Two choherent waves represented by `y_1 = A sin ((2pi x_1)/(lamda) - wt + pi/3) " and " y_2 = A sin ((2pi x_2)/(lamda) - wt + pi/6)` are superposed. The two waves will produce

Options

A. Constructive interference at `(x_1-x_2) = 2lamda`
B. Constructive interference at`(x_1 - x_2) = (23 lamda)/(24)`
C. destructive interference at `(x_1 - x_2) = 1.5 lamda`
D. destructive interference `(x_1 - x_2) = (11 lamda)/(24)`

Answer

To solve the problem involving the superposition of two coherent waves, we will follow these steps:

### Step 1: Identify the given waves
The two coherent waves are given as:
- \( y_1 = A \sin\left(\frac{2\pi x_1}{\lambda} - \omega t + \frac{\pi}{3}\right) \)
- \( y_2 = A \sin\left(\frac{2\pi x_2}{\lambda} - \omega t + \frac{\pi}{6}\right) \)

### Step 2: Determine the phase difference
The phase difference between the two waves can be calculated from their equations. The phase of \( y_1 \) is:
- \( \phi_1 = \frac{2\pi x_1}{\lambda} - \omega t + \frac{\pi}{3} \)

The phase of \( y_2 \) is:
- \( \phi_2 = \frac{2\pi x_2}{\lambda} - \omega t + \frac{\pi}{6} \)

The phase difference \( \Delta \phi \) is given by:
\[ \Delta \phi = \phi_1 - \phi_2 = \left(\frac{2\pi x_1}{\lambda} - \omega t + \frac{\pi}{3}\right) - \left(\frac{2\pi x_2}{\lambda} - \omega t + \frac{\pi}{6}\right) \]
This simplifies to:
\[ \Delta \phi = \frac{2\pi (x_1 - x_2)}{\lambda} + \left(\frac{\pi}{3} - \frac{\pi}{6}\right) \]
Calculating \( \frac{\pi}{3} - \frac{\pi}{6} \):
\[ \frac{\pi}{3} - \frac{\pi}{6} = \frac{2\pi}{6} - \frac{\pi}{6} = \frac{\pi}{6} \]
Thus, the phase difference becomes:
\[ \Delta \phi = \frac{2\pi (x_1 - x_2)}{\lambda} + \frac{\pi}{6} \]

### Step 3: Determine the path difference
The path difference \( \Delta x \) can be related to the phase difference by the formula:
\[ \Delta \phi = \frac{2\pi \Delta x}{\lambda} \]
Setting the two expressions for \( \Delta \phi \) equal gives:
\[ \frac{2\pi (x_1 - x_2)}{\lambda} + \frac{\pi}{6} = \frac{2\pi \Delta x}{\lambda} \]
Rearranging gives:
\[ \Delta x = (x_1 - x_2) + \frac{\lambda}{12} \]

### Step 4: Analyze for constructive and destructive interference
For constructive interference, the condition is:
\[ \Delta \phi = 2n\pi \quad (n = 0, 1, 2, \ldots) \]
For destructive interference, the condition is:
\[ \Delta \phi = (2n - 1)\pi \quad (n = 0, 1, 2, \ldots) \]

### Step 5: Substitute and solve for \( n \)
1. **Constructive Interference**:
\[ \frac{2\pi (x_1 - x_2)}{\lambda} + \frac{\pi}{6} = 2n\pi \]
Rearranging gives:
\[ \frac{2\pi (x_1 - x_2)}{\lambda} = 2n\pi - \frac{\pi}{6} \]
Simplifying:
\[ \frac{2(x_1 - x_2)}{\lambda} = 2n - \frac{1}{6} \]

2. **Destructive Interference**:
\[ \frac{2\pi (x_1 - x_2)}{\lambda} + \frac{\pi}{6} = (2n - 1)\pi \]
Rearranging gives:
\[ \frac{2\pi (x_1 - x_2)}{\lambda} = (2n - 1)\pi - \frac{\pi}{6} \]
Simplifying:
\[ \frac{2(x_1 - x_2)}{\lambda} = 2n - 1 - \frac{1}{6} \]

### Conclusion
The two coherent waves will produce regions of constructive and destructive interference based on the path difference and phase difference derived above.