JEE Main On 08 April 2018 Question 19
Question: An EM wave from air enters a medium. The electric fields are $ {{\overset{\to }{\mathop{E}},}_1}=E _{01}\widehat{x}\cos [ 2\pi v( \frac{z}{c}-t ) ] $ in air and $ {{\overset{\to }{\mathop{E}},}_2}=E _{02}\widehat{x}\cos [k(2z-ct)] $ in medium, where the wave number $ k $ and frequency $ v $ refer to their values in air. The medium is non-magnetic. If $ {\in _{r _1}} $ and $ {\in _{r _2}} $ refer to relative permittivitys of air and medium respectively, which of the following options is correct? [JEE Main Online 08-04-2018]
Options:
A) $ \frac{{\in _{r _1}}}{{\in _{r _2}}}=\frac{1}{4} $
B) $ \frac{{\in _{r _1}}}{{\in _{r _2}}}=\frac{1}{2} $
C) $ \frac{{\in _{r _1}}}{{\in _{r _2}}}=4 $
D) $ \frac{{\in _{r _1}}}{{\in _{r _2}}}=2 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \overrightarrow{{}E}=E _{01}\cos ( 2\pi v( \frac{z}{c}-t ) )\widehat{x}inair $ $ k=\frac{2\pi v}{c} $
$ \text{speed=c} $ $ {{\overrightarrow{{}E}}_2}=E _{02}\cos (k(2z-ct))\widehat{x} $
$ {{\overrightarrow{{}E}}_2}=E _{02}\cos ( \frac{2\pi v}{c}(2z-ct) )\widehat{x} $
$ \text{speed=}\frac{c}{2} $
$ c\propto \frac{1}{\sqrt{{\varepsilon_0}{\varepsilon _{r _1}}}} $
$ \frac{c}{2}\propto \frac{1}{\sqrt{{\varepsilon_0}{\varepsilon _{r _1}}}} $
$ 2=\sqrt{\frac{{\varepsilon _{r _2}}}{{\varepsilon _{r _1}}}}\Rightarrow \frac{{\varepsilon _{r _2}}}{{\varepsilon _{r _1}}}=4 $
$ \frac{{\varepsilon _{r _1}}}{{\varepsilon _{r _2}}}=\frac{1}{4} $
