JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Wave Optics

In a Young's double slit experiment, the intensity at some point on the screen is found to be

\frac{3}{4}

times of the maximum of the interference pattern. The path difference between the interfering waves at this point is

\frac{λ}{x}

where

λ

is wavelength of the incident light. The value of x is

____

Answer: 6

Solution

In Young's double slit experiment the intensity varies with the phase difference

ϕ

I = I_{ma x} cos^{2} \frac{ϕ}{2}

. Setting

I = \frac{3}{4} I_{ma x}

gives

cos^{2} \frac{ϕ}{2} = \frac{3}{4}

, so

cos \frac{ϕ}{2} = \frac{3}{2}

and

\frac{ϕ}{2} = \frac{π}{6}

, i.e.

ϕ = \frac{π}{3}

. The path difference corresponding to this phase difference is

Δ x = \frac{ϕ}{2 π} λ = \frac{π /3}{2 π} λ = \frac{λ}{6}

. Comparing with

\frac{λ}{x}

, the value of

x

6

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Integer answer

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