Two waves of equal amplitude a from two coherent sources `(S_1 & S_2)` interfere at a point R such that `S_2R-S_1R =1.5 lambda`The intensity at point R Would be

Options

A. Zero
B. `2a^2`
C. `5a^2`
D. `6a^2`

Answer

To solve the problem, we need to analyze the interference of two coherent waves at point R, given that the path difference \( S_2R - S_1R = 1.5 \lambda \).

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Amplitudes of the waves from sources \( S_1 \) and \( S_2 \) are equal, denoted as \( A \).
- The path difference is given as \( S_2R - S_1R = 1.5 \lambda \).

2. **Understanding Path Difference**:
- The path difference of \( 1.5 \lambda \) can be expressed as \( \frac{3}{2} \lambda \).
- This indicates that the waves are out of phase by \( 1.5 \lambda \).

3. **Determine the Type of Interference**:
- A path difference of \( \frac{3}{2} \lambda \) corresponds to an odd multiple of \( \frac{\lambda}{2} \) (specifically \( \frac{3}{2} \lambda \)).
- This results in destructive interference because the waves are out of phase by \( 180^\circ \) (or \( \pi \) radians).

4. **Calculate the Resultant Amplitude**:
- For destructive interference, the resultant amplitude \( A_R \) can be calculated using the formula:
\[ A_R = |A_1 - A_2| \]
- Since both amplitudes are equal (\( A_1 = A_2 = A \)):
\[ A_R = |A - A| = |0| = 0 \]

5. **Calculate the Intensity**:
- The intensity \( I \) of a wave is proportional to the square of the amplitude:
\[ I \propto A^2 \]
- Since the resultant amplitude \( A_R = 0 \):
\[ I_R = A_R^2 = 0^2 = 0 \]

6. **Conclusion**:
- The intensity at point R is \( 0 \).

### Final Answer:
The intensity at point R would be \( 0 \).