In a double slit experiment shown in figure, when light of wavelength \(400 \mathrm{~nm}\) is used, dark fringe is observed at \(P\). If \(D=0.2 \mathrm{~m}\). the minimum distance between the slits \(S_1\) and \(S_2\) is ____________ \(mm\).

\( 2 \sqrt{\mathrm{D}^2+\mathrm{d}^2}-2 \mathrm{D}=\frac{\lambda}{2} \)

\( \therefore \sqrt{\mathrm{D}^2+\mathrm{d}^2}-\mathrm{D}=\frac{\lambda}{4} \)

\( \therefore \sqrt{\mathrm{D}^2+\mathrm{d}^2}=\frac{\lambda}{4}+\mathrm{D} \)

\( \Rightarrow \mathrm{D}^2+\mathrm{d}^2=\mathrm{D}^2+\frac{\lambda^2}{16}+\frac{\mathrm{D} \lambda}{2} \)

\( \Rightarrow \mathrm{d}^2=\frac{\mathrm{D} \lambda}{2}+\frac{\lambda^2}{16}\)

\( \Rightarrow \mathrm{d}^2=\frac{0.2 \times 400 \times 10^{-9}}{2}+\frac{4 \times 10^{-14}}{4} \)

\( \Rightarrow \mathrm{d}^2 \approx 400 \times 10^{-10} \)

\( \therefore \mathrm{d}=20 \times 10^{-5} \)

\( \Rightarrow \mathrm{d}=0.20 \mathrm{~mm}\)