Q. At a point screen, the path difference between two interfering waves of equal intensities is $\frac{\lambda }{4}$ . Find the ratio of intensity at this point and that at the central fringe.
NTA AbhyasNTA Abhyas 2020
Solution:
Path difference between two interfering waves = $\frac{\lambda }{4}$
Phase difference is given by:
$\Delta \phi=k\Delta x$
$\Delta \phi=\frac{2 \left(\pi \right) \lambda }{\lambda \times 4}$
$\Delta \phi=\frac{\pi }{2}$
Resultant Intensity of equal intensities, $I=4I_{0}\left(cos\right)^{2}\left(\frac{\phi}{2}\right)$
$=4I_{0}\left(cos\right)^{2}\left(\frac{\pi }{2 \times 2}\right)$
At central fringe, $\phi=0$
$\frac{I_{1}}{I_{2}}=\frac{\left(cos\right)^{2} \left(\frac{_{\left(\phi\right)_{1}}}{1}\right)}{\left(cos\right)^{2} \left(\frac{\left(\phi\right)_{2}}{2}\right)}=\frac{\left(cos\right)^{2} \left(\frac{\pi }{4}\right)}{\left(cos\right)^{2} \left(\right. 0 \left.\right)}$
$\frac{I_{1}}{I_{2}}=\frac{1/\left(\left(\sqrt{2}\right)^{2}\right)}{1}$
$\frac{I_{1}}{I_{2}}=\frac{1}{2}$
Phase difference is given by:
$\Delta \phi=k\Delta x$
$\Delta \phi=\frac{2 \left(\pi \right) \lambda }{\lambda \times 4}$
$\Delta \phi=\frac{\pi }{2}$
Resultant Intensity of equal intensities, $I=4I_{0}\left(cos\right)^{2}\left(\frac{\phi}{2}\right)$
$=4I_{0}\left(cos\right)^{2}\left(\frac{\pi }{2 \times 2}\right)$
At central fringe, $\phi=0$
$\frac{I_{1}}{I_{2}}=\frac{\left(cos\right)^{2} \left(\frac{_{\left(\phi\right)_{1}}}{1}\right)}{\left(cos\right)^{2} \left(\frac{\left(\phi\right)_{2}}{2}\right)}=\frac{\left(cos\right)^{2} \left(\frac{\pi }{4}\right)}{\left(cos\right)^{2} \left(\right. 0 \left.\right)}$
$\frac{I_{1}}{I_{2}}=\frac{1/\left(\left(\sqrt{2}\right)^{2}\right)}{1}$
$\frac{I_{1}}{I_{2}}=\frac{1}{2}$