Modern Physics 7 Question 29
30. An EM wave from air enters a medium. The electric fields are $E _1=E _{01} \hat{x} \cos 2 \pi \nu \frac{z}{c}-t \quad$ in air and
$E _2=E _{02} \hat{X} \cos [k(2 z-c t)]$ in medium, where the wave number $k$ and frequency $v$ refer to their values in air. The medium is non-magnetic.
If $\varepsilon _{r _1}$ and $\varepsilon _{r _2}$ refer to relative permittivities of air and medium respectively, which of the following options is correct?
(a) $\frac{\varepsilon _{r _1}}{\varepsilon _{r _2}}=\frac{1}{2}$
(b) $\frac{\varepsilon _{r _1}}{\varepsilon _{r _2}}=4$
(c) $\frac{\varepsilon _{r _1}}{\varepsilon _{r _2}}=2$
(d) $\frac{\varepsilon _{r _1}}{\varepsilon _{r _2}}=\frac{1}{4}$
Show Answer
Solution:
- $v=\frac{\text { coefficient of } t}{\text { coefficient of } z}=\frac{1}{\sqrt{\varepsilon \mu}}$
where, $\varepsilon=\varepsilon _0 \varepsilon _r$ and $\mu=\mu _0 \mu _r$.
$$ \frac{v _{air}}{v _{med}}=\frac{c}{c / 2}=2=\frac{\sqrt{\mu _0 \varepsilon _0 \mu _{r _2} \varepsilon _{r _2}}}{\sqrt{\mu _0 \varepsilon _0 \mu _{r _1} \varepsilon _{r _1}}} \Rightarrow \frac{\varepsilon _{r _1}}{\varepsilon _{r _2}}=\frac{1}{4} $$
Note Medium is non-magnetic.
$$ \therefore \quad \mu _{r _1}=\mu _{r _2} $$
