EAMCET Medical EAMCET Medical Solved Paper-2007

  • question_answer
    The velocities of light in two different mediums are \[2\times {{10}^{8}}m{{s}^{-1}}\] and \[2.5\times {{10}^{8}}m{{s}^{-1}}\] respectively. The critical angle for these mediums is

    A)  \[{{\sin }^{-1}}\left( \frac{1}{5} \right)\]                               

    B)  \[{{\sin }^{-1}}\left( \frac{4}{5} \right)\]

    C)  \[{{\sin }^{-1}}\left( \frac{1}{2} \right)\]                               

    D)  \[{{\sin }^{-1}}\left( \frac{1}{4} \right)\]

    Correct Answer: B

    Solution :

                     Critical angle is given by \[\sin C=\mu =\frac{{{v}_{1}}}{{{v}_{2}}}\] \[=\frac{2\times {{10}^{8}}}{2.5\times {{10}^{8}}}=\frac{20}{25}=\frac{4}{5}\]                 or            \[C={{\sin }^{-1}}\left( \frac{4}{5} \right)\]


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