In a Young's double slit experiment, a light source emitting two wavelengths 500 nm and 600 nm is used. The minimum order of the bright fringe of the 500 nm wavelength that coincides exactly with a dark fringe of the 600 nm wavelength is

Step-by-step solution

Let the n^ th bright fringe of 500 nm coincide with the m^ th dark fringe of 600 nm . The position of the n^ th bright fringe is given by: y₁ = n ₁ D d The position of the m^ th dark fringe is given by: y₂ = (2m - 1) ₂ D 2d Equating the two positions for coincidence: n ₁ = (m - 1 2 ) ₂ n (500) = (m - 1 2 ) (600) 5n = 6m - 3 n = 6m - 3 5 We need the minimum positive integer value of n for which m is also an integer. For m = 1 , n = 3 5 (not an integer) For m = 2 , n = 9 5 (not an integer) For m = 3 , n = 15 5 = 3 Th