Q167 / 200 · MHT CET - Medical 2026 (25 April Shift 2) · Physics
The path difference between the two waves $y_1 = a_1\sin\left(\omega t - \dfrac{2\pi x}{\lambda} + \pi\right)$ and $y_2 = a_2\cos\left(\omega t - \dfrac{2\pi x}{\lambda} + \phi\right)$ is
- A$\dfrac{\lambda}{2\pi}\phi$
- Correct answerB$\dfrac{\lambda}{2\pi}\left(\phi - \dfrac{\pi}{2}\right)$
- C$\dfrac{\lambda}{2\pi}(\phi - \pi)$
- D$\dfrac{\lambda}{2\pi}\left(2\phi - \dfrac{\pi}{2}\right)$
Solution
The given wave equations are:
$y_1 = a_1\sin\left(\omega t - \dfrac{2\pi x}{\lambda} + \pi\right)$
$y_2 = a_2\cos\left(\omega t - \dfrac{2\pi x}{\lambda} + \phi\right)$
To compare the phases, we express $y_2$ in terms of the sine function:
$y_2 = a_2\sin\left(\omega t - \dfrac{2\pi x}{\lambda} + \phi + \dfrac{\pi}{2}\right)$
The phase of the first wave is $\Phi_1 = \omega t - \dfrac{2\pi x}{\lambda} + \pi$
The phase of the second wave is $\Phi_2 = \omega t - \dfrac{2\pi x}{\lambda} + \phi + \dfrac{\pi}{2}$
The phase difference $\Delta \Phi$ between the two waves is:
$\Delta \Phi = \Phi_2 - \Phi_1 = \left(\phi + \dfrac{\pi}{2}\right) - \pi = \phi - \dfrac{\pi}{2}$
The relation between path difference $\Delta x$ and phase difference $\Delta \Phi$ is given by:
$\Delta x = \dfrac{\lambda}{2\pi} \Delta \Phi$
Substituting the value of $\Delta \Phi$:
$\Delta x = \dfrac{\lambda}{2\pi}\left(\phi - \dfrac{\pi}{2}\right)$
Answer: $\dfrac{\lambda}{2\pi}\left(\phi - \dfrac{\pi}{2}\right)$
$y_1 = a_1\sin\left(\omega t - \dfrac{2\pi x}{\lambda} + \pi\right)$
$y_2 = a_2\cos\left(\omega t - \dfrac{2\pi x}{\lambda} + \phi\right)$
To compare the phases, we express $y_2$ in terms of the sine function:
$y_2 = a_2\sin\left(\omega t - \dfrac{2\pi x}{\lambda} + \phi + \dfrac{\pi}{2}\right)$
The phase of the first wave is $\Phi_1 = \omega t - \dfrac{2\pi x}{\lambda} + \pi$
The phase of the second wave is $\Phi_2 = \omega t - \dfrac{2\pi x}{\lambda} + \phi + \dfrac{\pi}{2}$
The phase difference $\Delta \Phi$ between the two waves is:
$\Delta \Phi = \Phi_2 - \Phi_1 = \left(\phi + \dfrac{\pi}{2}\right) - \pi = \phi - \dfrac{\pi}{2}$
The relation between path difference $\Delta x$ and phase difference $\Delta \Phi$ is given by:
$\Delta x = \dfrac{\lambda}{2\pi} \Delta \Phi$
Substituting the value of $\Delta \Phi$:
$\Delta x = \dfrac{\lambda}{2\pi}\left(\phi - \dfrac{\pi}{2}\right)$
Answer: $\dfrac{\lambda}{2\pi}\left(\phi - \dfrac{\pi}{2}\right)$
Hence, the correct answer is option: 'B'.