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Physics · Wave Optics

JEE Advanced 2025 — Paper 2 — 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=2 mmd=2 \mathrm{~mm} and the shortest distance between the slits and the screen is D=2 mD=2 \mathrm{~m}. Thickness of the combination of the wedges is t=12μ mt=12 \mu \mathrm{~m}. 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 \qquad

figure

Answer: 1.2

Numerical answer — enter this value.

Step-by-step solution

figure

x+y=12μmx+y=12 \mu m

412=1x\frac{4}{12}=\frac{1}{x}

x=3μ m\mathrm{x}=3 \mu \mathrm{~m}

y=6μmy=6 \mu m

∴Δ=ydD−(μB−1)x−(μA−1)y+(μB−1)y+(μA−1)x\therefore \Delta=\frac{y d}{D}-\left(\mu_{B}-1\right) x-\left(\mu_{A}-1\right) y+\left(\mu_{B}-1\right) y+\left(\mu_{A}-1\right) x

−ydD=−0.5×3−0.7×9+0.5×9+0.7×3\frac{-y d}{D}=-0.5 \times 3-0.7 \times 9+0.5 \times 9+0.7 \times 3

−ydD=−0.5×6−0.7×6\frac{-y d}{D}=-0.5 \times 6-0.7 \times 6

⇒−ydD=−1.2μm\Rightarrow \frac{-y d}{D}=-1.2 \mu m

⇒y=1.2×Dd=1.2×22×10−3×10−6\Rightarrow \mathrm{y}=\frac{1.2 \times \mathrm{D}}{\mathrm{d}}=\frac{1.2 \times 2}{2 \times 10^{-3}} \times 10^{-6}

=1.2 mm=1.2 \mathrm{~mm}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2025
Paper
Paper 2
Subject
Physics
Chapter
Wave Optics
Topic
Young's Double Slit Experiment and Its Modifications
In a Young's double slit experiment, a combination of two glass… | JEE Advanced 2025 PYQ with Solution · DhiX AI