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JEE Advanced 2025 · Paper 2, Physics question 14

In a Young's double slit experiment, a combination of two glass wedges AA and BB, 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=2d = 2 mm and the shortest distance between the slits and the screen is D=2D = 2 m. Thickness of the combination of the wedges is t=12 μt = 12\ \mum. The value of ll 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 _____.

A double-slit barrier (vertical line) with slits a distance d apart and the screen a distance D away; O is the point on the screen opposite the midpoint of the slits. In front of the slits stands a rectangular slab of thickness t made of two wedges joined along a diagonal: wedge A (top) and wedge B (bottom). The slab extends a distance l beyond each slit (above the upper slit and below the lower slit), so its total height is d + 2l; A is thickest at the top and B thickest at the bottom.
Answer and worked solution

Official answer: 1.2

Wedge thicknesses at each slit. The combination is a slab of thickness tt and height d+2l=4d+2l=4 mm. Wedge A's thickness falls linearly from tt at the top to 0 at the bottom.

  • Upper slit (1 mm from the top): A =34t=\frac34t, B =14t=\frac14t.
  • Lower slit: A =14t=\frac14t, B =34t=\frac34t.

Optical path difference.
Δ=(1.7⋅34+1.5⋅14)t−(1.7⋅14+1.5⋅34)t=0.1t=1.2 μm.\Delta=\left(1.7\cdot\tfrac34+1.5\cdot\tfrac14\right)t-\left(1.7\cdot\tfrac14+1.5\cdot\tfrac34\right)t=0.1t=1.2\ \mu\text{m}.

Shift of the central maximum.
y=ΔDd=1.2×10−6×22×10−3=1.2 mm.y=\frac{\Delta D}{d}=\frac{1.2\times10^{-6}\times2}{2\times10^{-3}}=1.2\ \text{mm}.
It shifts towards the upper slit, which has the longer optical path.

Topics:Interference and Young's double slit
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