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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Morning)

  • question_answer
    In a Young?s double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is \[\frac{1}{8}th\] of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

    A) 0.80                             

    B) 0.94  

    C) 0.85     

    D)                  0.74

    Correct Answer: C

    Solution :

    The phase difference between two waves is given as\[\Delta x\times \frac{2\pi }{\lambda }=\frac{\lambda }{8}\times \frac{2\pi }{\lambda }=\frac{\pi }{4}\] So, the intensity at this point is \[I={{I}_{0}}{{\cos }^{2}}\frac{\pi }{8}\] \[I={{I}_{0}}\left( \frac{1+\cos \frac{\pi }{4}}{2} \right)\] \[I={{I}_{0}}\left( \frac{1+\frac{1}{\sqrt{2}}}{2} \right)=0.85{{I}_{0}}\]


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