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In a Young's double slit experiment, the slits separated by 1 mm are illuminated by a mixture of two wavelengths lambda = 600 nm and lambda' = 750 nm.The distance of screen from slits is 1 m. The minimum distance from the common central bright fringe where the bright fringe of one interfernece pattern will coincide with the bright fringe of second interference pattern will be : |
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Answer» `0.3 cm` `y = (mDlambda)/(d)` and `y. = (m.Dlambda.)/(d)` Since `y = y.therefore (m)/(m.) = (lambda.)/(lambda) = (750)/(600) = 5/4` i.e., m = 5 and m. = 4 Now the position where 5th bright fringe of `lambda` pattern will coincide with 4th bright fringe of `lambda.` pattern. `y = (5 xx 1 x 600 xx 10^(-9))/(1 xx 10^(-3)) = 0.3 xx 10^(-3) m = 0.3 mm`. |
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