NEET-XII-Physics

exam-1 year:2016

with Solutions - page 2
  • #18
    The intensity at the maximum in a Young's double
    slit experiment is ``I_0``. Distance between two slits is
    d = 5 ``\lambda``, where ``\lambda`` is the wavelength of light used
    in the experiment. What will be the intensity in front
    of one of the slits on the screen placed at a distance
    D = 10 d ?
    (1) ``I_0``
    (2)``I_0``/4
    (3) 3/4 ``I_0``
    (4)``I_0``/2
    digAnsr:   4
    Ans : 4
    Sol. Path difference
    = S2P - S1P
    = + 2 2D d D
    S1
    S2
    O B
    P
    D
    d
    = D
     
    +  
     
    2
    2
    1 d
    1 D
    2 D
    = D
     
    +  = 
     
    2 2
    2
    d d
    1 1
    2D2D
    ▵x =
     
    = = =

    2d d 5
    2 10d 20 20 4
    ▵ =
      
    =

    2
    .
    4 2
    So, intensity at the desired point is
    I = I0cos2

    2
    = I0cos2

    4
    = 0
    I
    2