In a Young's double slit experiment, the intensity at some point on the screen is found to be 3 4 times of the maximum of the interference pattern.…
JEE Main 2026 — Physics Optics
2026integermedium
In a Young's double slit experiment, the intensity at some point on the screen is found to be 43 times of the maximum of the interference pattern. The path difference between the interfering waves at this point is xλ where λ is wavelength of the incident light. The value of x is _______.
Official previous-year question
Held on 2 Apr 2026 · Verified 6 Jul 2026.
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Solution
The intensity at a point in Young's double slit experiment is given by I=Imaxcos2(2ϕ)
Given I=43Imax
43Imax=Imaxcos2(2ϕ)
cos(2ϕ)=23
2ϕ=6π
ϕ=3π
The phase difference ϕ is related to the path difference Δx by ϕ=λ2πΔx
λ2πΔx=3π
Δx=6λ
Comparing with Δx=xλ, we get x=6
Answer: 6
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