The path difference between two waves $\mathrm{Y}_1=\mathrm{a}_1 \sin \left(\omega \mathrm{t}-\frac{2 \pi…

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
  1. $\frac{\lambda \phi}{2 \pi}$
  2. $\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)$
  3. $\frac{2 \pi}{\lambda}\left(\phi-\frac{\pi}{2}\right)$
  4. $\frac{2 \pi}{\lambda} \phi$

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, $\Delta \mathrm{x}=\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)$

Asked in: MHT CET 2024 (03 May Shift 1)

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