JEE Main 2023PhysicsWave OpticsHardNumerical

JEE Main 2023Wave Optics Question with Solution

JEE Main 2023 (06 Apr Shift 2)

Question

A beam of light consisting of two wavelengths 7000 Ao and 5500 Ao is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is 2.5 mm and the distance between the plane of slits and the screen is 150 cm. The least distance from the central fringe, where the bright fringes due to both the wavelengths coincide, is n×105 m. The value of n is _____.

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Show full solutionCorrect answer: 462
Correct answer
462

Step-by-step explanation

The given data is

λ1=7000 Ao

λ2=5500 Ao

d=2.5×10-3 m

D=1.5 m

The path difference is given by 

nλ1=mλ2

7n=5.5 m

14n=11 mn=11 and m=14

The formula for the distance of a bright fringe is

y=nλ1Dd

y=11×7×10-7×1.52.5×10-3

=46.2×10-4=462×10-5

It is given that 

n×10-5=462×10-5n=462

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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.