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Question on Young's double slit EXPERIMENT
  • In YDSE EXPERIMENT THE DISTANCE BETWEEN THE SLITS AND SCREEN IS a &DISTANCE BETWEEN TWO SLITS IS b.WAVELENGTH OF LIGHT USED IS LEMBDA. THE DISTANCE BETWEEN THE THIRD MAXIMA ON ONE SIDE AND FOURTH MINIMA ON OPPOSITE SIDE OF THE SCREEN IS :?:
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  • hi
    ans is 13(LAMBDA)a/2b.
    if correct thn ask for solution.
    if no thn give me correct ans i'll gt it solved
  • u gt the answer rite ... good work

    I share the solution:

    Maxima occurs under the condition :

    n(LAMBDA) = Xb/a

    Minima occurs under the condition:

    (n+1/2)(LAMBDA) = Xb/a

    Thus,

    For the 3rd Maxima :

    Xmax = 3(LAMBDA)a/b

    For 4th Minima :

    Xmin = 7(LAMBDA)a/2b

    Since they are on the opposite sides ,

    Required distance :

    X = Xmax + Xmin

    = 13(LAMBDA)a/2b

    add anything to the solution if u wish !!
  • mylie said:

    In YDSE EXPERIMENT THE DISTANCE BETWEEN THE SLITS AND SCREEN IS a &DISTANCE BETWEEN TWO SLITS IS b.WAVELENGTH OF LIGHT USED IS LEMBDA. THE DISTANCE BETWEEN THE THIRD MAXIMA ON ONE SIDE AND FOURTH MINIMA ON OPPOSITE SIDE OF THE SCREEN IS :?:




    answer is 13(LAMBDA)a/2b.

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