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Manipal Medical Manipal Medical Solved Paper-2015

  • question_answer
    In Young's double slit experiment, when wavelength used is \[6000\,\,\overset{\text{ }\!\!{}^\circ\!\!\text{ }}{\mathop{\text{A}}}\,\]and the screen is 40 cm from the slits, the fringes are 0.012 cm wide. What is the distance between the slits?

    A) 0.24 cm                   

    B) 0.2 cm

    C) 2.4 cm                                     

    D) 0.024 cm

    Correct Answer: B

    Solution :

    Fringe width, \[\beta =\frac{d\lambda }{d}\]                                \[d=\frac{D\lambda }{\beta }\]                                  \[=\frac{6000\times {{10}^{-10}}\times (40\times {{10}^{-2}})}{0.012\times {{10}^{-2}}}\]                                  \[=0.2\,\,cm\]


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