Two coherent waves represented by `y_(1) = A sin ((2 pi)/(lambda) x_(1) - omega t + (pi)/(4))` and `y_(2) = A sin (( 2pi)/(lambda) x_(2) - omega t + (pi)/(6))` are superposed. The two waves will produce

Options

A. constructive interference at `(x_(1) - x_(2)) = 2 lambda`
B. constructive interference at `(x_(1) - x_(2)) = 23//24 lambda`
C. destructive interference at `(x_(1) - x_(2)) = 1.5 lambda`
D. denstructive interference at `(x_(1) - x_(2)) = 11//24 lambda`

Explanation

To solve the problem of two coherent waves represented by the equations:

1. \( y_1 = A \sin\left(\frac{2\pi}{\lambda} x_1 - \omega t + \frac{\pi}{4}\right) \)
2. \( y_2 = A \sin\left(\frac{2\pi}{\lambda} x_2 - \omega t + \frac{\pi}{6}\right) \)

we need to determine the type of interference that occurs when these two waves are superposed.

### Step 1: Identify the phase of each wave
The phase of the first wave is:
\[ \phi_1 = \frac{2\pi}{\lambda} x_1 - \omega t + \frac{\pi}{4} \]
The phase of the second wave is:
\[ \phi_2 = \frac{2\pi}{\lambda} x_2 - \omega t + \frac{\pi}{6} \]

### Step 2: Calculate the phase difference
The phase difference \( \Delta \phi \) between the two waves is given by:
\[ \Delta \phi = \phi_1 - \phi_2 \]
Substituting the expressions for \( \phi_1 \) and \( \phi_2 \):
\[ \Delta \phi = \left(\frac{2\pi}{\lambda} x_1 - \omega t + \frac{\pi}{4}\right) - \left(\frac{2\pi}{\lambda} x_2 - \omega t + \frac{\pi}{6}\right) \]
This simplifies to:
\[ \Delta \phi = \frac{2\pi}{\lambda} (x_1 - x_2) + \left(\frac{\pi}{4} - \frac{\pi}{6}\right) \]

### Step 3: Simplify the phase difference
Now, we need to find a common denominator to simplify \( \frac{\pi}{4} - \frac{\pi}{6} \):
\[ \frac{\pi}{4} = \frac{3\pi}{12}, \quad \frac{\pi}{6} = \frac{2\pi}{12} \]
Thus:
\[ \frac{\pi}{4} - \frac{\pi}{6} = \frac{3\pi}{12} - \frac{2\pi}{12} = \frac{\pi}{12} \]
So, we have:
\[ \Delta \phi = \frac{2\pi}{\lambda} (x_1 - x_2) + \frac{\pi}{12} \]

### Step 4: Determine conditions for constructive interference
Constructive interference occurs when the phase difference is an integer multiple of \( 2\pi \):
\[ \Delta \phi = 2n\pi \quad (n \in \mathbb{Z}) \]
Setting the equation:
\[ \frac{2\pi}{\lambda} (x_1 - x_2) + \frac{\pi}{12} = 2n\pi \]
Rearranging gives:
\[ \frac{2\pi}{\lambda} (x_1 - x_2) = 2n\pi - \frac{\pi}{12} \]
Dividing through by \( 2\pi \):
\[ \frac{x_1 - x_2}{\lambda} = n - \frac{1}{24} \]
Thus:
\[ x_1 - x_2 = \lambda \left(n - \frac{1}{24}\right) \]

### Step 5: Determine conditions for destructive interference
Destructive interference occurs when the phase difference is an odd multiple of \( \pi \):
\[ \Delta \phi = (2n + 1)\pi \quad (n \in \mathbb{Z}) \]
Setting the equation:
\[ \frac{2\pi}{\lambda} (x_1 - x_2) + \frac{\pi}{12} = (2n + 1)\pi \]
Rearranging gives:
\[ \frac{2\pi}{\lambda} (x_1 - x_2) = (2n + 1)\pi - \frac{\pi}{12} \]
Dividing through by \( 2\pi \):
\[ \frac{x_1 - x_2}{\lambda} = \frac{(2n + 1)}{2} - \frac{1}{24} \]
Thus:
\[ x_1 - x_2 = \lambda \left(\frac{(2n + 1)}{2} - \frac{1}{24}\right) \]

### Conclusion
The two waves will produce both constructive and destructive interference depending on the values of \( n \). The specific conditions for each type of interference can be derived from the equations above.