Interference fringes were produced by Young's double slit method, the wavelength of light used being 6000 Å. The separation between the two slits is 2mm. The distance between the slits and screen is 10 cm. When a transparent plate of thickness 0.5 mm is placed over one of the slits, the fringe pattern is displaced by 5mm. The refractive index of the material of the plate is:

1.20

Here 2d = 2 mm = 2 × 10^-3 m, D = 10 cm = 0.10 m,

t = 0.5 mm = 0.5 × 10^-3 m, Δx = 5 mm = 5 × 10^-3 m, λ = 6 × 10^-7 m

As $x_0=\frac{D}{2d}(μ-1)t$

$∴μ-1=\frac{Δx(2d)}{Dt}=\frac{5 × 10^{-3}×2×10^{-3}}{0.10×0.5×10^{-3}}=0.2$ or $μ = 1 + 0.2 = 1.2$