In Young's double-slit experiment, a point source is placed on a solid slab of refractive index 6/5 at a distance of 2 mm from two slits spaced 3 mm apart as shown and at equal distance from both the slits. The screen is at a distance of 1 m from the slits. Wavelength of light used is 500 nm.

 

At P any point on the screen, optical path difference between SS₂P and SS₁P is zero.

       ${\left({\mathrm{SS}}_2\mathrm{P}-{\mathrm{SS}}_1\mathrm{P}\right)}_{\text{optical }}=0$

$\mathrm{y}=-{\mathrm{SS}}_2(\mathrm{\mu }-1)\frac{\mathrm{D}}{\mathrm{d}}=-2.5\left(\frac{1}{5}\right)\times \frac{{10}^3}{3}=\frac{-1}{6}\mathrm{m}\left(\text{ below }\mathrm{O}\right)$

Order of fringe at O is $\frac{\mathrm{y}}{\mathrm{\lambda D}/\mathrm{d}}=\frac{1\times 3\times {10}^{-3}}{6\times 500\times 1\times {10}^{-9}}=750$

If a film of refractive index $\mathrm{\mu }=1.8$ is to be placed in front of S₁, then the path difference at the position is 200th maxima.

$\begin{array}{l}\mathrm{Then}, \mathrm{t}(\mathrm{\mu }-1)=200\mathrm{\lambda }\Rightarrow \mathrm{t}=\frac{200\mathrm{\lambda }}{(\mathrm{\mu }-1)}\\ \mathrm{t}=\frac{200\times 500\times {10}^{-9}}{(1.8-1)}=1.25\times {10}^{-3}\mathrm{m}\\ \Rightarrow \mathrm{t}=1.25\mathrm{mm}\end{array}$