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 = (lamda D)/(d)` Where , D is the distance between the screen and slit and dis the distance between two slits.
When a film of thickness t and refractive index `mu` is
Placed over one of the slits, the fringe pattern is shifted by
Distance S and is given by
`s = (( mu - 1) t D)/( d)`
Given: `S = 20 beta`
From equation (i), (ii) and (iii) we get ,
`( mu -1) t = 20 lamda`
`or ( mu -1) = ( 20lamda )/( t) = (20 xx 5000 xx 10 ^(-8) cm)/( 2.5 xx 10 ^(-3) cm)`
`mu -1 = 0.4 or mu = 1.4`