JEE Main 2022PhysicsWave OpticsMediumNumerical

JEE Main 2022 — Wave Optics Question with Solution

JEE Main 2022 (29 Jun Shift 2)

Question

In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by 5×10-2 m towards the slits, the change in fringe width is 3×10-3 cm. If the distance between the slits is 1 mm, then the wavelength of the light will be____nm.

Enter your answer

Show full solutionCorrect answer: 600
Correct answer
600

Step-by-step explanation

We know that the fringe width in YDSE is β=λDd

⇒Δβ=λdΔD

⇒λ=ΔβdΔD=3×10-3×10-2×10-35×10-2=6×10-7 m=600 nm

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Wave Optics chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

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.