YDSE is conducted using light of wavelength 6000 Å to observe interference pattern. When a film of material `3.0 xx 10^(-3) cm` thick was placed over one of the slits, the fringe pattern shifted by a distance equal to 10 fringe widths. The refractive index of the material of the film is _ _ _ _ _ .

Answer

`1.2`

Explanation

Fringe width, `beta = (lambdaD)/d` . . . (i)
where, D: distance between screen and slit d: distance between two slits when a film of thickness t and refractive index iis placed over one of the slit, the fringe pattern is shift by distance S and is given by
`S = ((mu -1)+D)/d` . . . (ii)
Given: `S = 10 beta` . . . (iii)
From equations (i), (ii) and (iii), we get
`((mu-1)+D)/d = 10 (lambdaD)/d`
`implies mu -1 = (10 lambda)/t =(10xx6000xx10^(-9)cm)/(3xx10^(-3) cm)`
`implies mu-1 = 0.2`
`implies mu = 1.2`