RAJASTHAN ­ PET Rajasthan PET Solved Paper-2008

  • question_answer
    A ray of light is incident on the interface between water and glass at an angle i and retracted parallel to the water surface, then value of \[{{\mu }_{g}}\]will be

    A)  \[(4/3)\sin i\]

    B)  \[\frac{1}{\sin i}\]

    C)  \[4/3\]

    D)  \[1\]

    Correct Answer: B

    Solution :

     \[g{{\mu }_{w}}=\frac{\sin i}{\sin r}\] \[g{{\mu }_{a}}=\frac{\sin r}{\sin 90{}^\circ }\] \[\Rightarrow \] \[g{{\mu }_{w}}{{\times }_{w}}{{\mu }_{a}}=\frac{\sin i}{\sin r}\times \frac{\sin r}{\sin 90{}^\circ }=\sin i\] Or \[\frac{{{\mu }_{w}}}{{{\mu }_{g}}}\times \frac{{{\mu }_{a}}}{{{\mu }_{w}}}=\sin i\] \[\Rightarrow \] \[{{\mu }_{g}}=\frac{1}{\sin i}\]


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