JEE Advanced 2025 Paper 2 · Q14 · Wave Optics
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. The combination forms a slab of uniform total thickness $t = 12$ μm with a slanted interface. The separation between the slits is $d = 2$ mm and the shortest distance between the slits and the screen is $D = 2$ m. The vertical separation $l = 1$ mm shown in the figure determines how much of each wedge sits in front of each slit (similar-triangles geometry gives $3$ μm of A and $9$ μm of B in front of the upper slit, and $9$ μm of A and $3$ μm of B in front of the lower slit). Neglecting any refraction effect at the slanted interface of the wedges, by how many millimeters does the central maximum shift with respect to O due to the combination of the wedges? __________
Reveal answer + step-by-step solution
Correct answer:1.2
Solution
Optical path length contributed by the slab in front of slit 1 ($3$ μm of A + $9$ μm of B): $(n_A - 1) \times 3 + (n_B - 1) \times 9 = 0.7 \times 3 + 0.5 \times 9 = 2.1 + 4.5 = 6.6$ μm (extra). OPL contribution at slit 2 ($9$ μm of A + $3$ μm of B): $0.7 \times 9 + 0.5 \times 3 = 6.3 + 1.5 = 7.8$ μm (extra). Net optical path difference between the two slit-paths $= 7.8 - 6.6 = 1.2$ μm. Central maximum shifts by $\Delta y = (\Delta\,\text{OPL}) D/d = 1.2 \times 10^{-6}\text{ m} \times 2\text{ m}/(2 \times 10^{-3}\text{ m}) = 1.2 \times 10^{-3}$ m $= 1.2$ mm.
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