CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    If\[{{\varepsilon }_{0}}\]and\[{{\mu }_{0}}\]are the electric permittivity and magnetic permeability of free space and\[\varepsilon \] and\[\mu \]are the corresponding quantities in the medium, the index of refraction of the medium in terms of above parameter is:

    A)  \[\frac{\varepsilon \mu }{{{\varepsilon }_{0}}{{\mu }_{0}}}\]     

    B)         \[{{\left( \frac{\varepsilon \mu }{{{\varepsilon }_{0}}{{\mu }_{0}}} \right)}^{1/2}}\]

    C)  \[\left( \frac{{{\varepsilon }_{0}}{{\mu }_{0}}}{\varepsilon \mu } \right)\]           

    D)         \[{{\left( \frac{{{\varepsilon }_{0}}{{\mu }_{0}}}{\varepsilon \mu } \right)}^{1/2}}\]

    E)  none of these

    Correct Answer: B

    Solution :

    Velocity of light in vacuum \[c=\frac{1}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}\] Velocity of light in medium                 \[v=\frac{1}{\sqrt{\mu \varepsilon }}\] \[\therefore \]  \[\mu =\frac{c}{v}={{\left( \frac{\mu \varepsilon }{{{\mu }_{0}}{{\varepsilon }_{0}}} \right)}^{1/2}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner