Question 14

**Setting up the geometry** The two wedges together form a rectangular slab of thickness t = 12 μm whose height spans from l above the upper slit to l below the lower slit, i.e. total height H = d + 2l = 2 + 2(1) = 4 mm. Wedge A (n = 1.7) has full thickness t at the top and zero at the bottom; wedge B (n = 1.5) is the complement. So at height y measured from the bottom: - thickness of A: t·(y/H) - thickness of B: t·(1 − y/H) **Thicknesses at the slits** - Upper slit (y = 3 mm): A = 12·(3/4) = 9 μm, B = 3 μm - Lower slit (y = 1 mm): A = 3 μm, B = 9 μm **Optical path through the slab at each slit** - Upper slit: 1.7(9) + 1.5(3) = 15.3 + 4.5 = 19.8 μm - Lower slit: 1.7(3) + 1.5(9) = 5.1 + 13.5 = 18.6 μm Extra optical path at the upper slit: Δ = 19.8 − 18.6 = 1.2 μm. **Shift of central maximum** The central maximum moves toward the slit with the larger optical path, by y = (D/d)·Δ = (2 m / 2×10⁻³ m)(1.2×10⁻⁶ m) = 1.2×10⁻³ m = 1.2 mm.

No written reasoning was returned for this item.