JEE Main 2023PhysicsWave OpticsHardNumerical

JEE Main 2023 — Wave Optics Question with Solution

JEE Main 2023 (31 Jan Shift 2)

Question

Two light waves of wavelengths 800 and 600 nm are used in Young's double slit experiment to obtain interference fringes on a screen placed 7 m away from plane of slits. If the two slits are separated by 0.35 mm, then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelength coincide will be ______ mm.

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Show full solutionCorrect answer: 48
Correct answer
48

Step-by-step explanation

Fringe width for both cases can be written as.ω1=λ1Dd & ω2=λ2Dd.

Using values of wavelength, we get ω1=16 mm & ω2=12 mm.

Let y be the common distance of the bright fringes by the both wavelength, then

y=n1ω1=n2ω2.

As LCM ω1,ω2=48 mm, therefore at y=48 mm distance both bright fringes will be found.

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About this question

This is a previous-year question from JEE Main 2023, covering the Wave Optics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.