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BVP Medical BVP Medical Solved Paper-2015

  • question_answer
    In a baptism experiments, 5th dark fringe is obtained at a point. If a thin transparent film is placed in the path of one of waves, then 7th bright fringe is obtained at the same point. The thickness of the film in terms of wavelength A, and refractive index \[\frac{{{\gamma }_{1}}-{{\gamma }_{2}}}{3}+3\alpha \] will be

    A)  \[\frac{{{\gamma }_{1}}-{{\gamma }_{2}}}{3}+\alpha \]

    B)  \[{{E}_{I}}>{{E}_{II}}>{{E}_{III}}>{{E}_{IV}}>{{E}_{V}}\]

    C)  \[{{E}_{I}}>{{E}_{III}}={{E}_{V}}\,and\,\,\,{{E}_{II}}>{{E}_{IV}}\]

    D)  \[{{E}_{II}}={{E}_{IV}}={{E}_{V}}\,and\,\,\,{{E}_{I}}>{{E}_{III}}\]

    Correct Answer: D

    Solution :

                    For 5th dark fringe, \[{{x}_{1}}=(2n-1)\frac{\lambda }{2}\frac{D}{d}\] \[=\frac{9\lambda D}{2d}\] For 7th bright fringe, \[{{x}_{2}}=n\lambda \frac{D}{d}\]                                 \[=\frac{7\lambda D}{d}\] Now.         \[{{x}_{2}}-{{x}_{1}}=(\mu -1)t\frac{D}{d}\]                 \[\frac{\lambda D}{d}\left[ 7-\frac{9}{2} \right]=(\mu -1)t\frac{D}{d}\]                 \[t=\frac{25\lambda }{(\mu -1)}\]


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