In a Young’s double slit experiment, D equals the distance of screen from the slits and d is the separation between the slits. The distance of the nearest point from the central maximum where the intensity is same as that due to a single slit, is equal to
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a
λDd
b
λD2d
c
λD3d
d
2λDd
answer is C.
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Detailed Solution
I=4I0 cos2 πdλDyWhere y is the distance of a point on the screen measured from central bright fringe.Where I=I0, cos2 πdλDy=14⇒ cos πdλDy=±12⇒ πdλDy=π3, 2π−π3, 2π+π3etc∴ ymin = λD3d