JEE Main 2026 Physics question
The maximum intensity in a Young's double slit experiment is I₀. Distance between the slits (d) is 5λ, where λ is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at D=10d is .
Answer: (B)
Solution
y=d/2. Path difference δ=yd/D=(d²/2)/D=(25λ²/2)/(50λ)=λ/4, giving phase π/2. I=I₀cos²(π/4)=I₀/2.