Interference fringes were produced in Young's double slit experiment using light of wavelength 5000 Ã…. When a film of material `2.5xx 10^(-3) cm` thick was placed over one of the slits, the fringe pattern shifted by a distance equal to 20 fringe widths. The refractive index of the material of the film is

Options

A. `1.25`
B. `1.33`
C. `1.4`
D. `1.5`

Answer

C

Explanation

Fringe width, `beta=(lambdaD)/(d)` …(i)
where D is the distance the screen and slit and d is the distance between two slits.
When a film of thickness t and refractive index `mu` is placed over one of the slit, the fringe pattern is shifted by distance S and is given by
`S=((mu-1)tD)/(d)` ...(ii)
Given : `S=20 beta` ...(iii)
From equations (i), (ii) and (iii) we get,
`(mu-1)t=20lambda`
or `(mu-1)=(20lambda)/(t)=(20xx5000xx10^(-8))/(2.5xx10^(-3)cm)`
`mu-1=0.4 "or" mu=1.4`