Given here,
D=1m, λ=600×10−9m and n=5
Distance of nth bright fringe is given by, yn=dnλD, where, d is separation between the slits.
⇒d5×600×10−9×1=5×10−2
⇒d=5×10−25×600×10−9×1=60×10−6m
⇒d=60μm.
JEE Main 2023 — Physics Optics
In Young's double slits experiment, the position of 5th bright fringe from the central maximum is 5cm. The distance between slits and screen is 1m and wavelength of used monochromatic light is 600nm. The separation between the slits is:
Held on 25 Jan 2023 · Verified 6 Jul 2026.
60μm
48μm
12μm
36μm
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