A two slit Young's experiment is done with monochromatic light of wavelength 6000 Å. Slits are 2 mm apart and fringes are observed on a screen placed 10 cm away from the slits. If a transparent plate of thickness 0.5 mm is placed in front of one of the slit, interference pattern shifts by 5 mm. Then refractive index of transparent plate should be 

1.2

$ (\mu -1)t = \frac{xd}{D} $

$ \mu = 1 + \frac{xd}{tD} = 1 + \frac{0.5\times 0.2 }{10\times 0.05} = 0.2 $

$ \mu = 1.2$