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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Physics
Chapter
Wave Optics

JEE Advanced 2024 · Paper 2 · Physics · Wave Optics

In a Young's double slit experiment, each of the two slits A and B,...

In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their fixed center and with a mean separation of 0.8 mm0.8 \mathrm{~mm}. The distance between the slits at time tt is given by d=(0.8+0.04sin⁡ωt)mmd=(0.8+0.04 \sin \omega t) \mathrm{mm}, where ω=0.08rads−1\omega=0.08 \mathrm{rad} \mathrm{s}^{-1}. The distance of the screen from the slits is 1 m1 \mathrm{~m} and the wavelength of the light used to illuminate the slits is 6000A~…6000 Ã…. The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point OO. In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their The maximum speed in μm/s\mu \mathrm{m} / \mathrm{s} at which the 8th 8^{\text{th }} bright fringe will move is ________ .

Correct answer

0

Step-by-step solution

y=n⋅λDdv=dydt=−n⋅λ⋅dd2⋅d(d)dtd=0.8+0.04sin⁡ωtd(d)dt=0.04ωcos⁡ωt\begin{aligned} & \mathrm{y}=\mathrm{n} \cdot \frac{\lambda \mathrm{D}}{\mathrm{d}} \\ & \mathrm{v}=\frac{\mathrm{dy}}{\mathrm{dt}}=-\mathrm{n} \cdot \frac{\lambda \cdot \mathrm{d}}{\mathrm{d}^2} \cdot \frac{d(d)}{d t} \\ & d=0.8+0.04 \sin \omega t \\ & \frac{d(d)}{d t}=0.04 \omega \cos \omega t \end{aligned}\text{\text{ for } } \mathrm{v} \rightarrow \max \Rightarrow \frac{\mathrm{d}(\mathrm{d})}{\mathrm{dt}} \rightarrow \max \text{. }ForFor\frac{\mathrm{d}(\mathrm{d})}{\mathrm{dt}} \rightarrow \max..cos⁡ωt=1⇒sin⁡ωt=0⇒(d(d)dt)max⁡=0.04⇒d=0.8 mmvmax⁡=8×6000×10−10×1×0.04×0.080.8×0.8×10−6×10−3=24μm/s.\begin{aligned} & \cos \omega \mathrm{t}=1 \Rightarrow \sin \omega \mathrm{t}=0 \\ & \Rightarrow\left(\frac{\mathrm{d}(\mathrm{d})}{\mathrm{dt}}\right)_{\max }=0.04 \\ & \Rightarrow \mathrm{d}=0.8 \mathrm{~mm} \\ & \mathrm{v}_{\max }=\frac{8 \times 6000 \times 10^{-10} \times 1 \times 0.04 \times 0.08}{0.8 \times 0.8 \times 10^{-6} \times 10^{-3}}=24 \mu \mathrm{m} / \mathrm{s} . \end{aligned}$ (range is from 23.5 to 24.5)
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