Class 12 Physics Wave Optics | Destructive overlap |

**Destructive overlap**

- S
_{1}R-S_{2}R=-2.5λ or S_{2}R - S_{1}R =2.5λ - This means the waves from S
_{1}will take exactly 2.5λ cycles earlier than S_{2}. - It is non integral multiple of λ.
- Let the light wave emitted by wave at S
_{1}, y_{1}=a cosωt- Where a=amplitude of the wave,y
_{1}=displacement of the wave and cosωt =phase.

- Where a=amplitude of the wave,y
- Light wave emitted by S
_{2}, y_{2}=acos(ωt+5π) = -a cosωt- (Path difference)λ =>2π(phase difference),therefore 2.5λ=5 π
*.*

- (Path difference)λ =>2π(phase difference),therefore 2.5λ=5 π
- Therefore y=y
_{1}+ y_{2}=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

- Let the light wave emitted by wave at S

__Coherent & Incoherent Addition of waves__

- Consider 2 waves; first wave y
_{1}=a cosωt and second wave y_{2}=a cos (ωt+Φ)- WhereΦ =phase difference between the 2 waves.

- Intensity of both the waves = I
_{0}and I_{0}∝a^{2} - Resultant displacement y=y
_{1}+ y_{2}=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 ∝4a
^{2}cos^{2}(Φ /2) (becauseI ∝ (Amplitude)^{2}) **I =4I**_{0}cos^{2 }(**Φ****/2) where**a^{2}=I_{0}.

**Conclusion:-**

** **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…

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