• Tags: YDSE, glass-slab, shift
• Difficulty: Moderate

When a glass slab of thickness t and refractive index μ is placed over one slit (say $S_{1}$), the optical path through $S_{1}$ increases by (μ-1)t. The central fringe shifts toward the slab by $\Delta y$ = (μ-1)tD/d. The number of fringes shifted = (μ-1)t/λ. Important: the fringe width β does not change (it depends on λ, D, d — not the slab). The entire fringe pattern shifts as a whole. If slabs of different thicknesses and refractive indices are placed over both slits, the net shift depends on the difference in optical paths. If both slabs produce equal optical paths, the central fringe returns to the center.