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  • An EM wave from the air enters a medium. The electric fields are <img alt="\mathrm{\vec{E}_1=E_{01} \hat{x} \cos \left[2 \pi f\left(\frac{z}{c}-t\right)\right]}" src="https://entrancecorner.on

An EM wave from the air enters a medium. The electric fields are \mathrm{\vec{E}_1=E_{01} \hat{x} \cos \left[2 \pi f\left(\frac{z}{c}-t\right)\right]} in air and \mathrm{\vec{E}_2=E_{02} \hat{x} \cos [k(2 z-c t)]} in medium, where the wavenumber k and frequency f refer to their values in air. The medium is
non-magnetic. If  \mathrm{\varepsilon_{\mathrm{r} 1} \text { and } \varepsilon_{\mathrm{r} 2}} refer to relative permittivities of air and medium respectively, which of the following options is correct?

Option: 1

\mathrm{\varepsilon_{\mathrm{r} 1} / \varepsilon_{\mathrm{r} 2}=4}


Option: 2

\mathrm{\varepsilon_{\mathrm{r} 1} / \varepsilon_{\mathrm{r} 2}=2}


Option: 3

\mathrm{\varepsilon_{\mathrm{r} 1} / \varepsilon_{\mathrm{r} 2}=1/4}


Option: 4

\mathrm{\varepsilon_{\mathrm{r} 1} / \varepsilon_{\mathrm{r} 2}=1/2}


Answers (1)

best_answer

In the air, the EM wave is

\begin{aligned} & \vec{E}_1=E_{01} \hat{x} \cos \left[2 \pi f\left(\frac{z}{c}-t\right)\right] =\vec{E}_2=E_{02} \hat{x} \cos [k(z-c t)] \end{aligned}

In the medium, the EM wave is

\begin{aligned} & \mathrm{\vec{E}_2=E_{02} \hat{x} \cos [k(2 z-c t)]} \\ \\& \mathrm{\vec{E}_2=E_{02} \hat{x} \cos [2 k(z-(c / 2) t)]} \end{aligned}

During refraction, frequency remains unchanged, whereas the wavelength gets changed k’ = 2k (From equations)

\begin{aligned} & 2 \pi / \lambda^{\prime}=2\left(2 \pi / \lambda_0\right)=\lambda^{\prime}=\lambda_0 / 2 \text { Since, } v=c / 2 \end{aligned}

\mathrm{\frac{1}{\sqrt{\mu_0 \epsilon_{r 2}}}=\frac{1}{2} \times \frac{1}{\sqrt{\mu_0 \epsilon_{r 1}}}}

\mathrm{\varepsilon_{\mathrm{r} 1} / \varepsilon_{\mathrm{r} 2}=1 / 4}

Posted by

Kuldeep Maurya

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