BVP Medical BVP Medical Solved Paper-2012

  • question_answer
    In Youngs double slit experiment, the spacing between the slits is d and wavelength of light used is \[\frac{2v(n+1)}{n}\]. If the angular width of a fringe formed on a distant  screen is 1°, then value of d is

    A)  \[{{327}^{o}}C\]                              

    B)  \[{{427}^{o}}C\]

    C)  \[{{27}^{o}}C\]                                

    D)  \[0.52\]

    Correct Answer: C

    Solution :

                    \[\sin \theta =\theta =\frac{y}{D}\] So,        \[\Delta \theta =\frac{\Delta y}{D}\] Angular fringe width \[{{\theta }_{0}}=\Delta \theta \] (width. \[\Delta y=\beta \])                                 \[{{\theta }_{0}}=\frac{\beta }{D}=\frac{D\lambda }{d}\times \frac{1}{D}=\frac{\lambda }{d}\]                                 \[{{\theta }_{0}}={{1}^{o}}=\frac{\pi }{180}rad\] and        \[\lambda =6\times {{10}^{-7}}m\]                                 \[d=\frac{\lambda }{{{\theta }_{0}}}=\frac{180}{\pi }\times 6\times {{10}^{-7}}\]                                 \[=3.44\times {{10}^{-5}}m\]                                 \[=0.03mm\]


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