JEE Main 2024PhysicsWave OpticsMediumMCQ

JEE Main 2024 — Wave Optics Question with Solution

JEE Main 2024 (29 Jan Shift 2)

Question

In Young's double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is 7λ4. The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is:

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Show full solutionCorrect option: A
Correct answer
A12

Step-by-step explanation

Given the path difference is ∆x=7λ4

Hence, the phase difference can be found as follows:

ϕ=2πλ∆x=2πλ×7λ4=7π2

The formula to calculate the intensity is given by

I=Imaxcos2ϕ2   ...1

From equation (1), it follows that

IImax=cos2ϕ2=cos27π2×2=cos27π4=cos22π-π4=cos2π4=12

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About this question

This is a previous-year question from JEE Main 2024, 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.