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

Shift `= Delta beta=(D)/(d) (mu-1) t=(beta)/(lambda)(mu-1)t`
Thus, `20 beta=(beta)/(lambda)(mu-1) :.(20 lambda)/(t)=(mu-1)`
`:. mu=1+(20xx5x10^(-7))/(2.5xx10^(-5))=1.4`