In a YDSE,bichromatic light of wavelengths la | vedclass

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(D) The condition for the $n^{th}$ minima for wavelength $\lambda$ is given by $y_n = (2n-1) \frac{D\lambda}{2d}$.For two wavelengths to have overlapp

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$28$

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(D) The condition for the $n^{th}$ minima for wavelength $\lambda$ is given by $y_n = (2n-1) \frac{D\lambda}{2d}$.For two wavelengths to have overlapping dark fringes,their minima must coincide: $(2n-1) \frac{D\lambda_1}{2d} = (2m-1) \frac{D\lambda_2}{2d}$.This simplifies to $(2n-1) \lambda_1 = (2m-1) \lambda_2$.Substituting the values: $(2n-1) 400 = (2m-1) 560$.$\frac{2n-1}{2m-1} = \frac{560}{400} = \frac{7}{5}$.Thus,the first overlap occurs at $2n-1 = 7$ (i.e.,$n=4$) and $2m-1 = 5$ (i.e.,$m=3$).The position of this first overlap is $y_1 = 7 	imes \frac{1 	imes 400 	imes 10^{-6}}{2 	imes 0.1 	imes 10^{-3}} = 14 	imes 10^{-3}\, m = 14\, mm$.The next overlap occurs at the next odd multiple ratio,which is $\frac{21}{15}$ (where $2n-1=21$ and $2m-1=15$).The position of this second overlap is $y_2 = 21 	imes \frac{1 	imes 400 	imes 10^{-6}}{2 	imes 0.1 	imes 10^{-3}} = 42\, mm$.The distance between two successive regions of complete darkness is $y_2 - y_1 = 42\, mm - 14\, mm = 28\, mm$.

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