In a Young's double slit experiment using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 microns is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the slits and screen is doubled. It is found that the distance between successive maxima now is the same as observed fringe shift upon the introduced of the mica sheet . Calculate the wavelength of the monochromatic light used in the experiment .

Explanation

Let d be the separation between slits and D be distance of screen from plane of the slits . Let `lambda` is the wavelength of light used in the experiment.
Path difference introduced at the center of screen due to thin film can be written as follows :
`Delta x = (mu - 1)t`
If `Delta x = n lambda`, then
`n = ((mu - 1)t)/(lambda)`
Shift in the pattern will be n times fringes width. Hence pattern shift can be written as follows:
Shift = `n xx (lambda D)/d = ((mu - 1)t)/(lambda) xx (lambda D)/d = ((mu - 1)tD)/(d)`
New fringe width `omega = (lambda(2D))/(d) = (2 lambda D)/(d)`
It is given that
Shift = New fringe width
`((mu -1)tD)/(d) = (2 lambda D)/(d)`
`lambda = ((mu - 1)t)/2`
`lambda = ((mu - 1)t)/(2)`
`implies lambda = ((1.5 - 1) xx 2 xx 10^(-6))/(2) = 5 xx 10^(-7)m`
`implies lambda = 5000 Å`.