Question
What will be the path difference between two light waves $y_1=a_1 \sin \omega t$ and $y_2=a_2 \cos (\omega t+$ $\phi)$ ?
✓
Answer
Equation of first wave
$y_1=a_1 \sin \omega t$
Equation of second wave
$y_2=a_2 \cos (\omega t+\phi)$
or $y_2=\sin \left(\omega t+\phi+\frac{\pi}{2}\right)$
Phase difference between first and second wave
$\Delta \phi=\phi_2-\phi_1$
$=\left(\omega t+\phi+\frac{\pi}{2}\right)-\omega t$
or $\Delta \phi=\phi+\frac{\pi}{2}$
Phase difference due to path difference
$\Delta \phi=\frac{2 \pi}{\lambda} \times \Delta x$
Path difference
$\Delta x=\frac{\Delta \phi \times \lambda}{2 \lambda}$
$\therefore$ For given waves $I^{\text {st }}$ and $2^{\text {nd }}$, the path difference
$\Delta x=\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)$ $\qquad Ans. $
$y_1=a_1 \sin \omega t$
Equation of second wave
$y_2=a_2 \cos (\omega t+\phi)$
or $y_2=\sin \left(\omega t+\phi+\frac{\pi}{2}\right)$
Phase difference between first and second wave
$\Delta \phi=\phi_2-\phi_1$
$=\left(\omega t+\phi+\frac{\pi}{2}\right)-\omega t$
or $\Delta \phi=\phi+\frac{\pi}{2}$
Phase difference due to path difference
$\Delta \phi=\frac{2 \pi}{\lambda} \times \Delta x$
Path difference
$\Delta x=\frac{\Delta \phi \times \lambda}{2 \lambda}$
$\therefore$ For given waves $I^{\text {st }}$ and $2^{\text {nd }}$, the path difference
$\Delta x=\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)$ $\qquad Ans. $
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