In a Young double-slit experiment, the intensity at a point on the screen is 3 /4 of the maximum intensity. If the path difference at that point is λ /x, where λ is the wavelength, find x.
Physics · Wave Optics · Young's Double-Slit Experiment (YDSE)
JEE Mains 2026 2 April Evening Shift

Question

In a Young double-slit experiment, the intensity at a point on the screen is 3 /4 of the maximum intensity. If the path difference at that point is λ /x, where λ is the wavelength, find x.

G1 Physics · Accepted Answer

For two coherent waves of equal intensity, I = I(max) cos²(φ ÷ 2). Given, I = 3I(max) ÷ 4. Therefore, cos²(φ ÷ 2) = 3 ÷ 4. For the smallest positive phase difference, cos(φ ÷ 2) = √3 ÷ 2. Hence, φ ÷ 2 = π ÷ 6, so φ = π ÷ 3. Phase difference and path difference Δ are related by φ = 2πΔ ÷ λ. Thus, Δ = φλ ÷ (2π) = (π ÷ 3)λ ÷ (2π) = λ ÷ 6. Given, Δ = λ ÷ x. Therefore, x = 6.