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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 aperture screen distance is 2m. The fringe width is 1 mm. Light of 600 nm is used. If a thin plate of glass (m = 1.5) of thickness 0.06 mm is placed over one of the slits, then there will be a lateral displacement of the fringes by                                          [BCECE 2005]

    A)            0 cm                                         

    B)            5 cm

    C)            10 cm                                      

    D)            15 cm

    Correct Answer: B

    Solution :

                       Lateral displacement of fringes =\[\frac{\beta }{\lambda }(\mu -1)\,t\] \[=\frac{1\times {{10}^{-3}}}{600\times {{10}^{-9}}}(1.5-1)\times 0.06\times {{10}^{-3}}\]\[=\frac{1}{20}m\]\[=5\,cm.\]


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