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JEE Advanced 2024 · Paper 2, Physics question 15

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PARAGRAPH I

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=0.08 rad s−1\omega = 0.08\,\mathrm{rad\,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 6000 A˚6000\,\text{Å}. The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point O.

Young's double slit set-up: parallel light (arrows from the left) falls on a barrier with two slits, B (upper) and A (lower), separated by d (marked with a small vertical double arrow). A screen is 1 m to the right; the central point O on the screen lies on the dashed line through the midpoint of the slits.

The maximum speed in μ\mum/s at which the 8th8^{\text{th}} bright fringe will move is _____.

Answer and worked solution

Official answer: 23.5

Differentiate the fringe position:
dy8dt=−8λDd2dddt,dddt=0.04×10−3⋅0.08cos⁡ωt=3.2×10−6cos⁡ωt m/s.\frac{dy_8}{dt}=-\frac{8\lambda D}{d^2}\frac{dd}{dt},\qquad \frac{dd}{dt}=0.04\times10^{-3}\cdot0.08\cos\omega t=3.2\times10^{-6}\cos\omega t\ \mathrm{m/s}.

The speed is largest where ∣cos⁡ωt∣=1|\cos\omega t|=1, at which point d=0.8d=0.8 mm:
vmax≈4.8×10−6×3.2×10−6(0.8×10−3)2=2.4×10−5 m/s=24 μm/s.v_{max}\approx\frac{4.8\times10^{-6}\times3.2\times10^{-6}}{(0.8\times10^{-3})^2}=2.4\times10^{-5}\ \mathrm{m/s}=24\ \mu\mathrm{m/s}.

The small variation of dd moves this by less than 1%1\% (to about 24.124.1).

Topics:Interference and Young's double slitMotion in a straight line
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