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 the two slits is 0.1 mm and the wavelength of light used is \[4\times {{10}^{-7}}m\]. If the width of the fringe on the screen is 4 mm, the distance between screen and slit is [Bihar CMEET 1995]

    A)            0.1 mm                                   

    B)            1 cm

    C)            0.1 cm                                     

    D)            1 m

    Correct Answer: D

    Solution :

               \[\beta =\frac{\lambda \,D}{d}\Rightarrow (4\times {{10}^{-3}})=\frac{4\times {{10}^{-7}}\times D}{0.1\times {{10}^{-3}}}\Rightarrow D=1\,m\]


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