A double slit experiment is performed with light of wavelength 500 nm. A thin film of thickness 2 mm 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

Wavelength `lamda = 500 nm = 500 xx 10^(-9)m`
Thickness of the film, t= 2 `mum = 2xx 10^(-6)m`
Refractive index `mu=1.5`
When there is no thin placed in the path of any one of the two beams, the path difference between them is given by dy/D ( considering the two beams meetings at a point P). In case we put a thin film in the path of one of the beams , then the optical path of that beam gets longer. Now for the central maximum, path difference in absence of the film is `Deltax=0`. But when we put the film , the path difference becomes
`Delta = mut - t =(mu-1)t`
`=(1.5-1)xx2xx10^(-6)=10^(-6)m=1 mm`
Now, `Deltax=(dv)/(D) rArr y= D/d Delta x =D/dxx1 mum`
and the fringe width is given by
`W=(Dlamda)/(d) = D/d xx 500xx10^(-9) m = D/dxx0.5xx1mum`
`therefore y=D/dxx1mum=2xxD/dxx1/2xx1mum=2W`
As the film is placed in the path of the upper beam, the central maximum will shift upward by nearly two fringes.