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In a Young's double slit experiment, bi - chromatic light of wavelengths 400 nm and 560 nm are used.The distance between slits is 0.1 mm and distance between plane of slits and screen is 1 m.The minimum distance between two successive regions of complete darkness is : |
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Answer» 4 mm `therefore (2p - 1) (lambda_(1))/(2) = (2Q -1)(lambda_(2))/(2)` `(2p - 1) .(400)/(2) = (2q -1)(560)/(2)` `therefore (2p -1)/(2q -1) = (21)/(15) =(7)/(5)` i.e..4th minima of 400 nm coincides with 3rd minima of 560 nm POSITION of this minima is `y_(1) = (2p -1).(lambda_(1))/(2)(D)/(d)` ` = (2 xx 4 -1) ((400 xx10^(-9))/(2 xx 0.1 xx 10^(-3)) = 14 mm` Further 11th minima of 400 nm will coincide with 8th minima of 560nm(ratio (21)/(15)). Position of this minima is `y_(2) = (2 xx 11 -1)400 xx 10^(-9))/(2 xx 0.1 xx 10^(-3)) = 42 nm` `therefore y_(2) - y_(1) = 42 - 14 = 28 mm`. |
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