Modern Physics Ques 276
- An EM wave from air enters a medium. The electric fields are $E _1=E _{01} \hat{x} \cos [2 \pi \nu \frac{z}{c}-t ]$ 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) $\frac{\varepsilon _{r _1}}{\varepsilon _{r _2}}=\frac{1}{4}$
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Answer:
Correct Answer: 276.(d)
Solution:
Formula:
Relation Between The Magnetic Field Vector And The Electric Field Vector:
- $v=\frac{\text { coefficient of } t}{\text { coefficient of } z}=\frac{1}{\sqrt{\varepsilon \mu}}$
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} $
BETA
