The correct answer is: 600 nm
Distance between the slits, \(d = 0.28 mm = 0.28 × 10^{ -3} m\)
Distance between the slits and the screen, D = 1.4 m
Distance between the central fringe and the fourth (n = 4) fringe,
\(u = 1.2 cm = 1.2 × 10^{−2} m\)
In case of a constructive interference, we have the relation for the distance between the two fringes as:
\(u = nλ \frac{D}{d}\)
Where,
n = order of fringes = 4
\(λ\) = Wavelength of light used
\(∴ λ = \frac{ud}{nd}\)
\(= \frac{1.2 ×10^{-2} × 0.28 × 10^{-3} }{ 4 ×1.4}\)
\(= 6 ×10^{-7}\)
= 600 nm
Hence, the wavelength of light is 600 nm.