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In Young's double slit experiment, the wavelength of light used is ' $\lambda$ '. The intensity at a point is ' $I$ ' where path difference is $\left(\frac{\lambda}{4}\right)$. If $I_0$ denotes the maximum intensity, then the ratio $\left(\frac{\mathrm{I}}{\mathrm{I}_0}\right)$ is
$\left(\sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right)$
Options:
$\left(\sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right)$
Solution:
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Verified Answer
The correct answer is:
$\frac{1}{2}$
Phase difference, $\Delta \phi=\left(\frac{2 \pi}{\lambda}\right) \Delta l$
For path difference $\frac{\lambda}{4}$,
Phase difference $\Delta \phi=\frac{\pi}{2}$
Using, $\mathrm{I}=\mathrm{I}_0 \cos ^2 \frac{\phi}{2}$
$\begin{array}{ll}
\therefore & \frac{\mathrm{I}}{\mathrm{I}_0}=\cos ^2 \frac{\phi}{2}=\cos ^2\left(\frac{\pi}{4}\right) \\
\therefore & \frac{\mathrm{I}}{\mathrm{I}_0}=\frac{1}{2}
\end{array}$
For path difference $\frac{\lambda}{4}$,
Phase difference $\Delta \phi=\frac{\pi}{2}$
Using, $\mathrm{I}=\mathrm{I}_0 \cos ^2 \frac{\phi}{2}$
$\begin{array}{ll}
\therefore & \frac{\mathrm{I}}{\mathrm{I}_0}=\cos ^2 \frac{\phi}{2}=\cos ^2\left(\frac{\pi}{4}\right) \\
\therefore & \frac{\mathrm{I}}{\mathrm{I}_0}=\frac{1}{2}
\end{array}$
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