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JEE Main 2015 — Wave Optics Question with Solution

JEE Main 2015 (11 Apr Online)

Question

In a Young's double slit experiment with light of wavelength λ, the separation of slits is d and distance of screen is D such that D≫d≫λ . If the Fringe width is β , the distance from point of maximum intensity to the point where intensity falls to half of the maximum intensity on either side is:

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Show full solutionCorrect option: A
Correct answer
Aβ4

Step-by-step explanation

As Imax=4I

Then Inet=12 Imax=2I

⇒2I=2I1+cosϕ

⇒1+cosϕ=1

⇒cosϕ=0

⇒ϕ=π2,3π2,5π2, …

 ϕ=2πλΔx

⇒π2=2πλΔx

⇒Δx=λ4,3λ4,5λ4, …



Also path difference Δx=dsin⁡θ =λ4

At desired location

⇒d ynD=λ4

⇒ yn=λD4d=β4 ∵β=λDd

⇒ Separation between central maxima and desired y=β4

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About this question

This is a previous-year question from JEE Main 2015, 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.