JEE Main 2026 · 6 Apr · Shift 2 · Physics Q45

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PhysicsSingle correct+4 / -1Question 45 of 75

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.

Source S (λ = 6000 Å) on the axis lights slits S₁ and S₂; the screen is 50 cm from the slit plane, with the centre O′ on the axis and rays from S₁ and S₂ meeting at a point P below O′.

4000

3000

2000

1000

Answer and worked solution

Official answer: A

Condition. With two equal sources of intensity $I_1$ :
$I=2I_1(1+\cos\phi)=I_1\ \Rightarrow\ \cos\phi=-\tfrac12.$

Phase. $\phi=\tfrac{2\pi}3$ or $\tfrac{4\pi}3$ , so the path difference is $\tfrac\lambda3=2000$ Å or $\tfrac{2\lambda}3=4000$ Å.

Key: both (A) and (C) are accepted.

Topics:Interference and Young's double slit

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JEE Main 2026 · 6 Apr · Shift 2, Physics question 45