$(a)\;20\;mm \\ (b)\;40\;mm \\ (c)\;60\;mm \\ (d)\;80\;mm $

The fringes disappear when the maxima of $ \lambda_1$ fall over minima of $\lambda_2$.

$\large\frac{p}{\lambda_1}-\frac{p}{\lambda_2} =\frac{1}{2}$

Where p is the optical path difference at that point.

or $p=\large\frac{\lambda_1 \lambda_2}{2( \lambda_2 - \lambda_1)}$

Here $\lambda_1=4000\;A^{\circ}, \lambda_2= 4002 A^{\circ}$

$\therefore p=0.04\;cm$

in YDSE

$p= \large\frac{d y}{D}$

$\therefore y= \large\frac{D}{d}$$p$

$\qquad= \large\frac{(1000)}{10}$$(0.4) =40\;mm$

Hence b is the correct answer.

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