To solve this problem we will apply the concepts related to destructive interference from double-slit experiments. For this purpose we will define the path difference as,
[tex]\text{Path difference}= dsin\theta = (2n-1)\frac{\lambda}{2}[/tex]
Here,
[tex]\lambda[/tex] = Wavelength
[tex]\theta[/tex] = Angle when occurs the interference point of destructive interference
Our values are given as,
[tex]\text{Wavelength} = \lambda = 400nm = 4*10^{-7}m[/tex]
[tex]\text{Distance of Screen} = D = 1m[/tex]
Using the previous expression we have,
[tex]d \times \theta = \frac{\lambda}{2}[/tex]
[tex]d \times (0.1) = \frac{4*10^{-7}}{2}[/tex]
[tex]d = 2*10^{-6} m[/tex]
[tex]d = 2\mu m[/tex][tex]d = 2\mu m[/tex]
Therefore the distance between the two openings is [tex]2\mu m[/tex]
