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JEE Mains · Physics · STD 12 - 8. Electromagnetic waves

An EM wave from air enters a medium. The electric fields are \(\overrightarrow {{E_1}}  = {E_{01}}\hat x\;cos\left[ {2\pi v\left( {\frac{z}{c} - t} \right)} \right]\) in air and \(\overrightarrow {{E_2}}  = {E_{02}}\hat x\;cos\left[ {k\left( {2z - ct} \right)} \right]\) in medium, where the wave number \(k\) and frequency \(v\) refer to their values in air. The medium is nonmagnetic. If \(\varepsilon {_{{r_1}}}\) and \(\varepsilon {_{{r_2}}}\) refer to relative permittivities of air and medium respectively, which of the following options is correct?

  1. A \(\frac{{{_{{\epsilon r_1}}}}}{{{_{{\epsilon r_2}}}}} = 2\)
  2. B \(\frac{{{_{{\epsilon r_1}}}}}{{{_{{\epsilon r_2}}}}} = \frac{1}{4}\)
  3. C \(\frac{{{_{{\epsilon r_1}}}}}{{{_{{\epsilon r_2}}}}} = \frac{1}{2}\)
  4. D \(\frac{{{_{{\epsilon r_1}}}}}{{{_{{\epsilon r_2}}}}} = 4\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{{{_{{\epsilon r_1}}}}}{{{_{{\epsilon r_2}}}}} = \frac{1}{4}\)

Step-by-step Solution

Detailed explanation

Velocity of \(EM\) wave is given by \(\mathrm{v}=\frac{1}{\sqrt{\mu \epsilon}}\) Velocity in air \(=\frac{\omega}{\mathrm{k}}=\mathrm{C}\) Velocity in medium \(=\frac{\mathrm{C}}{2}\) Here, \(\mu_{1}=\mu_{2}=1\) as medium is non-magnetic…
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