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 by nearly two fringes
D. shift downward by ten fringes

Answer

C

Explanation

When a thin film is placed in the path of one of the beams, then the optical path of that beam gets longer.
The path difference is, `Delta x=(mu-1)t`
where, `mu` is refractive index and t is thickness.

Given, `mu=1.5, t=2xx10^(-6)m`
`therefore Delta x=(1.5-1)xx2xx10^(-6)=10^(-6)m=1mu m`
Also, path difference `Delta x=(dy)/(D) rArr y=(D)/(d)Delta x`
Also fringe width is given by
`beta=(D lambda)/(d)=(D)/(d)xx500xx10^(-9)=(D)/(d)xx0.5 mu m`
`rArr y=(D)/(d) xx1mu m= 2xx(D)/(d) xx (1)/(2)mu m=2beta`