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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2003

  • question_answer
    In an interference experiment, third bright fringe is obtained at a point on the screen with a light of 700 nm. What should be the wavelength of die light source in order to obtain 5th bright fringe at the same point?

    A)  630 nm                               

    B)  500 nm

    C)  420 nm                               

    D)  750 nm

    Correct Answer: C

    Solution :

    Here \[{{X}_{3}}={{X}_{5}}\]                 \[\frac{3D\lambda }{2d}=\frac{5D\lambda }{2d}\] \[\Rightarrow \]               \[3\lambda =5\lambda \] or \[\frac{\lambda }{\lambda }=\frac{3}{5}\]                 \[\lambda =\frac{3}{5}\times 700\,nm=420\,nm\]


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