In a Young's double slit experiment, a combination of two glass wedges A and B, having refractive indices 1.7 and 1.5, respectively, are placed in front of the slits, as shown in the figure. The separation between the slits is d=2 mm and the shortest distance between the slits and the screen is D=2 m. Thickness of the combination of the wedges is t=12 μm. The value of l as shown in the figure is 1 mm. Neglect any refraction effect at the slanted interface of the wedges. Due to the combination of the wedges, the central maximum shifts (in mm) with respect to O by ____
Answer: 1.2
Solution
From geometry, wedge A has thickness x=3 μm and wedge B has thickness y=9 μm at the slits.
Path difference introduced:
Δ=(μA−1)x+(μB−1)y−(μB−1)x−(μA−1)y
Δ=(μA−μB)(x−y)=(1.7−1.5)(3−9)=0.2×(−6)=−1.2 μm
Shift of central maximum:
y=d∣Δ∣D=2×10−31.2×10−6×2=1.2 mm
JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer
