In Young's double slit experiment, monochromatic light of wavelength 5000overset° A is used. The slits are 1.0 mm apart and screen is placed at 1.0m…
JEE Main 2024 — Physics Optics
2024integermedium
In Young's double slit experiment, monochromatic light of wavelength 5000A∘ is used. The slits are 1.0mm apart and screen is placed at 1.0m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is ______×10−6m.
Official previous-year question
Held on 1 Feb 2024 · Verified 6 Jul 2026.
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Solution
Let intensity of light on screen due to each slit is I0.
So internity at centre of screen is 4I0(as cosϕ=1).
Intensity at distance y from centre,
I=I0+I0+2I0I0cosϕ
As we know, Imax=4I0. Therefore,
⇒2Imax=2I0=2I0+2I0cosϕ
⇒cosϕ=0
⇒ϕ=2π
Hence, kΔx=2π
⇒λ2π(dsinθ)=2π
⇒λ2d×Dy=21
⇒y=4dλD=4×10−35×10−7×1
=125×10−6
=125
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