Q103 / 150 · MHT CET 2024 (03 May Shift 1) · Physics
The path difference between two waves $\mathrm{Y}_1=\mathrm{a}_1 \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}\right)$ and $\mathrm{Y}_2=\mathrm{a}_2 \cos \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\phi\right)$ is
- A$\frac{\lambda \phi}{2 \pi}$
- Correct answerB$\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)$
- C$\frac{2 \pi}{\lambda}\left(\phi-\frac{\pi}{2}\right)$
- D$\frac{2 \pi}{\lambda} \phi$
Solution
$\begin{aligned}
& y_1=a_1 \sin \left(\omega t-\frac{2 \pi x}{\lambda}\right) \text { and } \\
& y_2=a_2 \sin \left(\omega t-\frac{2 \pi x}{\lambda}+\phi+\frac{\pi}{2}\right)
\end{aligned}$
So phase difference, $\delta=\phi+\frac{\pi}{2}$ and Using, $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} . \delta$ we get, $\Delta \mathrm{x}=\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)$
So phase difference, $\delta=\phi+\frac{\pi}{2}$ and Using, $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} . \delta$ we get, $\Delta \mathrm{x}=\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)$
Hence, the correct answer is option: 'B'.