Position of the nth bright fringe w.r.t. central maxima in a YDSE is yn=ndλD. ⇒y10y10′=λλ′ or y10′=λλ′y10 =(600 nm660 nm)10 mm=11 mm
A double slit interference experiment performed with a light of wavelength 600 nm forms an interference fringe pattern on a screen with 10 th bright…
JEE Main 2025 — Physics Optics
A double slit interference experiment performed with a light of wavelength 600 nm forms an interference fringe pattern on a screen with 10 th bright fringe having its centre at a distance of 10 mm from the central maximum. Distance of the centre of the same 10th bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660 nm would be _________ mm .
Held on 28 Jan 2025 · Verified 6 Jul 2026.
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