JEE Main & Advanced Physics Wave Optics / तरंग प्रकाशिकी Question Bank Young's Double Slit Experiment and Biprism

  • question_answer
    In Young's double slit experiment, the distance between sources is 1 mm and distance between the screen and source is 1 m. If the fringe width on the screen is 0.06 cm, then l =                                [CPMT 1996]

    A) \[6000\,\,{AA}\]

    B) \[4000\,\,{AA}\]

    C) \[1200\,\,{AA}\]

    D) \[2400\,\,{AA}\]

    Correct Answer: A

    Solution :

    \[\beta =\frac{\lambda \,D}{d}\Rightarrow (0.06\times {{10}^{-2}})=\frac{\lambda \times 1}{1\times {{10}^{-3}}}\] \[\Rightarrow \lambda =6000\,{AA}\]


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