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Solve this problem - Electromagnetic Waves - JEE Main

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

\vec{E_{1}}= E_{01} \hat{x} cos [ 2\pi v\left ( \frac{z}{c} \ -t )\right ] in air and  \vec{E_{2}}= E_{02} \hat{x} cos [ k (2z-ct) ] in medium, where the wave number k and frequency ν refer to their values in air. The medium is non-magnetic. If  \epsilon {_{r1}} and \epsilon {_{r2}} refer to relative permittivities of air and medium respectively, which of the following options is correct ?
 

 

  • Option 1)

     \frac{\epsilon _{r_{1}}}{\epsilon _{r_{2}}}=\frac{1}{2}

  • Option 2)

    \frac{\epsilon _{r_{1}}}{\epsilon _{r_{2}}}=4

  • Option 3)

    \frac{\epsilon _{r_{1}}}{\epsilon _{r_{2}}}=2

  • Option 4)

    \frac{\epsilon _{r_{1}}}{\epsilon _{r_{2}}}=\frac{1}{4}

 
Answers (2)
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N neha

As we learnt that

 

Speed of light formula in vacuum -

c=\frac{1}{\sqrt{\mu_{o}\epsilon _{o}}}

c=2.99793 X 10^{8} m/s

- wherein

c = Speed of light in vacuum

\mu_{o} = Permeability of vacuum

\epsilon _{o} = Permittivity of vacuum

 

 \frac{V}{C}=\frac{1}{2}        (where v= speed in medium and c=speed of light in air)

For non magnetic medium; \mu =1

 

\frac{V}{C}=\sqrt{\frac{\epsilon 1}{\epsilon 2}}=\frac{1}{2}\: or\: {\frac{\epsilon 1}{\epsilon 2}} =\frac{1}{4}


Option 1)

 \frac{\epsilon _{r_{1}}}{\epsilon _{r_{2}}}=\frac{1}{2}

This is incorrect

Option 2)

\frac{\epsilon _{r_{1}}}{\epsilon _{r_{2}}}=4

This is incorrect

Option 3)

\frac{\epsilon _{r_{1}}}{\epsilon _{r_{2}}}=2

This is incorrect

Option 4)

\frac{\epsilon _{r_{1}}}{\epsilon _{r_{2}}}=\frac{1}{4}

This is correct

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