BVP Medical BVP Medical Solved Paper-2015

  • question_answer
    In Youngs double slit experiment, let \[\frac{\pi }{2}\times \frac{1}{\sqrt{15LC}}\] be the intensity of the central bright fringe and P be the fringe width. The intensity of a distance Y from the central bright fringe will be

    A)  \[\mu \]

    B)  \[\frac{15\lambda }{(\mu -1)}\]

    C)  \[15(\mu -1)\lambda \]

    D)  \[25(\mu -1)\lambda \]

    Correct Answer: B

    Solution :

                    Intensity at a point having phase difference of \[\phi \] is At the central maxima   \[{{I}_{0}}=4I\] At a distance y from the central fringe path difference        \[\Delta x=d\times \frac{y}{D}\] Phase difference  \[\phi =\frac{2\pi }{\lambda }(\Delta x)\]                 \[=\frac{2\pi }{\lambda }\left[ d\times \frac{y}{D} \right]=\frac{2\pi y}{\frac{D}{d}\lambda }=\frac{2\pi y}{\beta }\] \[\therefore \] The intensity at this point P                 \[I=4I{{\cos }^{2}}\left( \frac{\pi y}{\beta } \right)\]                 \[={{I}_{0}}{{\cos }^{2}}\left( \pi \frac{y}{\beta } \right)\]


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