An EM wave from air enters a medium. The electric fields are $\vec{E_{1}} = E_{01} \hat{x} \cos [2 π ν (\frac{z}{c} - t)]$ in air and $\vec{E_{2}} = E_{02} \hat{x} \cos [k (2 z - ct)]$ in medium, where the wave number $k$ and frequency $ν$ refer to their values in air. The medium is non-magnetic. If $ε_{r_{1}}$ and $ε_{r_{2}}$ refer to relative permittivity of air and medium respectively, which of the following options is correct?

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$\frac{ε_{r_{1}}}{ε_{r_{2}}} = 4$

$\frac{ε_{r_{1}}}{ε_{r_{2}}} = 2$

$\frac{ε_{r_{1}}}{ε_{r_{2}}} = \frac{1}{4}$

$\frac{ε_{r_{1}}}{ε_{r_{2}}} = \frac{1}{2}$

answer is C.

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Detailed Solution

In air, EM wave is $\vec{E_{1}} = E_{01} \hat{x} \cos [2 π ν (\frac{z}{c} - t)] = E_{01} \hat{x} \cos [k (z - ct)] (∵ k = \frac{2 π}{λ_{0}} = \frac{2 πν}{c})$

In medium, EM wave is $\vec{E_{2}} = E_{02} \hat{x} \cos [k (2 z - ct)] = E_{02} \hat{x} \cos [2 k (z - \frac{c}{2} t)]$

During refraction, frequency remains unchanged, whereas wavelength gets changed.

$∴ k' = 2 k$ (From equations)

$\frac{2 π}{λ'} = 2 (\frac{2 π}{λ_{0}})$ or $λ' = \frac{λ_{0}}{2}$

Since,

$ν = \frac{c}{2} \Rightarrow \frac{1}{\sqrt{μ_{0} ε_{0} ε_{r_{2}}}} = \frac{1}{2} \times \frac{1}{\sqrt{μ_{0} ε_{0} ε_{r_{1}}}} ∴ \frac{ε_{r_{1}}}{ε_{r_{2}}} = \frac{1}{4}$