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JEE Main & Advanced JEE Main Paper (Held On 19 April 2014)

  • question_answer
    A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is  \[{{\theta }_{iC}}\]and Brewster's angle of incidence is \[{{\theta }_{iB}},\] such that \[\sin {{\theta }_{iC}}/\sin {{\theta }_{iB}}\]\[=\eta =1.28.\] = h = 1.28. The relative refractive index of the two media is:     JEE Main Online Paper (Held On 19 April 2016)

    A) 0.2                                         

    B) 0.4

    C) 0.8         

    D) 0.9

    Correct Answer: C

    Solution :

                    Here, \[\sin {{\theta }_{ic}}/\sin {{\theta }_{iB}}=1.28\] As we know, \[\mu =\frac{\sin {{\theta }_{iB}}}{\sin \left( \frac{\pi }{2}-{{\theta }_{iB}} \right)}\]where, \[{{\theta }_{iB}}\]is Brewster?s angle of incidence, And, \[\mu =\frac{1}{\sin {{\theta }_{ic}}}\] On solving we get, relative refractive index of the two media.                                


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