Question

An EM wave from air enters a medium. The electric fields are ${\overrightarrow{E}}_1=E_{01}\hat{x}\cos \left[2\pi v(\frac{z}{c}-t)\right]$ in air and
${\overrightarrow{E}}_2=E_{02}\hat{x}\cos \left[k\right(2z-ct\left)\right]$ in medium, where the wave number $k$ and frequency v refer to their values in air. The medium is non-magnetic. If ${\epsilon }_{r_1}$ and ${\epsilon }_{r_2}$ refer to relative permitivities of air and medium respectively, which of the following options is correct?

• A. $\frac{{\epsilon }_{r_1}}{{\epsilon }_{r_2}}=\frac{1}{4}$
• B. $\frac{{\epsilon }_{r_1}}{{\epsilon }_{r_2}}=\frac{1}{2}$
• C. $\frac{{\epsilon }_{r_1}}{{\epsilon }_{r_2}}=4$
• D. $\frac{{\epsilon }_{r_1}}{{\epsilon }_{r_2}}=2$

Answer 1

The correct option is A $\frac{{\epsilon }_{r_1}}{{\epsilon }_{r_2}}=\frac{1}{4}$

Velocity of EM wave is given by $v=\frac{1}{\sqrt{\mu \epsilon }}$
Velocity in air $=\frac{\omega }{k}=c$
velocity in medium $=c/2$
Here, $\mu 1=\mu 2=1$ as medium is non magnetic
$\therefore \frac{\frac{1}{\sqrt{{\epsilon }_{r_1}}}}{\frac{1}{\sqrt{{\epsilon }_{r_2}}}}=\frac{c}{\left(\frac{c}{2}\right)}=2$
$\Rightarrow \frac{{\epsilon }_{r_1}}{{\epsilon }_{r_2}}=\frac{1}{4}$