Class 12 Physics Wave Optics Destructive overlap

Destructive overlap

  • 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.

Conclusion:-

            1.In case of constructive overlap path difference =nλ and phase difference

Φ=0, (+-)2π, (+-)4π…

  1. In case of destructive overlap path difference =(n+(1/2))nλ and phase difference =(+-) 3π,(+-) 5π…

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