A double slit experiment is performed with light of wavelength `500nm`. A thin film of thickness `2mum` and refractive index `1.5` is introduced in the path of the upper beam. The location of the central maximum will

Options

A. remain unshifted
B. shift downward by nearly two fringes
C. shift upward b nearly two fringes
D. shift downward by ten fringes

Answer

C

Explanation

(c ) When a thin film is placed in the path of one of the beams, the optical path of that beam gets longer.
The path difference is
`Delta x = (mu - 1) t`
Where `mu` is refractive index, `t` is thickness.
Given, `mu = 1.5, t = 2 xx 10^(-6) m`
`implies Delta x = (1.5 - 1) xx 2 xx 10^(-6) = 10^(-6) m = 1 mu m`
Also path difference `Delta x = (dy)/(D)`
`implies y = (D)/(d) Delta x`
Also fringe width is given by
`W = (D lambda)/(d)`
`= (D)/(d) xx 500 xx 10^(-9) = (D)/(d) xx 0.5 mu m`
`y = (D)/(d) xx 1 mu m`
`= 2 xx (D)/(d) xx (1)/(2) mu m = 2 W`
Hence, central maxima shifts upwards by nearly two fringes.
Note : When a thin film is introduced in the path of one of two interfering light beams, then the entire fringe pattern is displaced towards the beam in the path of which the film is introduced.