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JEE Advanced 2024 · Paper 2, Physics question 14

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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 8th8^{\text{th}} bright fringe above the point O oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer (μ\mum), is _____.

Answer and worked solution

Official answer: 598

The nn-th bright fringe is at yn=nλDdy_n=\dfrac{n\lambda D}{d}. For n=8n=8:
nλD=8×6×10−7×1=4.8×10−6 m2.n\lambda D=8\times6\times10^{-7}\times1=4.8\times10^{-6}\ \mathrm{m^2}.

dd swings between 0.760.76 mm and 0.840.84 mm, so
Δy=4.8×10−6(10.76×10−3−10.84×10−3)=4.8×10−3×0.1253 m≈601.5 μm.\Delta y=4.8\times10^{-6}\left(\frac{1}{0.76\times10^{-3}}-\frac{1}{0.84\times10^{-3}}\right)=4.8\times10^{-3}\times0.1253\ \mathrm m\approx601.5\ \mu\mathrm m.

Topics:Interference and Young's double slit
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