JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Morning)

  • question_answer
    A light wave is incident normally on a glass slab of refractive index 1.5. If 4% of light gets reflected and the amplitude of the electric field of the incident light is \[30V\,{{m}^{-1}},\]then the amplitude of the electric field for the wave propagating in the glass medium will be [JEE Main Online Paper Held On 12-Jan-2019 Morning]

    A) \[24V\,{{m}^{-1}}\]                            

    B) \[10V\,{{m}^{-1}}\]

    C) \[30V\,{{m}^{-1}}\]                

    D)   \[6V\,{{m}^{-1}}\]

    Correct Answer: A

    Solution :

    \[{{E}_{ot}}\] and \[{{E}_{oi}}\]be the perpendicular components of the electric field for transmitted and incident waves respectively. \[\frac{{{E}_{ot}}}{{{E}_{oi}}}=\frac{2{{\mu }_{1}}}{{{\mu }_{1}}+{{\mu }_{2}}}=\frac{2(1)}{1+1.5}=\frac{4}{5}\] \[\Rightarrow \]\[{{E}_{ot}}=30\times \frac{4}{5}=24V\,{{m}^{-1}}\]


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