JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    In Young's double slit experiment, first slit has width four times the width of the second slit. The ratio of the maximum intensity to the minimum intensity in the interference fringe system is :

    A)  \[2:1\]                          

    B)  \[4:1\]

    C)  \[9:1\]                          

    D)  \[8:1\]

    Correct Answer: C

    Solution :

    \[\frac{{{I}_{\max }}}{{{I}_{\min }}}\,=\frac{{{\left( \sqrt{{{I}_{1}}}+\,\sqrt{{{I}_{2}}} \right)}^{2}}}{{{\left( \sqrt{{{I}_{1}}}-\,\sqrt{{{I}_{2}}} \right)}^{2}}}\,=\frac{{{(\sqrt{4I}+\sqrt{I})}^{2}}}{{{(\sqrt{4I}-\sqrt{I})}^{2}}}=\frac{9}{1}\]

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