Class 12 Physics Wave Optics | Constructive Overlap |

**Constructive Overlap**

- Case 1:-
- Consider two coherent sources S
_{1}and S_{2}emitting light waves of same frequency and constant phase. - The wave fronts of both the sources will overlap with each other.
- Consider a point P as in the figure to calculate the intensity of disturbance.
- The distance of point P from S
_{1}and S_{2}is same. Therefore S_{1}P=S_{2} - 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 at S
_{2}y_{2}=acosωt - Intensity of both the waves =I
_{0}∝a^{2}(equation (1)) - Resultant displacement of the wave formed by the superposition of the waves
**y=y**_{1}+y_{2}=2acosωt

- Intensity I ∝ (Amplitude)
^{2} - I ∝ (2a)
^{2}=> I ∝4a^{2}where Amplitude=2a. **I=4 I**using equation(1)_{0}- This means the intensity at point P will be four times the intensity of the individual sources.

- Consider two coherent sources S

**Conclusion**: -- If a point is equidistant from two sources then the
- Amplitude as well as the intensity increases.

__Path difference__is defined as the difference in the paths from both the sources to a particular point.- This implies S
_{2}P - S_{1}P =0. - If the path difference is 0 then it will be constructive overlap.

- This implies S
- Case 2:- Considering a point Q which is not equidistant from the 2 sources and the path difference S
_{1}Q - S_{2}Q = 2λ(integral multiple)- => As S
_{1}Q > S_{2}Q therefore the waves originating from S_{1}have to travel a greater path than S_{2}. - Therefore waves from S
_{2}will reach exactly 2 cycles earlier than waves from S_{1}. Waves reach at S_{2}early by 2λ as compared to S_{1}. - One cycle corresponds to λ and two cycles correspond to 2λ.
- 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 at S
_{2}, y_{2}=acos(ωt -4 π) =a cosωt- (Path difference)λ =>2 π (phase difference),therefore 2λ=4π
*.* - This shows y
_{1}and y_{2}are in phase with each other. - Resultant y=y
_{1}+ y_{2}=2acosωt.

- (Path difference)λ =>2 π (phase difference),therefore 2λ=4π
- This shows constructive overlap happened when the path difference is 0 or when it is 2λ.
- The intensity
**I =4I**._{0} **Path difference =n λ**; where n=0, 1, 2, 3…

- => As S

- If a point is equidistant from two sources then the

.