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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 37

  • question_answer
    An EM wave from air enters a medium. The electric fields are \[{{\vec{E}}_{1}}={{E}_{01}}\hat{x}\,cos\,\,\left[ 2\pi \nu \left( \frac{z}{c}-t \right) \right]\] in air and \[{{\vec{E}}_{2}}={{E}_{02}}\hat{x}\,cos\,\left[ k(2z-ct) \right]\] in medium, where the wave number k and frequency v refer to their values in air. The medium is nonmagnetic. If \[{{\in }_{{{r}_{1}}}}and{{\in }_{{{r}_{2}}}}\] refer to relative permittivity?s of air and medium respectively, which of the following options is correct?

    A) \[\frac{{{\in }_{{{r}_{1}}}}}{{{\in }_{{{r}_{2}}}}}=4\]          

    B)        \[\frac{{{\in }_{{{r}_{1}}}}}{{{\in }_{{{r}_{2}}}}}=2\]

    C) \[\frac{{{\in }_{{{r}_{1}}}}}{{{\in }_{{{r}_{2}}}}}=\frac{1}{4}\]       

    D)        \[\frac{{{\in }_{{{r}_{1}}}}}{{{\in }_{{{r}_{2}}}}}=\frac{1}{2}\]

    Correct Answer: C

    Solution :

    Velocity of EM wave is given by \[v=\frac{1}{\sqrt{\mu \in }}\] Velocity in air \[=\,\,\frac{\omega }{k}=C\] Velocity in medium \[=\,\,\frac{C}{2}\] Here, \[{{\mu }_{1}}={{\mu }_{2}}=1\] as medium is non-magnetic \[\therefore \,\,\frac{\sqrt{\frac{1}{\sqrt{{{\in }_{{{r}_{1}}}}}}}}{\sqrt{\frac{1}{{{\in }_{{{r}_{2}}}}}}}=\frac{C}{\left( \frac{C}{2} \right)}=2\,\,\,\,\,\,\Rightarrow \,\,\,\frac{{{\in }_{{{r}_{1}}}}}{{{\in }_{{{r}_{2}}}}}=\frac{1}{4}\]


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