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7 June, 10:44

When 450-nm light is incident normally on a certain double-slit system the number of interference maxima within the central diffraction maximum is 5. When 900-nm light is incident on the same slit system the number is: a. 9b. 2c. 3d. 10e. 5

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  1. 7 June, 11:13
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    width of central diffraction maxima = 2 λD / d₁

    λ is wave length of light, D is screen distance and d₁ is slit width

    width of each interference fringe = λD / d₂, d₂ is slit separation.

    No of interference fringe in central diffraction fringe

    = width of central diffraction maxima / width of each interference fringe

    = 2 λD / d₁ x λDd₂ / λD

    No = 2 d₂ / d₁

    No = 5

    5 = 2 d₂ / d₁

    Since this number does not depend upon wavelength so it will remain the same

    No of required fringe will be 5.

    right option

    e) 5.
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