An EM wave from air enters a medium. The electric fields are → E _ 1 = E _ 01 ^ x [2 ((z)/(c) - t)] in air and → E _ 2 = E _ 02 ^ x [k(2z - ct)] in medium, where the wave number k and frequency v refer to their values

Correct answer: D. ϵ _ r _ 1 ϵ _ r _ 2 = (1)/(4) ∨ _ air ∨ _ med = (c)/(c / 2) = 2 = μ _ 0 ϵ _ 0 μ _ r _ 2 ϵ _ r _ 2 μ _ 0 ϵ _ 0 μ _ r _ 1 ϵ _ r _ 1 ϵ _ r _ 1 ϵ _ r _ 2 = (1)/(4)

JEE Main 20188 April

Exam: JEE Main
Year: 2018
Date: 8 April
Subject: Physics
Chapter: Electromagnetic Waves

JEE Main 2018 · 8 April · Physics · Electromagnetic Waves

An EM wave from air enters a medium. The electric fields are → E _...

An EM wave from air enters a medium. The electric fields are

{\overset{\to}{E}}_{1} = {E}_{01}\overset{^}{x}\cos \left[2\piv\left(\dfrac{z}{c} - t\right)\right]

in air and

{\overset{\to}{E}}_{2} = {E}_{02}\overset{^}{x}\cos \left[k\left(2z - ct\right)\right]

in medium, where the wave number k and frequency

v

refer to their values in the air. The medium is non-magnetic. If

{ϵ}_{{r}_{1}}

and

{ϵ}_{{r}_{2}}

refer to relative permittivities of air and medium respectively, which of the following options, is correct?

Options

A $\dfrac{{ϵ}_{{r}_{1}}}{{ϵ}_{{r}_{2}}} = \dfrac{1}{2}$
B $\dfrac{{ϵ}_{{r}_{1}}}{{ϵ}_{{r}_{2}}} = 4$
C $\dfrac{{ϵ}_{{r}_{1}}}{{ϵ}_{{r}_{2}}} = 2$
D $\dfrac{{ϵ}_{{r}_{1}}}{{ϵ}_{{r}_{2}}} = \dfrac{1}{4}$

Correct answer

D. $\dfrac{{ϵ}_{{r}_{1}}}{{ϵ}_{{r}_{2}}} = \dfrac{1}{4}$

Step-by-step solution

\dfrac{{∨}_{air}}{{∨}_{med}} = \dfrac{c}{c / 2} = 2 = \dfrac{\sqrt{{\mu}_{0}{ϵ}_{0}{\mu}_{{r}_{2}}{ϵ}_{{r}_{2}}}}{\sqrt{{\mu}_{0}{ϵ}_{0}{\mu}_{{r}_{1}}{ϵ}_{{r}_{1}}}}

\dfrac{{ϵ}_{{r}_{1}}}{{ϵ}_{{r}_{2}}} = \dfrac{1}{4}

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