Interference fringes were produced by Young's double slit method, the wavelength of light used being 6000Å. The separation between the two slits is 2 mm.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 5 min. Find the refractive index of the material of the plate.

Explanation

Here, `d=2, cm = 2xx10^(-3)m`
`D=10 cm = 0.10 m`,
`t=0.5 mm = 0.5xx10^(-3)m`
`Delta x = 5mm = 5xx10^(-3)m`
`lambda = 6xx10^(-7)m`
As `x_(0)=(D)/(d)(mu-1)t`
`therefore mu - 1=(x_(0)d)/(D xx t)=(5xx10^(-3)xx2xx10^(-3))/(0.10xx0.5xx10^(-3))=0.2`
or `mu = 1+0.2=1.2`