BCECE Medical BCECE Medical Solved Papers-2008

  • question_answer
    In a Youngs experiment, two coherent sources are placed 0.90 mm apart 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 monochromatic light used would be

    A)  \[60\times {{10}^{-4}}cm\]

    B)  \[10\times {{10}^{-4}}cm\]

    C)  \[10\times {{10}^{-5}}cm\]

    D)  \[6\times {{10}^{-5}}cm\]

    Correct Answer: D

    Solution :

    Distance of nth dark fringe from central fringe                 \[{{x}_{n}}\frac{(2n-1)\lambda D}{2d}\] \[\therefore \]  \[{{x}_{2}}=\frac{(2\times 2-1)\lambda D}{2d}\]                 \[=\frac{3\lambda d}{2d}\] \[\Rightarrow \] \[1\times {{10}^{-3}}=\frac{3\lambda \times 1}{2\times 0.9\times {{10}^{-3}}}\] \[\Rightarrow \] \[\lambda =6\times {{10}^{-5}}cm\]


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