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MGIMS WARDHA MGIMS WARDHA Solved Paper-2014

  • question_answer
    In Youngs  double     slit experiment, monochromatic light of wavelength 600 nm illuminates the pair of slits and produces an interference pattern in which two consecutive bright fringes are separated by 10 mm. Another source of monochromatic light produces the interference pattern in which of the two consecutive bright fringes are separated by 8 mm. The wavelength of light from the second source is

    A)  448 nm               

    B)  450 nm

    C)  480 nm               

    D)  580 nm

    Correct Answer: C

    Solution :

                    Fringe width\[(\beta )\]is given by \[\beta =\frac{D\lambda }{d}\]where, D is separation between slits and screen, d is separation between two slits and\[\lambda \]is wavelength used. For given value of d and D Here, \[\beta \,\infty \lambda \] Therefore, we have \[\frac{10\,mm}{8\,mm}=\frac{63\times {{10}^{-7}}m}{{{\lambda }_{2}}}\]\[{{\lambda }_{2}}=\frac{6.3\times {{10}^{-7}}}{1.25}=480\,mm\] The wavelength of another source of laser light \[=480nm\]


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