A double slit experiment is performed with light of wavelength 500 nm. 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 by nearly two fringes.
D. shift downward by 10 fringes

Answer

C

Explanation

Path difference produced due to film,
`Deltax= mut-t`
`=(nu-1)t`
`=(1.5-1)xx2xx10^(-6)`
`=1mu m`
Now `Deltax=(yd)/(D) y=(Deltax)/(d)`
`y=(D)/(d)xx1 mu m`
Width of fringe `W=(dlambda)/(d)`
`=(D)/(d)xx500xx10^(-9)m`
`=(D)/(d)xx0.5xx1 mum`
`:.y=(D)/(d)xx1 mum`
`=2xx(D)/(d)xx(1)/(2)xx1 mu =2W`
Hence, the central fringe will shift upward by nearly two fringes.