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In a Young's double slit experiment, the path difference, at a certain point on the screen, betwen two interfering waves is $\frac{1}{8}$ th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:
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The correct answer is:
0.85
Given, path difference, $\Delta x=\frac{\lambda}{8}$
Phase difference $(\Delta \phi)$ is given by
$\Delta \phi=\frac{2 \pi}{\lambda}(\Delta \mathrm{x})$
$\Delta \phi=\frac{(2 \pi)}{\lambda} \frac{\lambda}{8}=\frac{\pi}{4}$
For two sources in different phases,$$
\begin{array}{l}
I=I_{0} \cos ^{2}\left(\frac{\pi}{8}\right) \\
\frac{1}{I_{0}}=\cos ^{2}\left(\frac{\pi}{8}\right) \\
=\frac{1+\cos \frac{\pi}{4}}{2}=\frac{1+\frac{1}{\sqrt{2}}}{2}=0.85
\end{array}
$$
Phase difference $(\Delta \phi)$ is given by
$\Delta \phi=\frac{2 \pi}{\lambda}(\Delta \mathrm{x})$
$\Delta \phi=\frac{(2 \pi)}{\lambda} \frac{\lambda}{8}=\frac{\pi}{4}$
For two sources in different phases,$$
\begin{array}{l}
I=I_{0} \cos ^{2}\left(\frac{\pi}{8}\right) \\
\frac{1}{I_{0}}=\cos ^{2}\left(\frac{\pi}{8}\right) \\
=\frac{1+\cos \frac{\pi}{4}}{2}=\frac{1+\frac{1}{\sqrt{2}}}{2}=0.85
\end{array}
$$
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