- S1R-S2R=-2.5λ or S2R - S1R =2.5λ
- This means the waves from S1 will take exactly 2.5λ cycles earlier than S2.
- It is non integral multiple of λ.
- Let the light wave emitted by wave at S1 , y1=a cosωt
- Where a=amplitude of the wave,y1 =displacement of the wave and cosωt =phase.
- Light wave emitted by S2, y2=acos(ωt+5π) = -a cosωt
- (Path difference)λ =>2π(phase difference),therefore 2.5λ=5 π.
- Therefore y=y1+ y2 =a cosωt – a cosωt =0.
- Resultant Intensity I=0.
- This shows when the path difference is non integral of λ then destructive overlap takes place and resultant intensity is 0.
- Path difference=(n+(1/2)) λ
Coherent & Incoherent Addition of waves
- Consider 2 waves; first wave y1=a cosωt and second wave y2=a cos (ωt+Φ)
- WhereΦ =phase difference between the 2 waves.
- Intensity of both the waves = I0 and I0∝a2
- Resultant displacement y=y1+ y2 =a cosωt + a cos(ωt + Φ)
- =2acos(Φ/2) + cos(ωt+(Φ/2))
- [using cosA + cosB=2cos((A+B)/2) +cos((A-B)/2)]
- Where amplitude = 2acos(Φ/2)
- Also I ∝4a2 cos2(Φ /2) (becauseI ∝ (Amplitude)2)
- I =4I0 cos2 (Φ/2) where a2=I0.
1.In case of constructive overlap path difference =nλ and phase difference
Φ=0, (+-)2π, (+-)4π…
- In case of destructive overlap path difference =(n+(1/2))nλ and phase difference =(+-) 3π,(+-) 5π…