An EM wave from air enters a medium. The electric fields are E_ 1 =E_ 01 x [2 v ( z c -t…

Physics · JEE Main · NTA ExamsElectromagnetic Waves

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\(\overrightarrow{\mathrm{E}_{2}}=\mathrm{E}_{02}\) \(\hat{\mathrm{x}} \cos [\mathrm{k}(2 \mathrm{z}-\mathrm{ct})]\) in medium, where the wave number k and frequency v refer to their values in air. The medium is non-magnetic. If \(\epsilon_{\mathrm{r}_{1}} \text { and } \epsilon_{\mathrm{r}_{2}}\)refer to relative permittivities of air and medium respectively, which of the following options is correct ?
  1. \(\frac{\epsilon_{\mathrm{n}_{1}}}{\epsilon_{\mathrm{r}_{2}}}=\frac{1}{4}\)
  2. \(\frac{\epsilon_{\mathrm{r}_{1}}}{\epsilon_{\mathrm{r}_{2}}}=\frac{1}{2}\)
  3. \(\frac{\epsilon_{\mathrm{r}_{1}}}{\epsilon_{\mathrm{r}_{2}}}=4\)
  4. \(\frac{\epsilon_{\mathrm{r}_{1}}}{\epsilon_{\mathrm{r}_{2}}}=2\)

Answer

(C) _ r _ 1 _ r _ 2 =4

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