The path difference between the two waves
`y_(1)=a_(1) sin(omega t-(2pi x)/(lambda)) and y(2)=a_(2) cos(omega t-(2pi x)/(lambda)+phi)` is

Options

A. `((lambda)/(2pi))phi`
B. `lambda((phi+(pi/2))/(2pi))`
C. `(2pi)/(lambda)(phi - pi/2)`
D. `((2pi)/(lambda))phi`

Answer

To find the path difference between the two waves given by the equations:

1. \( y_1 = a_1 \sin(\omega t - \frac{2\pi x}{\lambda}) \)
2. \( y_2 = a_2 \cos(\omega t - \frac{2\pi x}{\lambda} + \phi) \)

we can follow these steps:

### Step 1: Rewrite the second wave in terms of sine
The cosine function can be expressed in terms of sine:
\[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \]
Thus, we can rewrite \( y_2 \):
\[ y_2 = a_2 \cos\left(\omega t - \frac{2\pi x}{\lambda} + \phi\right) = a_2 \sin\left(\omega t - \frac{2\pi x}{\lambda} + \phi + \frac{\pi}{2}\right) \]

### Step 2: Identify the phase terms
Now we can identify the phase terms in both wave functions:
- For \( y_1 \): The phase is \( \phi_1 = \omega t - \frac{2\pi x}{\lambda} \)
- For \( y_2 \): The phase is \( \phi_2 = \omega t - \frac{2\pi x}{\lambda} + \phi + \frac{\pi}{2} \)

### Step 3: Calculate the phase difference
The phase difference \( \Delta \phi \) between the two waves is given by:
\[ \Delta \phi = \phi_2 - \phi_1 = \left(\omega t - \frac{2\pi x}{\lambda} + \phi + \frac{\pi}{2}\right) - \left(\omega t - \frac{2\pi x}{\lambda}\right) \]
This simplifies to:
\[ \Delta \phi = \phi + \frac{\pi}{2} \]

### Step 4: Relate phase difference to path difference
The relationship between phase difference and path difference is given by:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
where \( \Delta x \) is the path difference. We can rearrange this to find \( \Delta x \):
\[ \Delta x = \frac{\lambda}{2\pi} \Delta \phi \]

### Step 5: Substitute the phase difference
Substituting \( \Delta \phi \) into the equation gives:
\[ \Delta x = \frac{\lambda}{2\pi} \left(\phi + \frac{\pi}{2}\right) \]

### Final Result
Thus, the path difference between the two waves is:
\[ \Delta x = \frac{\lambda}{2\pi} \left(\phi + \frac{\pi}{2}\right) \]