JEE Main 2022PhysicsWave OpticsMediumNumerical

JEE Main 2022 — Wave Optics Question with Solution

JEE Main 2022 (24 Jun Shift 1)

Question

Sodium light of wavelengths 650 nm and 655 nm is used to study diffraction at a single slit of aperture 0.5 mm. The distance between the slit and the screen is 2.0 m. The separation between the positions of the first maxima of diffraction pattern obtained in the two cases is _____ ×10-5 m

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Show full solutionCorrect answer: 3
Correct answer
3

Step-by-step explanation

The condition for maxima in diffraction is,

bsinθ=2n+1λ2.

 For the first maxima n=1. Therefore,

bsinθ=3λ2⇒sinθ=3λ2b=yDy=3λD2b

Now the difference between the positions of the maxima will be,

y1-y2=3D2bλ1-λ2⇒y1-y2=3×22×5×10-4×655-650×10-9=3×10-5 m

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About this question

This is a previous-year question from JEE Main 2022, 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.