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In a double slit experiment, the separation between slits is d = 0.25 cm and the distance of screen D = 120 cm from the slits.If I_(0) is intensity of central bright fringe, what is the intensity at distance x = 4.8 xx 10^(-5) m from central max ? Given lambda = 6000 Å : |
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Answer» `I_(0)` At the central fringe, `delta= 0` `I_(0) = I_(1) + I_(2) + 2I_(1) = 4I_(1)` At a distance x, PATH difference = `(xd)/(D)` Corresponding phase difference, `delta = (2pi)/(lambda).(xd)/(D)` `therefore I. = I_(1) + I_(2) + 2I_(1) cos (2pixd)/(lambda D)` ` = (I_(0))/(4) + (I_(0))/(4) + (2I_(0))/(4) cos(2pixd)/(lambda D)` `I. = (I_(0))/(2) [1 + cos (2pixd)/(lambda D)] = I_(0) cos^(2)(pixd)/(lambda D)` On solving = `I_(0) cos^(2)(pi)/(6) = (3I_(0))/(4)`. |
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