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In a double slit experiment, l_(o) is the intensity of the central bright fringe obtained with monochromatic light of lambda = 6000 A^(@). Determine the intensity at a distance of 4.8xx10^(-5) m from the central maximum if the separation between the slits is 0.25 cm and the distance between the screen and the double slit is 1.20 m. |
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Answer» SOLUTION :From the principle of superposition the resultant intensity l = `l_(1) + l_(2) + 2sqrt(l_(1)l_(2))cos delta`, where `delta` = phase difference =`(2pi)/lambda xx` path difference. At the central fringe, `delta` = 0, thus `l_(o) = l_(1)+l_(2)+2l_(1) = 4l_(1)` At a DISTANCE x from the central bright fringe, the path difference = `(xd)/D` and the corresponding Phase difference. `delta = (2pi)/lambda (xd)/D` . Hence, the resultant intensity at the assigned POSITION will be l=`l_(1)+l_(2)=2 sqrt(l_(1)l_(2)) cos delta = l_(0)/4 + l_(0)/4 + (2 l_(0))/4 cos delta` = `1/4 [1+1+2cos delta] = l_(0)/2 [1+ cos delta = l_(0) cos^(2) (delta/2)]` hence , `delta/2 = 1/2 (2pi)/lambda (xd)/D = pi/6` Thus l=`l_(0)/2 xx (sqrt(3)/2)^(2) = (3l_(0))/4 = 0.75 l_(0)` |
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