JEE Main 2026 Apr 6 Shift 2, Physics Q20: Young’s Double Slit Experiment

JEE Main2026Apr 6, Shift 2Physics

In a Young double slit experiment, the wavelength of incident light is 6000 Å, the separation between slits $S_{1}$ and $S_{2}$ is 5 cm and the distance between slits plane and screen is 50 cm, as shown in the figure below. If the resultant intensity at $P$ is equal to the intensity due to individual slits, the path difference between interfering waves is _______ Å.

A
4000
B
3000
C
2000
D
1000

Solution

Two-source interference: $I_{P} = 4 I_{0} cos^{2} (\frac{Δ ϕ}{2})$ , where $I_{0}$ is single-slit intensity.

Given $I_{P} = I_{0}$ , so $cos^{2} (\frac{Δ ϕ}{2}) = \frac{1}{4}$ $\Rightarrow$ $cos (\frac{Δ ϕ}{2}) = \frac{1}{2}$ $\Rightarrow$ $\frac{Δ ϕ}{2} = \frac{π}{3}$ .

$Δ ϕ = \frac{2 π}{3}$ , and using $Δ ϕ = \frac{2 π}{λ} Δ x$ :

$Δ x = \frac{λ}{3} = \frac{6000}{3} = 2000$ Å.

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JEE Main 2026 Apr 6 Shift 2, Physics Q20: Young’s Double Slit Experiment

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