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 when wavelength used is 6000 Å and the screen is 40 cm from the slits, the fringes are 0.012 cm wide. What is the distance between the slits [MP PMT 1995; Pb PET 2002]

    A)            0.024 cm                                

    B)            2.4 cm

    C)            0.24 cm                                   

    D)            0.2 cm

    Correct Answer: D

    Solution :

               \[\beta =\frac{\lambda D}{d}\]\[\Rightarrow d=\frac{\lambda D}{\beta }\]\[=\frac{6000\times {{10}^{-10}}\times (40\times {{10}^{-2}})}{0.012\times {{10}^{-2}}}=0.2\,cm.\]


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