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Q. In $YDSE$ , intensity at central maxima is $I_{0}$ . The ratio $\frac{I}{I_{0}}$ , at path difference $\frac{\lambda }{8}$ on the screen from central maxima, is closed to

NTA AbhyasNTA Abhyas 2020Wave Optics

Solution:

$\Delta \phi=\frac{2 \pi }{\lambda }\times \frac{\lambda }{8}=\frac{\pi }{4}$
$I=I_{0}cos^{2} \frac{\pi }{8}$
$=I_{0}\left(\frac{1 + cos \pi / 4}{2}\right)$
$=I_{0}\left(\frac{1 + \frac{1}{\sqrt{2}}}{2}\right)=0.85 \, I_{0}$