JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2018
- Medium
- 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?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\frac{{{_{{\epsilon r_1}}}}}{{{_{{\epsilon r_2}}}}} = \frac{1}{4}\)
Step-by-step Solution
Detailed explanation
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
\(\therefore \frac{\sqrt{\epsilon_{\eta}}}{\frac{1}{\sqrt{\epsilon_{\mathrm{r}_{2}}}}}=\frac{\mathrm{C}}{\left(\frac{\mathrm{C}}{2}\right)}=2 \quad \Rightarrow \quad \frac{\epsilon_{\mathrm{r}_{1}}}{\epsilon_{\mathrm{r}_{2}}}=\frac{1}{4}\)
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