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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 16

  • question_answer
    In the Young's double slit experiment using a monochromatic light of wavelength A, the path difference (in terms of an integer \[\lambda \]) corresponding to any point having half the peak intensity is

    A) \[(2n+1)\frac{\lambda }{2}\]    

    B)                    \[(2n+1)\frac{\lambda }{4}\]         

    C) \[(2n+1)\frac{\lambda }{8}\]         

    D)        \[(2n+1)\frac{\lambda }{16}\]

    Correct Answer: B

    Solution :

      [b] \[{{I}_{\max }}=4I\] \[2I=4I{{\cos }^{2}}\frac{\phi }{2}\] \[\cos \frac{\phi }{2}=\frac{1}{\sqrt{2}}\] \[\frac{\phi }{2}=(2n+1)\frac{\pi }{4}\] \[\frac{1}{2}.\frac{2\pi }{\lambda }.\Delta x=(2n+1)\frac{\pi }{4}\,\,\,\,\Rightarrow \,\,\,\,\Delta x=(2n+1)\frac{\lambda }{4}\]


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