JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Morning)

  • question_answer
    An electromagnetic wave of intensity 50 W \[{{m}^{-2}}\] enters in a medium of refractive index 'n' without any loss. The ratio of the magnitudes of electric fields, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

    A) \[\left( \sqrt{n},\frac{1}{\sqrt{n}} \right)\]                        

    B) \[\left( \sqrt{n},\sqrt{n} \right)\]

    C) \[\left( \frac{1}{\sqrt{n}},\frac{1}{\sqrt{n}} \right)\]         

    D) \[\left( \frac{1}{\sqrt{n}},\sqrt{n} \right)\]

    Correct Answer: A

    Solution :

    The intensity of the wave remain unchanged So, \[\frac{{{B}^{2}}}{{{\mu }_{0}}}c=\frac{B_{1}^{2}}{\mu }v\] For a non-magnetic medium; \[\mu ={{\mu }_{0}},\] \[\frac{{{B}_{1}}}{B}=\sqrt{n}\Rightarrow \frac{B}{{{B}_{1}}}=\frac{1}{\sqrt{n}}\]                             ?.(1) Also, \[\frac{E}{B}=c\]and\[\frac{{{E}_{1}}}{{{B}_{1}}}=v\Rightarrow \frac{E}{{{E}_{1}}}\frac{{{B}_{1}}}{B}=\frac{c}{v}=n\] \[\Rightarrow \frac{E}{{{E}_{1}}}=\frac{n}{\sqrt{n}}=\sqrt{n}[Using\,(i)]\]


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