can write for air and for the medium

$$k_1=\frac{\omega}{c}\sqrt{\epsilon_{r_1}}, \qquad k_2=\frac{\omega}{c}\sqrt{\epsilon_{r_2}}.$$

Taking the ratio and using $$k_2=2k_1$$ obtained above, we have

$$\frac{k_2}{k_1}=\frac{\dfrac{\omega}{c}\sqrt{\epsilon_{r_2}}}{\dfrac{\omega}{c}\sqrt{\epsilon_{r_1}}} =\frac{\sqrt{\epsilon_{r_2}}}{\sqrt{\epsilon_{r_1}}}=2.$$

Now squaring both sides:

$$\frac{\epsilon_{r_2}}{\epsilon_{r_1}}=4.$$

Inverting the fraction to obtain the ratio asked in the options:

$$\frac{\epsilon_{r_1}}{\epsilon_{r_2}}=\frac{1}{4}.$$

Hence, the correct answer is Option D.