An EM wave from air enters a medium. electric fields are overrightarrow {{E₁}} = {E

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8.Electromagnetic waves

medium

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?

$\frac{{{_{{\epsilon r_1}}}}}{{{_{{\epsilon r_2}}}}} = 2$

$\frac{{{_{{\epsilon r_1}}}}}{{{_{{\epsilon r_2}}}}} = \frac{1}{4}$

$\frac{{{_{{\epsilon r_1}}}}}{{{_{{\epsilon r_2}}}}} = \frac{1}{2}$

$\frac{{{_{{\epsilon r_1}}}}}{{{_{{\epsilon r_2}}}}} = 4$

(JEE MAIN-2018)

Solution

Velocity of $EM$ wave is given by $\mathrm{v}=\frac{1}{\sqrt{\mu \epsilon}}$

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

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

Standard 12

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Solution

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