Question

Transcribed Image Text:### Phasor Diagram Analysis
**Description:**
The diagram illustrates a basic setup for analyzing wave interference from three point sources of light. These sources are positioned linearly a distance \( D \) from a screen.
At point \( P \) on the screen, directly in front of the sources, the waves are in phase. This is depicted by a phasor diagram showing three phasors \( A_1 \), \( A_2 \), and \( A_3 \) combined to form a single resultant phasor \( A_{\text{net}} \).
#### Phasor Diagram Explanation:
- **Phasors \( A_1 \), \( A_2 \), \( A_3 \):** These represent the amplitude of each wave individually, all in phase at 0°.
- **Resultant Phasor \( A_{\text{net}} \):** The net amplitude \( A_{\text{net}} = 3A \), where \( A \) is the amplitude of each individual phasor. This occurs because the difference in phase between adjacent phasors is 0°.
### Tasks:
#### (a) Phasor Diagram with 45° Phase Difference
- **Diagram Needed:** Sketch a new phasor diagram where each adjacent phasor has a 45° phase difference.
- **Details to Include:** An arrow indicating \( A_{\text{net}} \) with numerical angle \( \theta \) from the first phasor.
#### (b) Phase Difference of 90°
- **Action:** Repeat step (a) for a 90° phase difference.
#### (c) Phase Difference of 120°
- **Action:** Repeat step (a) for a 120° phase difference.
#### (d) Phase Difference of 180°
- **Action:** Repeat step (a) for a 180° phase difference.
#### (e) Phase Difference of 240°
- **Action:** Repeat step (a) for a 240° phase difference.
This exercise helps in understanding the superposition principle and interference patterns resulting from different phase differences.
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