In a Young's experiment,two coherent sources | vedclass

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(D) The distance of the $n^{th}$ dark fringe from the central fringe is given by the formula: $x_n = \frac{(2n - 1) \lambda D}{2d}$ For the second dar

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$6 	imes 10^{-5} \, cm$

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(D) The distance of the $n^{th}$ dark fringe from the central fringe is given by the formula: $x_n = \frac{(2n - 1) \lambda D}{2d}$ For the second dark fringe $(n = 2)$: $x_2 = \frac{(2 	imes 2 - 1) \lambda D}{2d} = \frac{3 \lambda D}{2d}$ Given: $x_2 = 1 \, mm = 1 	imes 10^{-3} \, m$,$D = 1 \, m$,and $d = 0.90 \, mm = 0.9 	imes 10^{-3} \, m$. Substituting the values: $1 	imes 10^{-3} = \frac{3 	imes \lambda 	imes 1}{2 	imes 0.9 	imes 10^{-3}}$ $1 	imes 10^{-3} = \frac{3 \lambda}{1.8 	imes 10^{-3}}$ $3 \lambda = 1.8 	imes 10^{-6}$ $\lambda = 0.6 	imes 10^{-6} \, m = 6 	imes 10^{-7} \, m = 6 	imes 10^{-5} \, cm$.

In a Young's experiment,two coherent sources are placed $0.90 \, mm$ apart and the fringes are observed one metre away. If it produces the second dark fringe at a distance of $1 \, mm$ from the central fringe,the wavelength of monochromatic light used would be

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