The path difference between the two waves `y_1=a_1sin(omegat-(2pix)/(lamda))` and `y_2=a_2cos(omegat-(2pix)/(lamda)+phi)` is

Options

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

Explanation

To find the path difference between the two waves given by the equations \( y_1 = a_1 \sin(\omega t - \frac{2\pi x}{\lambda}) \) and \( y_2 = a_2 \cos(\omega t - \frac{2\pi x}{\lambda} + \phi) \), we can follow these steps:

### Step 1: Identify the phase of each wave
The first wave is given as:
\[ y_1 = a_1 \sin(\omega t - \frac{2\pi x}{\lambda}) \]
The phase of this wave is:
\[ \phi_1 = \omega t - \frac{2\pi x}{\lambda} \]

The second wave is given as:
\[ y_2 = a_2 \cos(\omega t - \frac{2\pi x}{\lambda} + \phi) \]
To express this in terms of sine, we can use the identity \( \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \):
\[ y_2 = a_2 \sin\left(\omega t - \frac{2\pi x}{\lambda} + \phi + \frac{\pi}{2}\right) \]
Thus, the phase of the second wave becomes:
\[ \phi_2 = \omega t - \frac{2\pi x}{\lambda} + \phi + \frac{\pi}{2} \]

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

### Step 3: Relate phase difference to path difference
The path difference \( \Delta x \) can be related to the phase difference using the formula:
\[ \Delta x = \frac{\Delta \phi \cdot \lambda}{2\pi} \]
Substituting \( \Delta \phi \) into this equation:
\[ \Delta x = \frac{\left(\phi + \frac{\pi}{2}\right) \lambda}{2\pi} \]
This can be simplified to:
\[ \Delta x = \frac{\phi \lambda}{2\pi} + \frac{\lambda}{4} \]

### Final Answer
Thus, the path difference between the two waves is:
\[ \Delta x = \frac{\phi \lambda}{2\pi} + \frac{\lambda}{4} \]

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