JEE Mains 06-Sep-2020 Shift 1 Solved Paper

For (A)xP−xQ =(d+2.5)−(d−2.5) =5m
Δϕ due to path difference =

2π

λ

(Δx)=

2π

20

(5)=

π

2

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 =

π

2

−

π

2

(due to P being ahead of Q by 90° ) =0
IA=I1+I2+2√I1√I2‌cos‌Δ‌ϕ=I+I+2√I√I‌cos(0)=4I
For C
xQ−xP=5m
Δϕ due to path difference =

2π

λ

(Δx) =

2π

20

(5)=

π

2

Total phase difference at C=

π

2

+

π

2

=π
Inet=I1+I2+2√I1√I2‌cos‌Δ‌ϕ=I+I+2√I√I‌cos(π)=0
For B
xP−xQ=0
Δϕ=

π

2

(Due to P being ahead of Q by 90°)
IB=I+I+2√I√I‌cos‌

π

2

=2I
IA:IB:IC=4I:2I:0 =2:1:0