Fringe width,
β=λDd ⇒ β∝ λ
When the wavelength is decreased from 600 nm to 400 nm, fringe width will also decrease by a factor of 46 (or) 23.
Hence, the number of fringes in the same segment will increase by a factor of 32.
∴ The number of fringes observed in the same segment =12×32=18