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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, carried out with light of wavelength l = 5000 Å, the distance between the slits is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at x =0. The third maximum (taking the central maximum as zeroth maximum) will be at x equal to [CBSE PMT 1992; MH CET 2002]

    A)            1.67 cm                                   

    B)            1.5 cm

    C)            0.5 cm                                     

    D)            5.0 cm

    Correct Answer: B

    Solution :

               Distance of third maxima from central maxima is \[x=\frac{3\lambda \,D}{d}=\frac{3\times 5000\times {{10}^{-10}}\times (200\times {{10}^{-2}})}{0.2\times {{10}^{-3}}}\] \[=1.5\,cm.\]


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