Question

In Youngs double slit experiment using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of $R.l.$ and thickness $1.964$ micron is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the slits and the screen is doubled, it is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. Calculate the wavelength of monochromatic light used in the experiment.

Answer 1

The fringe shift $\Delta s$ due to the introduction of any sheet of thickness $t$ and refractive index $n$ in the path of any of the interfering waves is given by,
$\Delta s=(n-1)t\frac{D}{2d}$
Due to change of distance of separation between the plane of the slits and screen, the fringe width is given by $\frac{\lambda \times 2D}{2d}$
According to the statement of the problem,
$\frac{\lambda \times 2D}{2d}=(n-1)t\frac{D}{2d}$
$\therefore \lambda =(n-1)t/2$
$=(1.6-1)\frac{1.964\times {10}^{-6}}{2}=5892\mathring{A}$