In a YDSE,bichromatic light of wavelengths la | vedclass

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(D) The condition for minima in $YDSE$ is given by $y = \frac{(2n-1) \lambda D}{2d}$.For two wavelengths to have a common dark fringe,their positions

Vedclass · Accepted Answer

$28$

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(D) The condition for minima in $YDSE$ is given by $y = \frac{(2n-1) \lambda D}{2d}$.For two wavelengths to have a common dark fringe,their positions must coincide: $y = \frac{(2n-1) \lambda_1 D}{2d} = \frac{(2m-1) \lambda_2 D}{2d}$.Substituting the values: $(2n-1) 	imes 400 = (2m-1) 	imes 560$.$\frac{2n-1}{2m-1} = \frac{560}{400} = \frac{7}{5}$.For the first coincidence,$2n-1 = 7 \Rightarrow n = 4$ and $2m-1 = 5 \Rightarrow m = 3$.The position of this first common dark fringe is $y_1 = \frac{7 	imes 400 	imes 10^{-9} 	imes 1}{2 	imes 0.1 	imes 10^{-3}} = 14 	imes 10^{-3} \, m = 14 \, mm$.The next coincidence occurs for the next odd multiples of the ratio $7/5$,which is $21/15$.For $2n-1 = 21 \Rightarrow n = 11$ and $2m-1 = 15 \Rightarrow m = 8$.The position of this second common dark fringe is $y_2 = \frac{21 	imes 400 	imes 10^{-9} 	imes 1}{2 	imes 0.1 	imes 10^{-3}} = 42 	imes 10^{-3} \, m = 42 \, mm$.The distance between these two successive regions of complete darkness is $\Delta y = y_2 - y_1 = 42 - 14 = 28 \, mm$.

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