Question

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

Answer 1

The minimum distance between the slits S₁ and S₂ is 0.20 mm

Path difference for minima at P

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

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

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

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

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

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

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

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

\(\Rightarrow d = 0.20\, mm\)