In young double slit experiment λ=500nm,d=1mm and D=1m .The minimum distance from the central maximum where intensity is half of the maximum intensity is (assume intensity from each slit is same)

Given Wavelength  λ=500 nm=5×10−7 mDistance between two slits  d=1 mm=10−3 mDistance between slits and screen  D=1 mMaximum intensity  =4I0I=4Iocos2ϕ2⇒2Io=4Iocos2ϕ2⇒cosϕ2=12∴ϕ=π2Path difference Δϕ=2πλΔx⇒Δx=λ4∴y=Δx.Dd=λD4d=5×10−7×14×10−3=1.25×10−4 m