CET Karnataka Medical CET - Karnataka Medical Solved Paper-2004

  • question_answer
    In a Youngs double slit experiment, the separation between the two slits is 0.9 mm and the fringes are observed one metre away. If it produces the second dark fringe at a distance of 1 mm from the central fringe, the wavelength of the monochromatic source of light used is:

    A)  450 nm

    B)  400 nm

    C)  500 nm

    D)  600 nm

    Correct Answer: D

    Solution :

    The distance between two consecutive dark or bright fringes is recognised as \[\beta \](fringe width) and that between central fringe and first dark fringe on either side is \[\beta /2\]. Given, spacing between second dark fringe and central fringe \[=\beta +\frac{\beta }{2}\] or \[\frac{3\beta }{2}=1mm\] or \[\beta =\frac{2}{3}\times 1mm\] or \[\frac{\lambda D}{d}=\frac{2}{3}mm\] \[\therefore \] \[\lambda =\frac{2}{3}\times {{10}^{-3}}\times \frac{0.9\times {{10}^{-3}}}{1}\] \[\Rightarrow \] \[\lambda =0.6\times {{10}^{-6}}m\] \[\therefore \] \[\lambda =600\times {{10}^{-9}}m\] \[=600nm\]


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