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

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

$\frac{ε_{r_{1}}}{ε_{r_{2}}} = 4$

$\frac{ε_{r_{1}}}{ε_{r_{2}}} = 2$

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

Solution

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

Velocity of EM wave is given by v = $\frac{1}{\sqrt{μ ε}}$

Velocity in air = $\frac{ω}{k} = C$

velocity in medium = C/2

Here, μ1 = μ2 =1 as medium is non magnetic

$∴ \frac{\frac{1}{\sqrt{ε_{r_{1}}}}}{\frac{1}{\sqrt{ε_{r_{2}}}}} = \frac{C}{(\frac{C}{2})} = 2 \Rightarrow \frac{ε_{r_{1}}}{ε_{r_{2}}} = \frac{1}{4}$

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