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

  • question_answer
    In the Young's double slit experiment, the spacing between two slits is 0.1 mm. If the screen is kept at a distance of 1.0 m from the slits and the wavelength of light is 5000 Å, then the fringe width is         [MP PMT 1993; RPET 1996]

    A)            1.0 cm                                     

    B)            1.5 cm

    C)            0.5 cm                                     

    D)            2.0 cm

    Correct Answer: C

    Solution :

               \[\beta =\frac{\lambda D}{d}=\frac{5000\times {{10}^{-10}}\times 1}{0.1\times {{10}^{-3}}}m=5\times {{10}^{-3}}m=0.5\ cm\].


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