CET Karnataka Engineering CET - Karnataka Engineering 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 \[\frac{\beta }{2}\] Given, spacing between second dark fringe and central fringe \[=\beta +\frac{\beta }{2}\] or            \[\frac{3\,\beta }{2}=1\,\,mm\] or            \[\beta =\frac{2}{3}\times 1\,mm\] or            \[\frac{\lambda \,D}{d}=\frac{2}{3}\,mm\] \[\therefore \]  \[\lambda =\frac{2}{3}\times {{10}^{-3}}\times \frac{0.9\times {{10}^{-3}}}{1}\] \[\Rightarrow \]               \[\lambda =0.6\times {{10}^{-6}}m\] \[\therefore \]  \[\lambda =600\times {{10}^{-9}}m\]                 \[=600\,\,nm\]


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