In Young's double-slit experiment, two glass wedges A and B with refractive indices 1.7 and 1.5 are placed in front of the slits as shown. Slit separation d = 2 mm, screen distance D = 2 m, total wedge thickness t = 12 μm, and l = 1 mm. Neglect refraction at the slanted interface. Find the shift of the central maximum in mm. Image
Physics · Circular Motion
JEE Advanced 2025 Paper 2

Question

In Young's double-slit experiment, two glass wedges A and B with refractive indices 1.7 and 1.5 are placed in front of the slits as shown. Slit separation d = 2 mm, screen distance D = 2 m, total wedge thickness t = 12 μm, and l = 1 mm. Neglect refraction at the slanted interface. Find the shift of the central maximum in mm. Image

Physics · Circular Motion

JEE Advanced 2025 Paper 2

G1 Physics · Accepted Answer

From the wedge geometry, the difference in thicknesses of the two materials intercepted by the two slit rays is Δt = t(l/d) = 12 μm × (1/2) = 6 μm. The optical path difference is δ = (1.7 − 1.5)(6 μm) = 1.2 μm. Central-fringe shift: y = Dδ/d = (2)(1.2 × 10⁻⁶)/(2 × 10⁻³) = 1.2 × 10⁻³ m = 1.2 mm.