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

  • question_answer
    In an interference experiment, third bright fringe is obtained at a point on the screen with a light of 700 nm. What should be the wavelength of the light source in order to obtain 5th bright fringe at the same point         [KCET 2003]

    A)            500 nm

    B)            630 nm 

    C)            750 nm

    D)            420 nm

    Correct Answer: D

    Solution :

               \[{{n}_{1}}{{\lambda }_{1}}={{n}_{2}}{{\lambda }_{2}}\Rightarrow 3\times 700=5\times {{\lambda }_{2}}\Rightarrow {{\lambda }_{2}}=420\ nm\]


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