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JEE Main & Advanced JEE Main Solved Paper-2016

  • question_answer
    In an experiment for determination of refractive index of glass of a prism by \[i-\delta ,\] plot, it was found that a ray incident at angle \[35{}^\circ \], suffers a deviation of \[40{}^\circ \] and that it emerges at angle \[79{}^\circ \]. In that case which of the following is closest to the maximum possible value of the refractive index? [JEE Main Solved Paper-2016 ]

    A) 1.8                                         

    B) 1.5

    C) 1.6                                         

    D) 1.7

    Correct Answer: B

    Solution :

                    \[i={{35}^{o}},\delta ={{40}^{o}},e={{79}^{o}}\]                 \[\delta =i+e-A\]             \[{{40}^{o}}={{35}^{o}}+{{79}^{o}}-A\] \[A={{74}^{o}}\]and\[{{r}_{1}}+{{r}_{2}}=A={{74}^{o}}\] solving these, we get \[\mu =\text{1}.\text{5}\] \[\mu <\frac{\sin \left( \frac{74+40}{2} \right)}{\sin 37}\]\[{{\mu }_{\max }}=1.44\]


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