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 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.