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=(lambda D)/(d) " " ` … (i)
where D is the distance between 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=20beta " " ` ... (iii)
From equations (i), (ii) and (iii) we get,
`(mu-1)t = 20lambda`
or `(mu-1)=(20lambda)/(t)=(20xx5000xx10^(-8)cm)/(2.5xx10^(-3)cm)`
`mu-1 =0.4 " or " mu=1.4`