MCQ
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 $\overrightarrow {{E_2}}  = {E_{02}}\hat x\;cos\left[ {k\left( {2z - ct} \right)} \right]$ in medium, where the wave number $k$ and frequency $v$ refer to their values in air. The medium is nonmagnetic. If $\varepsilon {_{{r_1}}}$ and $\varepsilon {_{{r_2}}}$ refer to relative permittivities of air and medium respectively, which of the following options is correct?
  • A
    $\frac{{{_{{\in r_1}}}}}{{{_{{\in r_2}}}}} = 2$
  • $\frac{{{_{{\in r_1}}}}}{{{_{{\in r_2}}}}} = \frac{1}{4}$
  • C
    $\frac{{{_{{\in r_1}}}}}{{{_{{\in r_2}}}}} = \frac{1}{2}$
  • D
    $\frac{{{_{{\in r_1}}}}}{{{_{{\in r_2}}}}} = 4$

Answer

Correct option: B.
$\frac{{{_{{\in r_1}}}}}{{{_{{\in r_2}}}}} = \frac{1}{4}$
b
Velocity of $EM$ wave is given by $\mathrm{v}=\frac{1}{\sqrt{\mu \in}}$

Velocity in air $=\frac{\omega}{\mathrm{k}}=\mathrm{C}$

Velocity in medium $=\frac{\mathrm{C}}{2}$

Here, $\mu_{1}=\mu_{2}=1$ as medium is non-magnetic

$\therefore \frac{\sqrt{\in_{\eta}}}{\frac{1}{\sqrt{\in_{\mathrm{r}_{2}}}}}=\frac{\mathrm{C}}{\left(\frac{\mathrm{C}}{2}\right)}=2 \quad \Rightarrow \quad \frac{\in_{\mathrm{r}_{1}}}{\in_{\mathrm{r}_{2}}}=\frac{1}{4}$

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