xP−xQ=(d+2.5)−(d−2.5)=5mΔϕ due to path difference =λ2π(Δx)=202π(5)=2π At A,Q is ahead of P by path, as wave emitted by Q reaches before wave emitted by P. Total phase difference at A=2π−2π (due to P being ahead of Q by 90∘ ) =0IA=I1+I2+2I1I2cosΔϕ=I+I+2IIcos(0)=4I For C xQ−xP=5mΔϕ due to path difference =λ2π(Δx)=202π(5)=2π Total phase difference at C=2π+2π=πInet=I1+I2+2I1I2cosΔϕ=I+I+2IIcos(π)=0 For B xP−xQ=0Δϕ=2π (Due to P being ahead of Q by 90∘) IB=I+I+2IIcos2π=2IIA:IB:IC=4I:2I:0=2:1:0
