To determine the path difference between the two given wave functions, we need to analyze the expressions for the waves:
The first wave is given by: \(y_1 = a_1 \sin\left(\omega t - \frac{2\pi}{\lambda} x\right)\)
The second wave is given by: \(y_2 = a_2 \cos\left(\omega t - \frac{2\pi}{\lambda} x + \phi\right)\)
Let's convert the second wave from cosine to sine form because it is easier to compare waves in the same trigonometric form. We can use the trigonometric identity:
\(\cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right)\)
Thus, the second wave can be rewritten as:
\(y_2 = a_2 \sin\left(\omega t - \frac{2\pi}{\lambda} x + \phi + \frac{\pi}{2}\right)\)
Now, comparing the phase differences in the sine functions, the phase difference between the two waves is:
\(\Delta\phi = \left(\phi + \frac{\pi}{2}\right)\)
The path difference \(\Delta x\) is related to the phase difference through the formula:
\(\Delta x = \frac{\lambda}{2\pi} \Delta\phi\)
Substituting the value of \(\Delta\phi\):
\(\Delta x = \frac{\lambda}{2\pi} \left(\phi + \frac{\pi}{2}\right)\)
Thus, the correct answer is: \(\frac{\lambda}{2\pi}\left(\phi + \frac{\pi}{2}\right)\), which matches the provided correct answer option.