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In Young's double slit experiment (d)/(D) = 10^(-4) (d = distance between slits, D =distance of screen from the slits) At a point P on the screen the resulting intensity is equal to the intensity due to individual slit I_(0).Then the distance of point P from the central maximum is (lambda = 6000 Å) : |
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Answer» 2 mm As `I = I_(0)` `THEREFORE I_(0) = 4I_(0) cos^(2)((phi)/(2))` `therefore cos((phi)/(2)) = 1/2 Rightarrow (phi)/(2) = (pi)/(3)` `phi = (2pi)/(3) = (2 pi)/(lambda).Deltax` `therefore 1/3 = 1/(lambda).Deltax` But `Deltax = (yd)/(D)` `therefore 1/3 = 1/(lambda).(yd)/(D)` `therefore y = (lambda)/(3((d)/(D)) = 2 xx 10^(-3)` m `y = 2mm`. |
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