JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Wave Optics

In a Young double slit experiment, the wavelength of incident light is

6000 \overset{˚}{A}

, the separation between slits

S_{1}

and

S_{2}

is 5 cm, 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✓ Correct
(D)1000

Answer: (C)

Solution

For two equal-intensity beams the resultant intensity is

I = 4 I_{0} cos^{2} \frac{Δ ϕ}{2}

. Setting this equal to the intensity of a single slit,

I = I_{0}

, gives

cos \frac{Δ ϕ}{2} = \frac{1}{2}

, so

\frac{Δ ϕ}{2} = \frac{π}{3}

and

Δ ϕ = \frac{2 π}{3}

. The path difference is

Δ x = \frac{λ}{2 π} Δ ϕ = \frac{λ}{3} = \frac{6000}{3} = 2000 \overset{˚}{A}

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Wave Optics

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